Mixing
$int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}} cdot frac{x^{k_2}}{left(x^{n_2} + a_2...
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Recently I was able to find a result to a common definite integral:
begin{equation}
J(n_1, k_1, m_1) = int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}}:dx = frac{a^{frac{k_1 + 1}{n_1} - m_1}}{n_1}Gammaleft(m_1 - frac{k_1 + 1}{n_1}right)Gammaleft(frac{k_1 + 1}{n_1}right)
end{equation}
The advantage with this integral is that if you can introduce a free parameter into your integral (ala Feynman's Trick) then under a given transformation using that parameter (i.e. derivatives, Laplace Transform, Fourier Transforms) that you can isolate out the parameter. This makes taking the inverse transform quite easy in most cases.
I'm hoping to expand this method to cater to more difficult integrals where multiple parameters are introduced. In doing so, I've come this very similar but much more complicated integral.
begin{equation}
Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}} cdot frac{x^{k_2}}{left(x^{n_2} + a_2 right)^{m_2}}:dx
end{equation}
Where $a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2 in mathbb{R}^{+}$
Unfortunately I'm stuck with this one and am interested to find out if anyone has encountered this form before and if so, if they know a method to find a solution expressed in terms of elementary or non-elementary functions.
Any starting points would be greatly appreciated.
Also, if the solution can be represented with certain restrictions on the parameters, please post up. The only absolute conditions are $a_{1},a_{2},m_{1},m_{2} gt 0$. Keen on all solutions restricted or not.
Edit - Just thinking now that the one thing that can be done to simplify the integral is to let $u = x^{n_1}$ or $u = x^{n_2}$. Here I will use the later to render the integral as
begin{align}
&Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{left( u^{frac{1}{n_2}}right)^{k_1}}{left(left( u^{frac{1}{n_2}}right)^{n_1} + a_1 right)^{m_1}} cdot frac{left( u^{frac{1}{n_2}}right)^{k_2}}{left(u + a_2 right)^{m_2}}frac{1}{n_2}u^{frac{1 - n_2}{n_2}}:du\
quad& = frac{1}{n_2}int_0^{infty} frac{u^{frac{k_1 + k_2 + 1 - n_2}{n_2}}}{left( u^{frac{n_1}{n_2}} + a_1 right)^{m_1}} frac{1}{left(u + a_2right)^{m_1}}:du = frac{1}{n_2}int_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du
end{align}
integration definite-integrals
$endgroup$
|
show 1 more comment
$begingroup$
Recently I was able to find a result to a common definite integral:
begin{equation}
J(n_1, k_1, m_1) = int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}}:dx = frac{a^{frac{k_1 + 1}{n_1} - m_1}}{n_1}Gammaleft(m_1 - frac{k_1 + 1}{n_1}right)Gammaleft(frac{k_1 + 1}{n_1}right)
end{equation}
The advantage with this integral is that if you can introduce a free parameter into your integral (ala Feynman's Trick) then under a given transformation using that parameter (i.e. derivatives, Laplace Transform, Fourier Transforms) that you can isolate out the parameter. This makes taking the inverse transform quite easy in most cases.
I'm hoping to expand this method to cater to more difficult integrals where multiple parameters are introduced. In doing so, I've come this very similar but much more complicated integral.
begin{equation}
Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}} cdot frac{x^{k_2}}{left(x^{n_2} + a_2 right)^{m_2}}:dx
end{equation}
Where $a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2 in mathbb{R}^{+}$
Unfortunately I'm stuck with this one and am interested to find out if anyone has encountered this form before and if so, if they know a method to find a solution expressed in terms of elementary or non-elementary functions.
Any starting points would be greatly appreciated.
Also, if the solution can be represented with certain restrictions on the parameters, please post up. The only absolute conditions are $a_{1},a_{2},m_{1},m_{2} gt 0$. Keen on all solutions restricted or not.
Edit - Just thinking now that the one thing that can be done to simplify the integral is to let $u = x^{n_1}$ or $u = x^{n_2}$. Here I will use the later to render the integral as
begin{align}
&Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{left( u^{frac{1}{n_2}}right)^{k_1}}{left(left( u^{frac{1}{n_2}}right)^{n_1} + a_1 right)^{m_1}} cdot frac{left( u^{frac{1}{n_2}}right)^{k_2}}{left(u + a_2 right)^{m_2}}frac{1}{n_2}u^{frac{1 - n_2}{n_2}}:du\
quad& = frac{1}{n_2}int_0^{infty} frac{u^{frac{k_1 + k_2 + 1 - n_2}{n_2}}}{left( u^{frac{n_1}{n_2}} + a_1 right)^{m_1}} frac{1}{left(u + a_2right)^{m_1}}:du = frac{1}{n_2}int_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du
end{align}
integration definite-integrals
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2
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Just as a joke : $a_2=0$ !
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– Claude Leibovici
Feb 3 at 9:56
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Haha oh, I wish I could set it that way :-)
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– user150203
Feb 3 at 10:14
1
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maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
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– clathratus
Feb 5 at 1:40
1
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yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
$begingroup$
Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34
|
show 1 more comment
$begingroup$
Recently I was able to find a result to a common definite integral:
begin{equation}
J(n_1, k_1, m_1) = int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}}:dx = frac{a^{frac{k_1 + 1}{n_1} - m_1}}{n_1}Gammaleft(m_1 - frac{k_1 + 1}{n_1}right)Gammaleft(frac{k_1 + 1}{n_1}right)
end{equation}
The advantage with this integral is that if you can introduce a free parameter into your integral (ala Feynman's Trick) then under a given transformation using that parameter (i.e. derivatives, Laplace Transform, Fourier Transforms) that you can isolate out the parameter. This makes taking the inverse transform quite easy in most cases.
I'm hoping to expand this method to cater to more difficult integrals where multiple parameters are introduced. In doing so, I've come this very similar but much more complicated integral.
begin{equation}
Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}} cdot frac{x^{k_2}}{left(x^{n_2} + a_2 right)^{m_2}}:dx
end{equation}
Where $a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2 in mathbb{R}^{+}$
Unfortunately I'm stuck with this one and am interested to find out if anyone has encountered this form before and if so, if they know a method to find a solution expressed in terms of elementary or non-elementary functions.
Any starting points would be greatly appreciated.
Also, if the solution can be represented with certain restrictions on the parameters, please post up. The only absolute conditions are $a_{1},a_{2},m_{1},m_{2} gt 0$. Keen on all solutions restricted or not.
