Mixing
“Milk” the integral $int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$
$begingroup$
I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
begin{align}
int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx&= int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=frac12int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rmathrm dx\
&=sqrt{frac{pi(a+1)}{2}}frac{Gamma(r-frac12)}{(2a+2)^rGamma(r)}
,end{align}
Which works for $r>frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}mathrm dx=frac{sqrt{pi},Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
Taking $frac{d}{dr}$ on both sides,
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}logleft(frac{x}{x^2+1}right)mathrm dx=frac{sqrt{pi}}{2}frac{d}{dr}frac{Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}log^nleft[frac{x}{x^2+1}right]frac{mathrm dx}{(x^2+1)^2}=frac{sqrtpi}{2^{n+4}}left(frac{d}{dr}right)^nfrac{Gamma(r+frac32)}{4^rGamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
integration soft-question big-list
edited Jan 28 at 18:06
klirk
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
$endgroup$
add a comment |
$begingroup$
I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
begin{align}
int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx&= int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=frac12int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rmathrm dx\
&=sqrt{frac{pi(a+1)}{2}}frac{Gamma(r-frac12)}{(2a+2)^rGamma(r)}
,end{align}
Which works for $r>frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}mathrm dx=frac{sqrt{pi},Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
Taking $frac{d}{dr}$ on both sides,
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}logleft(frac{x}{x^2+1}right)mathrm dx=frac{sqrt{pi}}{2}frac{d}{dr}frac{Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}log^nleft[frac{x}{x^2+1}right]frac{mathrm dx}{(x^2+1)^2}=frac{sqrtpi}{2^{n+4}}left(frac{d}{dr}right)^nfrac{Gamma(r+frac32)}{4^rGamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
integration soft-question big-list
edited Jan 28 at 18:06
klirk
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
$endgroup$
$begingroup$
What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
2
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23
add a comment |
$begingroup$
I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
begin{align}
int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx&= int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=frac12int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rmathrm dx\
&=sqrt{frac{pi(a+1)}{2}}frac{Gamma(r-frac12)}{(2a+2)^rGamma(r)}
,end{align}
Which works for $r>frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}mathrm dx=frac{sqrt{pi},Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
Taking $frac{d}{dr}$ on both sides,
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}logleft(frac{x}{x^2+1}right)mathrm dx=frac{sqrt{pi}}{2}frac{d}{dr}frac{Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}log^nleft[frac{x}{x^2+1}right]frac{mathrm dx}{(x^2+1)^2}=frac{sqrtpi}{2^{n+4}}left(frac{d}{dr}right)^nfrac{Gamma(r+frac32)}{4^rGamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
integration soft-question big-list
edited Jan 28 at 18:06
klirk
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
$endgroup$
I found the following integral in chapter $13$ of Irresistible Integrals, and I would like to see which conclusions you can reach from it. My goal in asking this question is to see which methods I can employ in the future to generalize/"milk" cool integrals like this. I admit this post is very similar to the original "Integral Miking" post, but since this post is concerning a specific integral, it is not a duplicate.
begin{align}
int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx&= int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=frac12int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}mathrm dx\
&=int_0^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rmathrm dx\
&=sqrt{frac{pi(a+1)}{2}}frac{Gamma(r-frac12)}{(2a+2)^rGamma(r)}
,end{align}
Which works for $r>frac12$ and all(?) $s$, because as the authors showed, the integral is independent of $s$.
This question wouldn't be complete without my attempts:
Setting $a=1$, we have
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}mathrm dx=frac{sqrt{pi},Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
Taking $frac{d}{dr}$ on both sides,
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}logleft(frac{x}{x^2+1}right)mathrm dx=frac{sqrt{pi}}{2}frac{d}{dr}frac{Gamma(r-frac12)}{2^{2r}Gamma(r)}.$$
And it can be shown, in a somewhat similar way, that
$$int_0^inftyleft(frac{x}{x^2+1}right)^{2r}log^nleft[frac{x}{x^2+1}right]frac{mathrm dx}{(x^2+1)^2}=frac{sqrtpi}{2^{n+4}}left(frac{d}{dr}right)^nfrac{Gamma(r+frac32)}{4^rGamma(r+2)}.$$
Unfortunately, I feel as if my creative well has run dry, and I would like to see what you can get from this integral. Have fun!
