Mixing
An Irresistible integral: $int_0^infty frac{x^{2m}}{(ax^2+b)^n}mathrm...
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According to the author of Irresistible Integrals,
$$I(n,m;a,b)=int_0^infty frac{x^{2m}mathrm dx}{(ax^2+b)^n}=frac{pi}{2a^mb^{n-m-1}sqrt{ab}}frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$$
Which I am attempting to prove. My progress:
$$J(n;a,b)=I(n,0;a,b)=int_0^infty frac{mathrm dx}{(ax^2+b)^n}$$
As I have shown in other posts of mine, this integral satisfies the recurrence
$$J(n;a,b)=frac{2n-3}{2bn}J(n-1;a,b)$$
And has the base case $$J(1;a,b)=fracpi{2sqrt{ab}}$$
So
$$J(n;a,b)=frac{pi}{2^{2n-1}b^{n-1}sqrt{ab}}{2n-2choose n-1}$$
Then I noticed that
$$frac{partial}{partial a}J(n;a,b)=-nI(n+1,1;a,b)$$
$$frac{partial^2}{partial a^2}J(n;a,b)=n(n+1)I(n+2,2;a,b)$$
$$...$$
$$frac{partial^k}{partial a^k}J(n;a,b)=(-1)^k(n)_kI(n+k,k;a,b)=frac{pisqrtpi}{2^{2n-1}b^{n-1}a^{k}sqrt{ab},Gamma(frac{1-2k}2)}{2n-2choose n-1}$$
Here $(n)_k=frac{Gamma(n+k)}{Gamma(n)}$. Yet I am confused as to how to proceed. Could I have some help?
real-analysis calculus integration
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
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add a comment |
$begingroup$
According to the author of Irresistible Integrals,
$$I(n,m;a,b)=int_0^infty frac{x^{2m}mathrm dx}{(ax^2+b)^n}=frac{pi}{2a^mb^{n-m-1}sqrt{ab}}frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$$
Which I am attempting to prove. My progress:
$$J(n;a,b)=I(n,0;a,b)=int_0^infty frac{mathrm dx}{(ax^2+b)^n}$$
As I have shown in other posts of mine, this integral satisfies the recurrence
$$J(n;a,b)=frac{2n-3}{2bn}J(n-1;a,b)$$
And has the base case $$J(1;a,b)=fracpi{2sqrt{ab}}$$
So
$$J(n;a,b)=frac{pi}{2^{2n-1}b^{n-1}sqrt{ab}}{2n-2choose n-1}$$
Then I noticed that
$$frac{partial}{partial a}J(n;a,b)=-nI(n+1,1;a,b)$$
$$frac{partial^2}{partial a^2}J(n;a,b)=n(n+1)I(n+2,2;a,b)$$
$$...$$
$$frac{partial^k}{partial a^k}J(n;a,b)=(-1)^k(n)_kI(n+k,k;a,b)=frac{pisqrtpi}{2^{2n-1}b^{n-1}a^{k}sqrt{ab},Gamma(frac{1-2k}2)}{2n-2choose n-1}$$
Here $(n)_k=frac{Gamma(n+k)}{Gamma(n)}$. Yet I am confused as to how to proceed. Could I have some help?
real-analysis calculus integration
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
$endgroup$
1
$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
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– Frank W.
Jan 22 at 21:05
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@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
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Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50
add a comment |
$begingroup$
According to the author of Irresistible Integrals,
$$I(n,m;a,b)=int_0^infty frac{x^{2m}mathrm dx}{(ax^2+b)^n}=frac{pi}{2a^mb^{n-m-1}sqrt{ab}}frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$$
Which I am attempting to prove. My progress:
$$J(n;a,b)=I(n,0;a,b)=int_0^infty frac{mathrm dx}{(ax^2+b)^n}$$
As I have shown in other posts of mine, this integral satisfies the recurrence
$$J(n;a,b)=frac{2n-3}{2bn}J(n-1;a,b)$$
And has the base case $$J(1;a,b)=fracpi{2sqrt{ab}}$$
So
$$J(n;a,b)=frac{pi}{2^{2n-1}b^{n-1}sqrt{ab}}{2n-2choose n-1}$$
Then I noticed that
$$frac{partial}{partial a}J(n;a,b)=-nI(n+1,1;a,b)$$
$$frac{partial^2}{partial a^2}J(n;a,b)=n(n+1)I(n+2,2;a,b)$$
$$...$$
$$frac{partial^k}{partial a^k}J(n;a,b)=(-1)^k(n)_kI(n+k,k;a,b)=frac{pisqrtpi}{2^{2n-1}b^{n-1}a^{k}sqrt{ab},Gamma(frac{1-2k}2)}{2n-2choose n-1}$$
Here $(n)_k=frac{Gamma(n+k)}{Gamma(n)}$. Yet I am confused as to how to proceed. Could I have some help?