Edit - Just thinking now that the one thing that can be done to simplify the integral is to let $u = x^{n_1}$ or $u = x^{n_2}$. Here I will use the later to render the integral as
begin{align}
&Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{left( u^{frac{1}{n_2}}right)^{k_1}}{left(left( u^{frac{1}{n_2}}right)^{n_1} + a_1 right)^{m_1}} cdot frac{left( u^{frac{1}{n_2}}right)^{k_2}}{left(u + a_2 right)^{m_2}}frac{1}{n_2}u^{frac{1 - n_2}{n_2}}:du\
quad& = frac{1}{n_2}int_0^{infty} frac{u^{frac{k_1 + k_2 + 1 - n_2}{n_2}}}{left( u^{frac{n_1}{n_2}} + a_1 right)^{m_1}} frac{1}{left(u + a_2right)^{m_1}}:du = frac{1}{n_2}int_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du
end{align}
integration definite-integrals
$endgroup$
Recently I was able to find a result to a common definite integral:
begin{equation}
J(n_1, k_1, m_1) = int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}}:dx = frac{a^{frac{k_1 + 1}{n_1} - m_1}}{n_1}Gammaleft(m_1 - frac{k_1 + 1}{n_1}right)Gammaleft(frac{k_1 + 1}{n_1}right)
end{equation}
The advantage with this integral is that if you can introduce a free parameter into your integral (ala Feynman's Trick) then under a given transformation using that parameter (i.e. derivatives, Laplace Transform, Fourier Transforms) that you can isolate out the parameter. This makes taking the inverse transform quite easy in most cases.
I'm hoping to expand this method to cater to more difficult integrals where multiple parameters are introduced. In doing so, I've come this very similar but much more complicated integral.
begin{equation}
Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{x^{k_1}}{left(x^{n_1} + a_1 right)^{m_1}} cdot frac{x^{k_2}}{left(x^{n_2} + a_2 right)^{m_2}}:dx
end{equation}
Where $a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2 in mathbb{R}^{+}$
Unfortunately I'm stuck with this one and am interested to find out if anyone has encountered this form before and if so, if they know a method to find a solution expressed in terms of elementary or non-elementary functions.
Any starting points would be greatly appreciated.
Also, if the solution can be represented with certain restrictions on the parameters, please post up. The only absolute conditions are $a_{1},a_{2},m_{1},m_{2} gt 0$. Keen on all solutions restricted or not.
Edit - Just thinking now that the one thing that can be done to simplify the integral is to let $u = x^{n_1}$ or $u = x^{n_2}$. Here I will use the later to render the integral as
begin{align}
&Hleft(a_1, a_2,n_1,n_2,k_1,k_2,m_1,m_2right)= int_0^{infty} frac{left( u^{frac{1}{n_2}}right)^{k_1}}{left(left( u^{frac{1}{n_2}}right)^{n_1} + a_1 right)^{m_1}} cdot frac{left( u^{frac{1}{n_2}}right)^{k_2}}{left(u + a_2 right)^{m_2}}frac{1}{n_2}u^{frac{1 - n_2}{n_2}}:du\
quad& = frac{1}{n_2}int_0^{infty} frac{u^{frac{k_1 + k_2 + 1 - n_2}{n_2}}}{left( u^{frac{n_1}{n_2}} + a_1 right)^{m_1}} frac{1}{left(u + a_2right)^{m_1}}:du = frac{1}{n_2}int_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du
end{align}
integration definite-integrals
integration definite-integrals
edited Feb 5 at 23:36
asked Feb 3 at 9:24
user150203
2
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Just as a joke : $a_2=0$ !
$endgroup$
– Claude Leibovici
Feb 3 at 9:56
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Haha oh, I wish I could set it that way :-)
$endgroup$
– user150203
Feb 3 at 10:14
1
$begingroup$
maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
$endgroup$
– clathratus
Feb 5 at 1:40
1
$begingroup$
yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
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Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34
|
show 1 more comment
2
$begingroup$
Just as a joke : $a_2=0$ !
$endgroup$
– Claude Leibovici
Feb 3 at 9:56
$begingroup$
Haha oh, I wish I could set it that way :-)
$endgroup$
– user150203
Feb 3 at 10:14
1
$begingroup$
maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
$endgroup$
– clathratus
Feb 5 at 1:40
1
$begingroup$
yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
$begingroup$
Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34
2
2
$begingroup$
Just as a joke : $a_2=0$ !
$endgroup$
– Claude Leibovici
Feb 3 at 9:56
$begingroup$
Just as a joke : $a_2=0$ !
$endgroup$
– Claude Leibovici
Feb 3 at 9:56
$begingroup$
Haha oh, I wish I could set it that way :-)
$endgroup$
– user150203
Feb 3 at 10:14
$begingroup$
Haha oh, I wish I could set it that way :-)
$endgroup$
– user150203
Feb 3 at 10:14
1
1
$begingroup$
maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
$endgroup$
– clathratus
Feb 5 at 1:40
$begingroup$
maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
$endgroup$
– clathratus
Feb 5 at 1:40
1
1
$begingroup$
yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
$begingroup$
yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
$begingroup$
Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34
$begingroup$
Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34
|
show 1 more comment
4 Answers
4
active
oldest
votes
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To long for the comment.
Maybe it's too trite, but
In the case $n_1=n_2$ this integral can be rewrite through the hypergeometric function $_2F_1$.
I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form
$$
I= alphaint_0^{infty} frac{u^{k_3}}{left(u + 1right)^{m_1}} frac{1}{left(u + cright)^{m_2}}:du
$$
like it was done by DavidG.