real-analysis calculus integration
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
$endgroup$
According to the author of Irresistible Integrals,
$$I(n,m;a,b)=int_0^infty frac{x^{2m}mathrm dx}{(ax^2+b)^n}=frac{pi}{2a^mb^{n-m-1}sqrt{ab}}frac{(2m-1)!!(2n-2m-3)!!}{(2m-2)!!}$$
Which I am attempting to prove. My progress:
$$J(n;a,b)=I(n,0;a,b)=int_0^infty frac{mathrm dx}{(ax^2+b)^n}$$
As I have shown in other posts of mine, this integral satisfies the recurrence
$$J(n;a,b)=frac{2n-3}{2bn}J(n-1;a,b)$$
And has the base case $$J(1;a,b)=fracpi{2sqrt{ab}}$$
So
$$J(n;a,b)=frac{pi}{2^{2n-1}b^{n-1}sqrt{ab}}{2n-2choose n-1}$$
Then I noticed that
$$frac{partial}{partial a}J(n;a,b)=-nI(n+1,1;a,b)$$
$$frac{partial^2}{partial a^2}J(n;a,b)=n(n+1)I(n+2,2;a,b)$$
$$...$$
$$frac{partial^k}{partial a^k}J(n;a,b)=(-1)^k(n)_kI(n+k,k;a,b)=frac{pisqrtpi}{2^{2n-1}b^{n-1}a^{k}sqrt{ab},Gamma(frac{1-2k}2)}{2n-2choose n-1}$$
Here $(n)_k=frac{Gamma(n+k)}{Gamma(n)}$. Yet I am confused as to how to proceed. Could I have some help?
real-analysis calculus integration
real-analysis calculus integration
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
edited Jan 23 at 0:55
clathratus
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
asked Jan 22 at 18:26
clathratusclathratus
4,8871338
4,8871338
1
$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
$endgroup$
– Frank W.
Jan 22 at 21:05
$begingroup$
@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
$begingroup$
Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50
add a comment |
1
$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
$endgroup$
– Frank W.
Jan 22 at 21:05
$begingroup$
@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
$begingroup$
Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50
1
1
$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
$endgroup$
– Frank W.
Jan 22 at 21:05
$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
$endgroup$
– Frank W.
Jan 22 at 21:05
$begingroup$
@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
$begingroup$
@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
$begingroup$
Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50
$begingroup$
Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50
add a comment |
2 Answers
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Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand.
For a positive, continuous function $beta (x)$ observe that for $p > 0$
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta(x)} , du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n in mathbb{N}$ such that $n > m$. Choosing $beta (x) = ax^2 + b$ and $p = n$ we have
$$frac{1}{(ax^2 + b)^n} = frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-u(ax^2 + b)} ,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
begin{align}
I(m,n; a,b) &= frac{1}{Gamma (n)} int_0^infty int_0^infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} , du , dx\
&= frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-ub} int_0^infty x^{2m} e^{-uax^2} , dx , du,
end{align}
after the order of integration has been changed.