After that one can do the change of variables $y=u/(1+u)$
$$
I=frac{alpha}{c^{m_2}}int_0^{1} frac{y^{k_3}}{left(1 - yright)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}:du=frac{alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)
$$
I am not sure about the existance of the answer in the general case.
answered Feb 7 at 7:40
PeterPeter
510213
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Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
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– user150203
Feb 7 at 7:43
add a comment |
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This is an integral of two Meijer G-functions which gives a Fox H-function. Let $nu_1 = 1/n_1$, $,nu_2 = 1/n_2$, $,sigma = k_1 + k_2 + 1$. The Mellin transform is
$$mathcal M_{x to p}[(x + a)^{-m}] =
frac {Gamma(p) Gamma(m - p)} {Gamma(m)} a^{p -m}, \
mathcal M[(x^{n_1} + a_1)^{-m_1}] =
nu_1 mathcal M[(x + a_1)^{-m_1}](nu_1 p), \
mathcal M[x^{-sigma} (x^{-n_2} + a_2)^{-m_2}] =
nu_2 mathcal M[(x + a_2)^{-m_2}](nu_2 (sigma - p)).$$
The convolution is mapped to the product:
$$mathcal M[f * g] =
mathcal M {left[int_0^infty f(x)
,g {left( frac omega x right)} ,frac {dx} x right]} =
mathcal M[f] mathcal M[g].$$
The product is again a rational combination of linear gamma functions times $z^p$, the inverse transform of which is, by definition, the H-function. Evaluating it at $1$ ($omega = 1$ above) gives
$$int_0^infty
frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}}
frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx =
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)} times \
mathcal M^{-1}[
Gamma(m_1 - nu_1 p) Gamma(nu_2 sigma - nu_2 p)
Gamma(nu_1 p) Gamma(m_2 - nu_2 sigma + nu_2 p)
(a_1^{-nu_1} a_2^{nu_2})^{-p}](1) = \
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)}
H_{2, 2}^{2, 2} {left( a_1^{-nu_1} a_2^{nu_2} middle|
{(1 - m_1, nu_1), (1 - nu_2 sigma, nu_2) atop
(0, nu_1), (m_2 - nu_2 sigma, nu_2)} right)}.$$
The integral exists if $sigma < n_1 m_1 + n_2 m_2$.
answered Feb 8 at 6:11
MaximMaxim
6,3081221
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Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
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– user150203
Feb 8 at 10:31
add a comment |
$begingroup$
HINT
Let us start from the OP result
begin{align}
&H_1left(a_1, a_2, m_1, m_2, n_3, k_3right)
= frac{1}{n_2}intlimits_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du.tag1
end{align}
Taking in account, that identity
begin{align} &frac{x^n}{left(x^n + aright)^m}
=frac{x^n+a-a}{left(u^n + aright)^m}
=frac{1}{left(x^n + aright)^{m-1}}
-frac{a}{left(x^n + aright)^m}tag2 end{align}
allows to manage the degree in the numerator both decreasing and increasing,
assume WLOG $mathbf{k_3<m_1, k<m_2}.$
Also, due to the scaling substitution, let WLOG $mathbf{a_1=1}.$
Under these conditions, enough to consider the integral
begin{align}
&I(a,k,n,p,q)
= intlimits_0^{infty} frac{u^k}{left(u^n + 1right)^p} frac{1}{left(u + aright)^{q}}:du,quad k<p,quad k<q, quad a>1.tag3
end{align}
Is known the binomial decomposition
begin{align}
&(1-z)^{-d-1}=sumlimits_{j=d}^inftybinom{j}{d}z^{j-d}
= sumlimits_{j=d}^inftydfrac{Gamma(j+1)}{Gamma(d+1)Gamma(j-d+1)}z^{j-d},quad|z|<1.tag4
end{align}
If $|u|<1$ then from $(4)$ should Maclaurin series
begin{align}
&(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1}dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{n(j-q+1)}.tag5
end{align}
If $|u|>1$ then from $(4)$ can be obtained Laurent series
begin{align}
&(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1} dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{-n(j-q+1)}.tag6
end{align}
At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function
$$int_0^1dfrac{x^k}{(x+a)^m}:dx=dfrac{a^{-m}}{k+1},{_2F_1left(k+1,m,k+2,-dfrac1aright)}tag7$$
(see also Wolfram Alpha),
$$int_1^inftydfrac{x^k}{(x+a)^m}:dx=dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}tag8$$
(see also Wolfram Alpha).
Splitting the issue integral to the intervals $(0,1)$ and $(1,infty)$ and applying solutions $(5)-(8),$ one can present the result $I(a,k,n,p,q)$ as the sum of the pair of series, using gamma function and hypergeometric functions.
Thanks to the other members of the discussion for their interesting ideas.
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
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Thank you very much for your solution - very much appreciated.
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– user150203
Feb 8 at 11:09
add a comment |
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This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $ldots$
The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta (x)} , du,$$
for suitable conditions on the function $beta (x)$.
Let us use this parametrisation on your first integral $J(a_1, n_1, m_1, k_1)$. From Schwinger parametrisation we observe that
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} , du,$$
(here $beta (x) = x^{n_1} + a_1$ and $p = m_1$). Your integral for $J$ then becomes
$$J(a_1, n_1, m_1, k_1) = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du int_0^infty x^{k_1} e^{-u x^{n_1}} , dx,$$
where a change in the order of integration has been made. Now if we let $t = u x^{n_1}$ we have
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du frac{1}{n_1} u^{-frac{(k_1 + 1)}{n_1}} int_0^infty t^{frac{k_1 + 1}{n_1} - 1} e^{-t} , dt\
&= frac{1}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} , du.
end{align}
Enforcing a substitution of $u mapsto u/a_1$ leads to
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} , du\[2ex]
&= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (m_1 - frac{k_1 + 1}{n_1} right ) Gamma left (frac{k_1 + 1}{n_1} right ).
end{align}
Now attempting such an approach on the integral $H$. Here the following two Schwinger parametrisations are used:
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} , dy,$$
and
$$frac{1}{(x^{n_2} + a_2)^{m_2}} = frac{1}{Gamma (m_2)} int_0^infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} , dz.$$
So the integral for $H$, after a change in the order of integration has been made, becomes
$$H = frac{1}{Gamma (m_1) Gamma (m_2)} int_0^infty y^{m_1 - 1} e^{-y a_1} , dy int_0^infty z^{m_2 - 1} e^{-z a_2} , dz int_0^infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} , dx.$$
The roadblock is what is to be done with that inner $x$-integral if $n_1 neq n_2$?
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
$endgroup$
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
add a comment |
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$begingroup$
To long for the comment.
Maybe it's too trite, but
In the case $n_1=n_2$ this integral can be rewrite through the hypergeometric function $_2F_1$.
I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form
$$
I= alphaint_0^{infty} frac{u^{k_3}}{left(u + 1right)^{m_1}} frac{1}{left(u + cright)^{m_2}}:du
$$
like it was done by DavidG.
After that one can do the change of variables $y=u/(1+u)$
$$
I=frac{alpha}{c^{m_2}}int_0^{1} frac{y^{k_3}}{left(1 - yright)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}:du=frac{alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)
$$
I am not sure about the existance of the answer in the general case.
answered Feb 7 at 7:40
PeterPeter
510213
$endgroup$
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
add a comment |
$begingroup$
To long for the comment.
Maybe it's too trite, but
In the case $n_1=n_2$ this integral can be rewrite through the hypergeometric function $_2F_1$.
I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form
$$
I= alphaint_0^{infty} frac{u^{k_3}}{left(u + 1right)^{m_1}} frac{1}{left(u + cright)^{m_2}}:du
$$
like it was done by DavidG.
After that one can do the change of variables $y=u/(1+u)$
$$
I=frac{alpha}{c^{m_2}}int_0^{1} frac{y^{k_3}}{left(1 - yright)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}:du=frac{alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)
$$
I am not sure about the existance of the answer in the general case.
answered Feb 7 at 7:40
PeterPeter
510213
$endgroup$
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
add a comment |
$begingroup$
To long for the comment.
Maybe it's too trite, but
In the case $n_1=n_2$ this integral can be rewrite through the hypergeometric function $_2F_1$.
I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form
$$
I= alphaint_0^{infty} frac{u^{k_3}}{left(u + 1right)^{m_1}} frac{1}{left(u + cright)^{m_2}}:du
$$
like it was done by DavidG.
After that one can do the change of variables $y=u/(1+u)$
$$
I=frac{alpha}{c^{m_2}}int_0^{1} frac{y^{k_3}}{left(1 - yright)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}:du=frac{alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)
$$
I am not sure about the existance of the answer in the general case.
answered Feb 7 at 7:40
PeterPeter
510213
$endgroup$
To long for the comment.
Maybe it's too trite, but
In the case $n_1=n_2$ this integral can be rewrite through the hypergeometric function $_2F_1$.
I do not want to pay attention to the coefficient. I this case the integral can be rewrite in the following form
$$
I= alphaint_0^{infty} frac{u^{k_3}}{left(u + 1right)^{m_1}} frac{1}{left(u + cright)^{m_2}}:du
$$
like it was done by DavidG.
After that one can do the change of variables $y=u/(1+u)$
$$
I=frac{alpha}{c^{m_2}}int_0^{1} frac{y^{k_3}}{left(1 - yright)^{k_3+2-m_1-m_2} (1-y(1-1/c))^{m_2}}:du=frac{alpha}{c^{m_2}}Г(k_3+1) Г(m_1+m_2-k_3-1) _2F_1(m_2,k_3+1,m_1+m_2,1-1/c)
$$
I am not sure about the existance of the answer in the general case.
answered Feb 7 at 7:40
PeterPeter
510213
edited Feb 7 at 7:59
answered Feb 7 at 7:40
PeterPeter
510213
answered Feb 7 at 7:40
PeterPeter
510213
answered Feb 7 at 7:40
PeterPeter
510213
510213
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
add a comment |
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
$begingroup$
Appreciate the comment and your solution for this special case. I have formed that solution as well. Sadly for the applications I have for this integral $n_1 = n_2$ is a rarity. Again, appreciate the comment. Please keep at it!
$endgroup$
– user150203
Feb 7 at 7:43
add a comment |
$begingroup$
This is an integral of two Meijer G-functions which gives a Fox H-function. Let $nu_1 = 1/n_1$, $,nu_2 = 1/n_2$, $,sigma = k_1 + k_2 + 1$. The Mellin transform is
$$mathcal M_{x to p}[(x + a)^{-m}] =
frac {Gamma(p) Gamma(m - p)} {Gamma(m)} a^{p -m}, \
mathcal M[(x^{n_1} + a_1)^{-m_1}] =
nu_1 mathcal M[(x + a_1)^{-m_1}](nu_1 p), \
mathcal M[x^{-sigma} (x^{-n_2} + a_2)^{-m_2}] =
nu_2 mathcal M[(x + a_2)^{-m_2}](nu_2 (sigma - p)).$$
The convolution is mapped to the product:
$$mathcal M[f * g] =
mathcal M {left[int_0^infty f(x)
,g {left( frac omega x right)} ,frac {dx} x right]} =
mathcal M[f] mathcal M[g].$$
The product is again a rational combination of linear gamma functions times $z^p$, the inverse transform of which is, by definition, the H-function. Evaluating it at $1$ ($omega = 1$ above) gives
$$int_0^infty
frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}}
frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx =
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)} times \
mathcal M^{-1}[
Gamma(m_1 - nu_1 p) Gamma(nu_2 sigma - nu_2 p)
Gamma(nu_1 p) Gamma(m_2 - nu_2 sigma + nu_2 p)
(a_1^{-nu_1} a_2^{nu_2})^{-p}](1) = \
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)}
H_{2, 2}^{2, 2} {left( a_1^{-nu_1} a_2^{nu_2} middle|
{(1 - m_1, nu_1), (1 - nu_2 sigma, nu_2) atop
(0, nu_1), (m_2 - nu_2 sigma, nu_2)} right)}.$$
The integral exists if $sigma < n_1 m_1 + n_2 m_2$.
answered Feb 8 at 6:11
MaximMaxim
6,3081221
$endgroup$
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
add a comment |
$begingroup$
This is an integral of two Meijer G-functions which gives a Fox H-function. Let $nu_1 = 1/n_1$, $,nu_2 = 1/n_2$, $,sigma = k_1 + k_2 + 1$. The Mellin transform is
$$mathcal M_{x to p}[(x + a)^{-m}] =
frac {Gamma(p) Gamma(m - p)} {Gamma(m)} a^{p -m}, \
mathcal M[(x^{n_1} + a_1)^{-m_1}] =
nu_1 mathcal M[(x + a_1)^{-m_1}](nu_1 p), \
mathcal M[x^{-sigma} (x^{-n_2} + a_2)^{-m_2}] =
nu_2 mathcal M[(x + a_2)^{-m_2}](nu_2 (sigma - p)).$$
The convolution is mapped to the product:
$$mathcal M[f * g] =
mathcal M {left[int_0^infty f(x)
,g {left( frac omega x right)} ,frac {dx} x right]} =
mathcal M[f] mathcal M[g].$$
The product is again a rational combination of linear gamma functions times $z^p$, the inverse transform of which is, by definition, the H-function. Evaluating it at $1$ ($omega = 1$ above) gives
$$int_0^infty
frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}}
frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx =
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)} times \
mathcal M^{-1}[
Gamma(m_1 - nu_1 p) Gamma(nu_2 sigma - nu_2 p)
Gamma(nu_1 p) Gamma(m_2 - nu_2 sigma + nu_2 p)
(a_1^{-nu_1} a_2^{nu_2})^{-p}](1) = \
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)}
H_{2, 2}^{2, 2} {left( a_1^{-nu_1} a_2^{nu_2} middle|
{(1 - m_1, nu_1), (1 - nu_2 sigma, nu_2) atop
(0, nu_1), (m_2 - nu_2 sigma, nu_2)} right)}.$$
The integral exists if $sigma < n_1 m_1 + n_2 m_2$.