Enforcing a substitution of $x mapsto sqrt{dfrac{x}{ua}}$ leads to
begin{align}
I(m,n;a,b) &= frac{1}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m - 3/2} e^{-ub} int_0^infty x^{m - 1/2} e^{-x} , dx\
&= frac{Gamma (m + 1/2)}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-ub} , du.
end{align}
Next, enforcing a substitution of $u mapsto u/b$ yields
begin{align}
I(m,n;a,b) &= frac{Gamma (m + 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-u} , du\
&= frac{Gamma (m + 1/2) Gamma (n - m - 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
end{align}
or in terms of the beta function
$$I(m,n;a,b) = frac{1}{2b^n} left (frac{b}{a} right )^{m + frac{1}{2}} operatorname{B} left (n - m - frac{1}{2}, m + frac{1}{2} right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = frac{n pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} sqrt{ab}} frac{binom{2m}{m} binom{2n - 2m}{n - m}}{binom{n}{m}}.$$
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
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Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
add a comment |
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You may substitute $ax^2 + b = b/u$ to obtain
begin{align*}
int_{0}^{infty} frac{x^{2m}}{(ax^2+b)^n} , mathrm{d}x
&= int_{0}^{1} frac{left( frac{b}{a}(frac{1}{u}-1) right)^m}{left(frac{b}{u}right)^n} , frac{sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} , mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} mathrm{B}left(n-m-tfrac{1}{2}, m + tfrac{1}{2}right).
end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
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Does this reduce to the answer presented in Irresistible Integrals?
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– clathratus
Jan 22 at 18:48
1
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@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
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– Sangchul Lee
Jan 22 at 19:02
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@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
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– DavidG
Jan 23 at 11:52
add a comment |
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2 Answers
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Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand.
For a positive, continuous function $beta (x)$ observe that for $p > 0$
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta(x)} , du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n in mathbb{N}$ such that $n > m$. Choosing $beta (x) = ax^2 + b$ and $p = n$ we have
$$frac{1}{(ax^2 + b)^n} = frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-u(ax^2 + b)} ,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
begin{align}
I(m,n; a,b) &= frac{1}{Gamma (n)} int_0^infty int_0^infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} , du , dx\
&= frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-ub} int_0^infty x^{2m} e^{-uax^2} , dx , du,
end{align}
after the order of integration has been changed.
Enforcing a substitution of $x mapsto sqrt{dfrac{x}{ua}}$ leads to
begin{align}
I(m,n;a,b) &= frac{1}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m - 3/2} e^{-ub} int_0^infty x^{m - 1/2} e^{-x} , dx\
&= frac{Gamma (m + 1/2)}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-ub} , du.
end{align}
Next, enforcing a substitution of $u mapsto u/b$ yields
begin{align}
I(m,n;a,b) &= frac{Gamma (m + 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-u} , du\
&= frac{Gamma (m + 1/2) Gamma (n - m - 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
end{align}
or in terms of the beta function
$$I(m,n;a,b) = frac{1}{2b^n} left (frac{b}{a} right )^{m + frac{1}{2}} operatorname{B} left (n - m - frac{1}{2}, m + frac{1}{2} right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = frac{n pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} sqrt{ab}} frac{binom{2m}{m} binom{2n - 2m}{n - m}}{binom{n}{m}}.$$
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
$endgroup$
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
add a comment |
$begingroup$
Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand.
For a positive, continuous function $beta (x)$ observe that for $p > 0$
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta(x)} , du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n in mathbb{N}$ such that $n > m$. Choosing $beta (x) = ax^2 + b$ and $p = n$ we have
$$frac{1}{(ax^2 + b)^n} = frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-u(ax^2 + b)} ,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
begin{align}
I(m,n; a,b) &= frac{1}{Gamma (n)} int_0^infty int_0^infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} , du , dx\
&= frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-ub} int_0^infty x^{2m} e^{-uax^2} , dx , du,
end{align}
after the order of integration has been changed.
Enforcing a substitution of $x mapsto sqrt{dfrac{x}{ua}}$ leads to
begin{align}
I(m,n;a,b) &= frac{1}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m - 3/2} e^{-ub} int_0^infty x^{m - 1/2} e^{-x} , dx\
&= frac{Gamma (m + 1/2)}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-ub} , du.
end{align}
Next, enforcing a substitution of $u mapsto u/b$ yields
begin{align}
I(m,n;a,b) &= frac{Gamma (m + 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-u} , du\
&= frac{Gamma (m + 1/2) Gamma (n - m - 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
end{align}
or in terms of the beta function
$$I(m,n;a,b) = frac{1}{2b^n} left (frac{b}{a} right )^{m + frac{1}{2}} operatorname{B} left (n - m - frac{1}{2}, m + frac{1}{2} right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = frac{n pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} sqrt{ab}} frac{binom{2m}{m} binom{2n - 2m}{n - m}}{binom{n}{m}}.$$
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
$endgroup$
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
add a comment |
$begingroup$
Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand.