answered Feb 8 at 6:11
MaximMaxim
6,3081221
$endgroup$
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
add a comment |
$begingroup$
This is an integral of two Meijer G-functions which gives a Fox H-function. Let $nu_1 = 1/n_1$, $,nu_2 = 1/n_2$, $,sigma = k_1 + k_2 + 1$. The Mellin transform is
$$mathcal M_{x to p}[(x + a)^{-m}] =
frac {Gamma(p) Gamma(m - p)} {Gamma(m)} a^{p -m}, \
mathcal M[(x^{n_1} + a_1)^{-m_1}] =
nu_1 mathcal M[(x + a_1)^{-m_1}](nu_1 p), \
mathcal M[x^{-sigma} (x^{-n_2} + a_2)^{-m_2}] =
nu_2 mathcal M[(x + a_2)^{-m_2}](nu_2 (sigma - p)).$$
The convolution is mapped to the product:
$$mathcal M[f * g] =
mathcal M {left[int_0^infty f(x)
,g {left( frac omega x right)} ,frac {dx} x right]} =
mathcal M[f] mathcal M[g].$$
The product is again a rational combination of linear gamma functions times $z^p$, the inverse transform of which is, by definition, the H-function. Evaluating it at $1$ ($omega = 1$ above) gives
$$int_0^infty
frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}}
frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx =
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)} times \
mathcal M^{-1}[
Gamma(m_1 - nu_1 p) Gamma(nu_2 sigma - nu_2 p)
Gamma(nu_1 p) Gamma(m_2 - nu_2 sigma + nu_2 p)
(a_1^{-nu_1} a_2^{nu_2})^{-p}](1) = \
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)}
H_{2, 2}^{2, 2} {left( a_1^{-nu_1} a_2^{nu_2} middle|
{(1 - m_1, nu_1), (1 - nu_2 sigma, nu_2) atop
(0, nu_1), (m_2 - nu_2 sigma, nu_2)} right)}.$$
The integral exists if $sigma < n_1 m_1 + n_2 m_2$.
answered Feb 8 at 6:11
MaximMaxim
6,3081221
$endgroup$
This is an integral of two Meijer G-functions which gives a Fox H-function. Let $nu_1 = 1/n_1$, $,nu_2 = 1/n_2$, $,sigma = k_1 + k_2 + 1$. The Mellin transform is
$$mathcal M_{x to p}[(x + a)^{-m}] =
frac {Gamma(p) Gamma(m - p)} {Gamma(m)} a^{p -m}, \
mathcal M[(x^{n_1} + a_1)^{-m_1}] =
nu_1 mathcal M[(x + a_1)^{-m_1}](nu_1 p), \
mathcal M[x^{-sigma} (x^{-n_2} + a_2)^{-m_2}] =
nu_2 mathcal M[(x + a_2)^{-m_2}](nu_2 (sigma - p)).$$
The convolution is mapped to the product:
$$mathcal M[f * g] =
mathcal M {left[int_0^infty f(x)
,g {left( frac omega x right)} ,frac {dx} x right]} =
mathcal M[f] mathcal M[g].$$
The product is again a rational combination of linear gamma functions times $z^p$, the inverse transform of which is, by definition, the H-function. Evaluating it at $1$ ($omega = 1$ above) gives
$$int_0^infty
frac {x^{k_1}} {(x^{n_1} + a_1)^{m_1}}
frac {x^{k_2}} {(x^{n_2} + a_2)^{m_2}} dx =
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)} times \
mathcal M^{-1}[
Gamma(m_1 - nu_1 p) Gamma(nu_2 sigma - nu_2 p)
Gamma(nu_1 p) Gamma(m_2 - nu_2 sigma + nu_2 p)
(a_1^{-nu_1} a_2^{nu_2})^{-p}](1) = \
frac
{nu_1 nu_2 a_1^{-m_1} a_2^{nu_2 sigma - m_2}}
{Gamma(m_1) Gamma(m_2)}
H_{2, 2}^{2, 2} {left( a_1^{-nu_1} a_2^{nu_2} middle|
{(1 - m_1, nu_1), (1 - nu_2 sigma, nu_2) atop
(0, nu_1), (m_2 - nu_2 sigma, nu_2)} right)}.$$
The integral exists if $sigma < n_1 m_1 + n_2 m_2$.
answered Feb 8 at 6:11
MaximMaxim
6,3081221
answered Feb 8 at 6:11
MaximMaxim
6,3081221
answered Feb 8 at 6:11
MaximMaxim
6,3081221
answered Feb 8 at 6:11
MaximMaxim
6,3081221
6,3081221
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
add a comment |
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
$begingroup$
Absolutely brilliant. Thank you very much. From the solution, this was well beyond my 'pay grade'. I look forward going through this in detail.
$endgroup$
– user150203
Feb 8 at 10:31
add a comment |
$begingroup$
HINT
Let us start from the OP result
begin{align}
&H_1left(a_1, a_2, m_1, m_2, n_3, k_3right)
= frac{1}{n_2}intlimits_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du.tag1
end{align}
Taking in account, that identity
begin{align} &frac{x^n}{left(x^n + aright)^m}
=frac{x^n+a-a}{left(u^n + aright)^m}
=frac{1}{left(x^n + aright)^{m-1}}
-frac{a}{left(x^n + aright)^m}tag2 end{align}
allows to manage the degree in the numerator both decreasing and increasing,
assume WLOG $mathbf{k_3<m_1, k<m_2}.$
Also, due to the scaling substitution, let WLOG $mathbf{a_1=1}.$
Under these conditions, enough to consider the integral
begin{align}
&I(a,k,n,p,q)
= intlimits_0^{infty} frac{u^k}{left(u^n + 1right)^p} frac{1}{left(u + aright)^{q}}:du,quad k<p,quad k<q, quad a>1.tag3
end{align}
Is known the binomial decomposition
begin{align}
&(1-z)^{-d-1}=sumlimits_{j=d}^inftybinom{j}{d}z^{j-d}
= sumlimits_{j=d}^inftydfrac{Gamma(j+1)}{Gamma(d+1)Gamma(j-d+1)}z^{j-d},quad|z|<1.tag4
end{align}
If $|u|<1$ then from $(4)$ should Maclaurin series
begin{align}
&(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1}dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{n(j-q+1)}.tag5
end{align}
If $|u|>1$ then from $(4)$ can be obtained Laurent series
begin{align}
&(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1} dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{-n(j-q+1)}.tag6
end{align}
At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function
$$int_0^1dfrac{x^k}{(x+a)^m}:dx=dfrac{a^{-m}}{k+1},{_2F_1left(k+1,m,k+2,-dfrac1aright)}tag7$$
(see also Wolfram Alpha),
$$int_1^inftydfrac{x^k}{(x+a)^m}:dx=dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}tag8$$
(see also Wolfram Alpha).
Splitting the issue integral to the intervals $(0,1)$ and $(1,infty)$ and applying solutions $(5)-(8),$ one can present the result $I(a,k,n,p,q)$ as the sum of the pair of series, using gamma function and hypergeometric functions.
Thanks to the other members of the discussion for their interesting ideas.
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
$endgroup$
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
add a comment |
$begingroup$
HINT
Let us start from the OP result
begin{align}
&H_1left(a_1, a_2, m_1, m_2, n_3, k_3right)
= frac{1}{n_2}intlimits_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du.tag1
end{align}
Taking in account, that identity
begin{align} &frac{x^n}{left(x^n + aright)^m}
=frac{x^n+a-a}{left(u^n + aright)^m}
=frac{1}{left(x^n + aright)^{m-1}}
-frac{a}{left(x^n + aright)^m}tag2 end{align}
allows to manage the degree in the numerator both decreasing and increasing,
assume WLOG $mathbf{k_3<m_1, k<m_2}.$
Also, due to the scaling substitution, let WLOG $mathbf{a_1=1}.$
Under these conditions, enough to consider the integral
begin{align}
&I(a,k,n,p,q)
= intlimits_0^{infty} frac{u^k}{left(u^n + 1right)^p} frac{1}{left(u + aright)^{q}}:du,quad k<p,quad k<q, quad a>1.tag3
end{align}
Is known the binomial decomposition
begin{align}
&(1-z)^{-d-1}=sumlimits_{j=d}^inftybinom{j}{d}z^{j-d}
= sumlimits_{j=d}^inftydfrac{Gamma(j+1)}{Gamma(d+1)Gamma(j-d+1)}z^{j-d},quad|z|<1.tag4
end{align}
If $|u|<1$ then from $(4)$ should Maclaurin series
begin{align}
&(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1}dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{n(j-q+1)}.tag5
end{align}
If $|u|>1$ then from $(4)$ can be obtained Laurent series
begin{align}
&(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1} dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{-n(j-q+1)}.tag6
end{align}
At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function
$$int_0^1dfrac{x^k}{(x+a)^m}:dx=dfrac{a^{-m}}{k+1},{_2F_1left(k+1,m,k+2,-dfrac1aright)}tag7$$
(see also Wolfram Alpha),
$$int_1^inftydfrac{x^k}{(x+a)^m}:dx=dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}tag8$$
(see also Wolfram Alpha).
Splitting the issue integral to the intervals $(0,1)$ and $(1,infty)$ and applying solutions $(5)-(8),$ one can present the result $I(a,k,n,p,q)$ as the sum of the pair of series, using gamma function and hypergeometric functions.
Thanks to the other members of the discussion for their interesting ideas.
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
$endgroup$
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
add a comment |
$begingroup$
HINT
Let us start from the OP result
begin{align}
&H_1left(a_1, a_2, m_1, m_2, n_3, k_3right)
= frac{1}{n_2}intlimits_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du.tag1
end{align}
Taking in account, that identity
begin{align} &frac{x^n}{left(x^n + aright)^m}
=frac{x^n+a-a}{left(u^n + aright)^m}
=frac{1}{left(x^n + aright)^{m-1}}
-frac{a}{left(x^n + aright)^m}tag2 end{align}
allows to manage the degree in the numerator both decreasing and increasing,
assume WLOG $mathbf{k_3<m_1, k<m_2}.$
Also, due to the scaling substitution, let WLOG $mathbf{a_1=1}.$
Under these conditions, enough to consider the integral
begin{align}
&I(a,k,n,p,q)
= intlimits_0^{infty} frac{u^k}{left(u^n + 1right)^p} frac{1}{left(u + aright)^{q}}:du,quad k<p,quad k<q, quad a>1.tag3
end{align}
Is known the binomial decomposition
begin{align}
&(1-z)^{-d-1}=sumlimits_{j=d}^inftybinom{j}{d}z^{j-d}
= sumlimits_{j=d}^inftydfrac{Gamma(j+1)}{Gamma(d+1)Gamma(j-d+1)}z^{j-d},quad|z|<1.tag4
end{align}
If $|u|<1$ then from $(4)$ should Maclaurin series
begin{align}
&(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1}dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{n(j-q+1)}.tag5
end{align}
If $|u|>1$ then from $(4)$ can be obtained Laurent series
begin{align}
&(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1} dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{-n(j-q+1)}.tag6
end{align}
At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function
$$int_0^1dfrac{x^k}{(x+a)^m}:dx=dfrac{a^{-m}}{k+1},{_2F_1left(k+1,m,k+2,-dfrac1aright)}tag7$$
(see also Wolfram Alpha),
$$int_1^inftydfrac{x^k}{(x+a)^m}:dx=dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}tag8$$
(see also Wolfram Alpha).
Splitting the issue integral to the intervals $(0,1)$ and $(1,infty)$ and applying solutions $(5)-(8),$ one can present the result $I(a,k,n,p,q)$ as the sum of the pair of series, using gamma function and hypergeometric functions.
Thanks to the other members of the discussion for their interesting ideas.