For a positive, continuous function $beta (x)$ observe that for $p > 0$
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta(x)} , du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n in mathbb{N}$ such that $n > m$. Choosing $beta (x) = ax^2 + b$ and $p = n$ we have
$$frac{1}{(ax^2 + b)^n} = frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-u(ax^2 + b)} ,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
begin{align}
I(m,n; a,b) &= frac{1}{Gamma (n)} int_0^infty int_0^infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} , du , dx\
&= frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-ub} int_0^infty x^{2m} e^{-uax^2} , dx , du,
end{align}
after the order of integration has been changed.
Enforcing a substitution of $x mapsto sqrt{dfrac{x}{ua}}$ leads to
begin{align}
I(m,n;a,b) &= frac{1}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m - 3/2} e^{-ub} int_0^infty x^{m - 1/2} e^{-x} , dx\
&= frac{Gamma (m + 1/2)}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-ub} , du.
end{align}
Next, enforcing a substitution of $u mapsto u/b$ yields
begin{align}
I(m,n;a,b) &= frac{Gamma (m + 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-u} , du\
&= frac{Gamma (m + 1/2) Gamma (n - m - 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
end{align}
or in terms of the beta function
$$I(m,n;a,b) = frac{1}{2b^n} left (frac{b}{a} right )^{m + frac{1}{2}} operatorname{B} left (n - m - frac{1}{2}, m + frac{1}{2} right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = frac{n pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} sqrt{ab}} frac{binom{2m}{m} binom{2n - 2m}{n - m}}{binom{n}{m}}.$$
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
$endgroup$
Here is an alternative method where I will make use of a so-called Schwinger parametrisation. Such a parametrisation is particularly suited to integrals where a polynomial raised to a power appears in the denominator of the integrand.
For a positive, continuous function $beta (x)$ observe that for $p > 0$
$$frac{1}{beta^p (x)} = frac{1}{Gamma (p)} int_0^infty u^{p - 1} e^{-u beta(x)} , du.$$
It is this observation that is known as the Schwinger parametrisation.
I will assume $a,b > 0$ and $m,n in mathbb{N}$ such that $n > m$. Choosing $beta (x) = ax^2 + b$ and $p = n$ we have
$$frac{1}{(ax^2 + b)^n} = frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-u(ax^2 + b)} ,du.$$
The integral $I(m, n; a, b)$ can thus be rewritten as
begin{align}
I(m,n; a,b) &= frac{1}{Gamma (n)} int_0^infty int_0^infty x^{2m} u^{n - 1} e^{-u(ax^2 + b)} , du , dx\
&= frac{1}{Gamma (n)} int_0^infty u^{n - 1} e^{-ub} int_0^infty x^{2m} e^{-uax^2} , dx , du,
end{align}
after the order of integration has been changed.
Enforcing a substitution of $x mapsto sqrt{dfrac{x}{ua}}$ leads to
begin{align}
I(m,n;a,b) &= frac{1}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m - 3/2} e^{-ub} int_0^infty x^{m - 1/2} e^{-x} , dx\
&= frac{Gamma (m + 1/2)}{2 Gamma (n) a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-ub} , du.
end{align}
Next, enforcing a substitution of $u mapsto u/b$ yields
begin{align}
I(m,n;a,b) &= frac{Gamma (m + 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}} int_0^infty u^{n - m -3/2} e^{-u} , du\
&= frac{Gamma (m + 1/2) Gamma (n - m - 1/2)}{2 Gamma (n) b^{n - m - 1/2} a^{m + 1/2}},
end{align}
or in terms of the beta function
$$I(m,n;a,b) = frac{1}{2b^n} left (frac{b}{a} right )^{m + frac{1}{2}} operatorname{B} left (n - m - frac{1}{2}, m + frac{1}{2} right ),$$
in agreement with the result given by @Sangchul Lee. Granted, the substitution Sangchul Lee uses is pretty slick, and gets one to this point a lot quicker than making use of a Schwinger parametrisation, but it at least shows you an alternative approach to reaching the same point.