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
$endgroup$
HINT
Let us start from the OP result
begin{align}
&H_1left(a_1, a_2, m_1, m_2, n_3, k_3right)
= frac{1}{n_2}intlimits_0^{infty} frac{u^{k_3}}{left(u^{n_3} + a_1right)^{m_1}} frac{1}{left(u + a_2right)^{m_2}}:du.tag1
end{align}
Taking in account, that identity
begin{align} &frac{x^n}{left(x^n + aright)^m}
=frac{x^n+a-a}{left(u^n + aright)^m}
=frac{1}{left(x^n + aright)^{m-1}}
-frac{a}{left(x^n + aright)^m}tag2 end{align}
allows to manage the degree in the numerator both decreasing and increasing,
assume WLOG $mathbf{k_3<m_1, k<m_2}.$
Also, due to the scaling substitution, let WLOG $mathbf{a_1=1}.$
Under these conditions, enough to consider the integral
begin{align}
&I(a,k,n,p,q)
= intlimits_0^{infty} frac{u^k}{left(u^n + 1right)^p} frac{1}{left(u + aright)^{q}}:du,quad k<p,quad k<q, quad a>1.tag3
end{align}
Is known the binomial decomposition
begin{align}
&(1-z)^{-d-1}=sumlimits_{j=d}^inftybinom{j}{d}z^{j-d}
= sumlimits_{j=d}^inftydfrac{Gamma(j+1)}{Gamma(d+1)Gamma(j-d+1)}z^{j-d},quad|z|<1.tag4
end{align}
If $|u|<1$ then from $(4)$ should Maclaurin series
begin{align}
&(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1}dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{n(j-q+1)}.tag5
end{align}
If $|u|>1$ then from $(4)$ can be obtained Laurent series
begin{align}
&(1+u^n)^{-q} = u^{-nq}(1+u^n)^{-q} = sumlimits_{j=q-1}^infty(-1)^{j-q+1} dfrac{Gamma(j+1)}{Gamma(q)Gamma(j-q+2)}u^{-n(j-q+1)}.tag6
end{align}
At the same time, obtained integrals on the both intervals can be expressed via hypergeometric function
$$int_0^1dfrac{x^k}{(x+a)^m}:dx=dfrac{a^{-m}}{k+1},{_2F_1left(k+1,m,k+2,-dfrac1aright)}tag7$$
(see also Wolfram Alpha),
$$int_1^inftydfrac{x^k}{(x+a)^m}:dx=dfrac{_2F_1(m,-k+m-1,m-k,-a)}{-k+m-1}tag8$$
(see also Wolfram Alpha).
Splitting the issue integral to the intervals $(0,1)$ and $(1,infty)$ and applying solutions $(5)-(8),$ one can present the result $I(a,k,n,p,q)$ as the sum of the pair of series, using gamma function and hypergeometric functions.
Thanks to the other members of the discussion for their interesting ideas.
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
edited Feb 8 at 3:53
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
answered Feb 8 at 2:04
Yuri NegometyanovYuri Negometyanov
12.7k1729
12.7k1729
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
add a comment |
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
$begingroup$
Thank you very much for your solution - very much appreciated.
$endgroup$
– user150203
Feb 8 at 11:09
add a comment |
$begingroup$
This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $ldots$
The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta (x)} , du,$$
for suitable conditions on the function $beta (x)$.
Let us use this parametrisation on your first integral $J(a_1, n_1, m_1, k_1)$. From Schwinger parametrisation we observe that
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} , du,$$
(here $beta (x) = x^{n_1} + a_1$ and $p = m_1$). Your integral for $J$ then becomes
$$J(a_1, n_1, m_1, k_1) = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du int_0^infty x^{k_1} e^{-u x^{n_1}} , dx,$$
where a change in the order of integration has been made. Now if we let $t = u x^{n_1}$ we have
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du frac{1}{n_1} u^{-frac{(k_1 + 1)}{n_1}} int_0^infty t^{frac{k_1 + 1}{n_1} - 1} e^{-t} , dt\
&= frac{1}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} , du.
end{align}
Enforcing a substitution of $u mapsto u/a_1$ leads to
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} , du\[2ex]
&= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (m_1 - frac{k_1 + 1}{n_1} right ) Gamma left (frac{k_1 + 1}{n_1} right ).
end{align}
Now attempting such an approach on the integral $H$. Here the following two Schwinger parametrisations are used:
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} , dy,$$
and
$$frac{1}{(x^{n_2} + a_2)^{m_2}} = frac{1}{Gamma (m_2)} int_0^infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} , dz.$$
So the integral for $H$, after a change in the order of integration has been made, becomes
$$H = frac{1}{Gamma (m_1) Gamma (m_2)} int_0^infty y^{m_1 - 1} e^{-y a_1} , dy int_0^infty z^{m_2 - 1} e^{-z a_2} , dz int_0^infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} , dx.$$
The roadblock is what is to be done with that inner $x$-integral if $n_1 neq n_2$?
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
$endgroup$
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
add a comment |
$begingroup$
This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $ldots$
The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta (x)} , du,$$
for suitable conditions on the function $beta (x)$.
Let us use this parametrisation on your first integral $J(a_1, n_1, m_1, k_1)$. From Schwinger parametrisation we observe that
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} , du,$$
(here $beta (x) = x^{n_1} + a_1$ and $p = m_1$). Your integral for $J$ then becomes
$$J(a_1, n_1, m_1, k_1) = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du int_0^infty x^{k_1} e^{-u x^{n_1}} , dx,$$
where a change in the order of integration has been made. Now if we let $t = u x^{n_1}$ we have
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du frac{1}{n_1} u^{-frac{(k_1 + 1)}{n_1}} int_0^infty t^{frac{k_1 + 1}{n_1} - 1} e^{-t} , dt\
&= frac{1}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} , du.
end{align}
Enforcing a substitution of $u mapsto u/a_1$ leads to
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} , du\[2ex]
&= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (m_1 - frac{k_1 + 1}{n_1} right ) Gamma left (frac{k_1 + 1}{n_1} right ).
end{align}
Now attempting such an approach on the integral $H$. Here the following two Schwinger parametrisations are used:
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} , dy,$$
and
$$frac{1}{(x^{n_2} + a_2)^{m_2}} = frac{1}{Gamma (m_2)} int_0^infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} , dz.$$
So the integral for $H$, after a change in the order of integration has been made, becomes
$$H = frac{1}{Gamma (m_1) Gamma (m_2)} int_0^infty y^{m_1 - 1} e^{-y a_1} , dy int_0^infty z^{m_2 - 1} e^{-z a_2} , dz int_0^infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} , dx.$$
The roadblock is what is to be done with that inner $x$-integral if $n_1 neq n_2$?