Note that in terms of central binomial coefficients the result can be expressed as
$$I(m,n; a, b) = frac{n pi}{(2n - 2m - 1) 2^{2n} a^m b^{n - m - 1} sqrt{ab}} frac{binom{2m}{m} binom{2n - 2m}{n - m}}{binom{n}{m}}.$$
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
answered Feb 10 at 4:00
omegadotomegadot
6,3972829
6,3972829
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
add a comment |
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
$begingroup$
Very very impressive! Thank you very much!
$endgroup$
– clathratus
Feb 10 at 4:01
add a comment |
$begingroup$
You may substitute $ax^2 + b = b/u$ to obtain
begin{align*}
int_{0}^{infty} frac{x^{2m}}{(ax^2+b)^n} , mathrm{d}x
&= int_{0}^{1} frac{left( frac{b}{a}(frac{1}{u}-1) right)^m}{left(frac{b}{u}right)^n} , frac{sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} , mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} mathrm{B}left(n-m-tfrac{1}{2}, m + tfrac{1}{2}right).
end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
1
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
add a comment |
$begingroup$
You may substitute $ax^2 + b = b/u$ to obtain
begin{align*}
int_{0}^{infty} frac{x^{2m}}{(ax^2+b)^n} , mathrm{d}x
&= int_{0}^{1} frac{left( frac{b}{a}(frac{1}{u}-1) right)^m}{left(frac{b}{u}right)^n} , frac{sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} , mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} mathrm{B}left(n-m-tfrac{1}{2}, m + tfrac{1}{2}right).
end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
1
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
add a comment |
$begingroup$
You may substitute $ax^2 + b = b/u$ to obtain
begin{align*}
int_{0}^{infty} frac{x^{2m}}{(ax^2+b)^n} , mathrm{d}x
&= int_{0}^{1} frac{left( frac{b}{a}(frac{1}{u}-1) right)^m}{left(frac{b}{u}right)^n} , frac{sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} , mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} mathrm{B}left(n-m-tfrac{1}{2}, m + tfrac{1}{2}right).
end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
$endgroup$
You may substitute $ax^2 + b = b/u$ to obtain
begin{align*}
int_{0}^{infty} frac{x^{2m}}{(ax^2+b)^n} , mathrm{d}x
&= int_{0}^{1} frac{left( frac{b}{a}(frac{1}{u}-1) right)^m}{left(frac{b}{u}right)^n} , frac{sqrt{b/a}}{2u^{3/2}(1-u)^{1/2}} mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} int_{0}^{1} u^{n-m-3/2}(1-u)^{m-1/2} , mathrm{d}u \
&= frac{(b/a)^{m+1/2}}{2 b^n} mathrm{B}left(n-m-tfrac{1}{2}, m + tfrac{1}{2}right).
end{align*}
Hope the computation is correct as I hurried a bit when writing things down...
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
answered Jan 22 at 18:42
Sangchul LeeSangchul Lee
95.6k12171279
95.6k12171279
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
1
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
add a comment |
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
1
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
$begingroup$
Does this reduce to the answer presented in Irresistible Integrals?
$endgroup$
– clathratus
Jan 22 at 18:48
1
1
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@clathratus The gamma function of half-integers reduces to double-factorial multiplied by $sqrt{pi}/2^{[text{some integer}]}$. So, if I computed everything correctly, yes.
$endgroup$
– Sangchul Lee
Jan 22 at 19:02
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
$begingroup$
@SangchulLee - Is there a method (i.e. Half Tangent Substitution) with your substitution? or was it something you saw yourself? Very cool! It compressed my method of 3 subs to one in a single go! math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:52
add a comment |
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$begingroup$
I believe you’re missing a $mathrm dx$ in the title. Just nitpicking.
$endgroup$
– Frank W.
Jan 22 at 21:05
$begingroup$
@FrankW. Thanks for catching that :)
$endgroup$
– clathratus
Jan 23 at 0:57
$begingroup$
Already done :-) Just need to drag the $a$ out the front. math.stackexchange.com/questions/3057298/…
$endgroup$
– DavidG
Jan 23 at 11:50