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
$endgroup$
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
add a comment |
$begingroup$
This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $ldots$
The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta (x)} , du,$$
for suitable conditions on the function $beta (x)$.
Let us use this parametrisation on your first integral $J(a_1, n_1, m_1, k_1)$. From Schwinger parametrisation we observe that
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} , du,$$
(here $beta (x) = x^{n_1} + a_1$ and $p = m_1$). Your integral for $J$ then becomes
$$J(a_1, n_1, m_1, k_1) = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du int_0^infty x^{k_1} e^{-u x^{n_1}} , dx,$$
where a change in the order of integration has been made. Now if we let $t = u x^{n_1}$ we have
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du frac{1}{n_1} u^{-frac{(k_1 + 1)}{n_1}} int_0^infty t^{frac{k_1 + 1}{n_1} - 1} e^{-t} , dt\
&= frac{1}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} , du.
end{align}
Enforcing a substitution of $u mapsto u/a_1$ leads to
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} , du\[2ex]
&= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (m_1 - frac{k_1 + 1}{n_1} right ) Gamma left (frac{k_1 + 1}{n_1} right ).
end{align}
Now attempting such an approach on the integral $H$. Here the following two Schwinger parametrisations are used:
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} , dy,$$
and
$$frac{1}{(x^{n_2} + a_2)^{m_2}} = frac{1}{Gamma (m_2)} int_0^infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} , dz.$$
So the integral for $H$, after a change in the order of integration has been made, becomes
$$H = frac{1}{Gamma (m_1) Gamma (m_2)} int_0^infty y^{m_1 - 1} e^{-y a_1} , dy int_0^infty z^{m_2 - 1} e^{-z a_2} , dz int_0^infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} , dx.$$
The roadblock is what is to be done with that inner $x$-integral if $n_1 neq n_2$?
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
$endgroup$
This is not a solution. It is an approach that shows where some of the potential roadblocks lay. Of course, this is not to say they are not insurmountable, but $ldots$
The approach to be use will make use of the so-called Schwinger parametrisation that makes use of the well-known observation of
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta (x)} , du,$$
for suitable conditions on the function $beta (x)$.
Let us use this parametrisation on your first integral $J(a_1, n_1, m_1, k_1)$. From Schwinger parametrisation we observe that
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u (x^{n_1} + a_1)} , du,$$
(here $beta (x) = x^{n_1} + a_1$ and $p = m_1$). Your integral for $J$ then becomes
$$J(a_1, n_1, m_1, k_1) = frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du int_0^infty x^{k_1} e^{-u x^{n_1}} , dx,$$
where a change in the order of integration has been made. Now if we let $t = u x^{n_1}$ we have
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{1}{Gamma (m_1)} int_0^infty u^{m_1 - 1} e^{-u a_1} , du frac{1}{n_1} u^{-frac{(k_1 + 1)}{n_1}} int_0^infty t^{frac{k_1 + 1}{n_1} - 1} e^{-t} , dt\
&= frac{1}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{m_1 n_1 - k_1 - 1}{n_1} - 1} e^{-u a_1} , du.
end{align}
Enforcing a substitution of $u mapsto u/a_1$ leads to
begin{align}
J(a_1, n_1, m_1, k_1) &= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (frac{k_1 + 1}{n_1} right ) int_0^infty u^{frac{n_1 m_1 - k_1 - 1}{n_1} - 1} e^{-u} , du\[2ex]
&= frac{a_1^{frac{k_1 +1}{n_1} - m_1}}{n_1 Gamma (m_1)} Gamma left (m_1 - frac{k_1 + 1}{n_1} right ) Gamma left (frac{k_1 + 1}{n_1} right ).
end{align}
Now attempting such an approach on the integral $H$. Here the following two Schwinger parametrisations are used:
$$frac{1}{(x^{n_1} + a_1)^{m_1}} = frac{1}{Gamma (m_1)} int_0^infty y^{m_1 - 1} e^{-y (x^{n_1} + a_1)} , dy,$$
and
$$frac{1}{(x^{n_2} + a_2)^{m_2}} = frac{1}{Gamma (m_2)} int_0^infty z^{m_2 - 1} e^{-z (x^{n_2} + a_2)} , dz.$$
So the integral for $H$, after a change in the order of integration has been made, becomes
$$H = frac{1}{Gamma (m_1) Gamma (m_2)} int_0^infty y^{m_1 - 1} e^{-y a_1} , dy int_0^infty z^{m_2 - 1} e^{-z a_2} , dz int_0^infty x^{k_1 + k_2} e^{-(y x^{n_1} + z x^{n_2})} , dx.$$
The roadblock is what is to be done with that inner $x$-integral if $n_1 neq n_2$?
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
answered Feb 7 at 22:30
omegadotomegadot
6,2692829
6,2692829
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
add a comment |
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
1
1
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
Fantastic post. I did not know about the Schwinger parameterisation!!! Thanks heaps, this will serve greatly
$endgroup$
– user150203
Feb 8 at 1:04
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
What are the restrictions on such a function $beta$ for that property to hold?
$endgroup$
– clathratus
Feb 8 at 4:24
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
$begingroup$
@clathratus that is a good question. To be quite honest with you I am not exactly sure. Admittedly, suitable conditions on the function $beta (x)$ is pretty vague. If I manage to find this out I will add it to my answer.
$endgroup$
– omegadot
Feb 8 at 7:46
add a comment |
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2
$begingroup$
Just as a joke : $a_2=0$ !
$endgroup$
– Claude Leibovici
Feb 3 at 9:56
$begingroup$
Haha oh, I wish I could set it that way :-)
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– user150203
Feb 3 at 10:14
1
$begingroup$
maybe preform $umapsto a_2 u$ then use a series for $frac1{(u+1)^{m_2}}$
$endgroup$
– clathratus
Feb 5 at 1:40
1
$begingroup$
yeah break up the integral with $int_0^infty=int_0^1 +int_1^infty $ then use $$(x+1)^alpha=sum_{ngeq0}{alphachoose n}x^n,qquad xin(-1,1)$$ and $$left(frac{x+1}xright)^alpha=sum_{ngeq0}{alphachoose n}x^{-n},qquad xinBbb Rsetminus [-1,1]$$ after you do the substitution I mentioned above.
$endgroup$
– clathratus
Feb 5 at 1:45
$begingroup$
Thanks heaps for those pointers. I will apply tonight.
$endgroup$
– user150203
Feb 5 at 7:34