Edit: Context
The authors of Irresistible Integrals called this integral a "Master Formula" because it apparently could produce a plethora of identities. I would like to see which identities you can derive from said integral.
integration soft-question big-list
integration soft-question big-list
edited Jan 28 at 18:06
klirk
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
edited Jan 28 at 18:06
klirk
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
edited Jan 28 at 18:06
klirk
2,288631
edited Jan 28 at 18:06
klirk
2,288631
edited Jan 28 at 18:06
klirk
2,288631
2,288631
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
asked Jan 28 at 16:52
clathratusclathratus
5,0011438
5,0011438
$begingroup$
What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
2
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23
add a comment |
$begingroup$
What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
2
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23
$begingroup$
What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
2
2
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23
add a comment |
2 Answers
2
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oldest
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First I will like to give some steps and maybe some more insight of this integral.$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx+int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$$
With $xrightarrow frac{1}{x}$ in the second one we get:
$$int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx=int_0^1 left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2left(frac{1}{x^s}+1right)}mathrm dx$$
Now if we add with first part of the integral that was splited using: $displaystyle{frac{1}{x^s+1}+frac{1}{frac{1}{x^s}+1}=1},$ that is way the $s$ doesn't affect our integral. Anyway we get indeed by adding :
$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
Again via $xrightarrow frac{1}{x}$ we get:
$$I=int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
$$Rightarrow I=frac12int_0^infty left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx=frac12 int_{-infty}^infty left(frac{1}{x^2+frac{1}{x^2}+2a}right)^rleft(1+frac{1}{x^2}right)dx$$
And now by writting $displaystyle{x^2+frac{1}{x^2}=left(x-frac{1}{x}right)^2+2}$ and do a $x-frac{1}{x}=t$ we get: $$I=frac12 int_{-infty}^infty frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=xsqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=frac12 int_{-infty}^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx= frac12 int_{-infty}^infty left(frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}right)^r frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $xrightarrow x-frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $int_{-infty}^infty f(x)dx=int_{-infty}^infty f(x-cot x)dx$ to get :$$I=frac12 int_{-infty}^infty frac{1}{(x^2+2(a+1))^r}dx=frac12 int_{-infty}^infty frac{1}{((x-cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=frac12$ and $r=2$ to get: $$int_0^infty frac{1}{((x-cot x)^2 +1)^2}dx=sqrt 2 pi$$
Of course one can do the same thing when there is $x^4$ in the denominator, but that's quite evilish $smile$.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=frac12int_0^infty x^{r-1} frac{1}{(x^2+2ax+1)^r}left(sqrt x+frac{1}{sqrt x}right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem. And maybe I'll add another time other ideas.
answered Jan 29 at 18:35
ZackyZacky
7,77511062
$endgroup$
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
add a comment |
$begingroup$
This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}int_n...int (a+1)^{1-r} , da^n=frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}left( frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}},left(frac{d}{da}right)^n (a+1)^{1-r} =frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}} left( -(1-r-1) (1-r-2)...(1-r-n) ,(a+1)^{1-n-r} right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
First I will like to give some steps and maybe some more insight of this integral.$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx+int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$$
With $xrightarrow frac{1}{x}$ in the second one we get:
$$int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx=int_0^1 left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2left(frac{1}{x^s}+1right)}mathrm dx$$
Now if we add with first part of the integral that was splited using: $displaystyle{frac{1}{x^s+1}+frac{1}{frac{1}{x^s}+1}=1},$ that is way the $s$ doesn't affect our integral. Anyway we get indeed by adding :
$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
Again via $xrightarrow frac{1}{x}$ we get:
$$I=int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
$$Rightarrow I=frac12int_0^infty left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx=frac12 int_{-infty}^infty left(frac{1}{x^2+frac{1}{x^2}+2a}right)^rleft(1+frac{1}{x^2}right)dx$$
And now by writting $displaystyle{x^2+frac{1}{x^2}=left(x-frac{1}{x}right)^2+2}$ and do a $x-frac{1}{x}=t$ we get: $$I=frac12 int_{-infty}^infty frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=xsqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=frac12 int_{-infty}^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx= frac12 int_{-infty}^infty left(frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}right)^r frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $xrightarrow x-frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $int_{-infty}^infty f(x)dx=int_{-infty}^infty f(x-cot x)dx$ to get :$$I=frac12 int_{-infty}^infty frac{1}{(x^2+2(a+1))^r}dx=frac12 int_{-infty}^infty frac{1}{((x-cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=frac12$ and $r=2$ to get: $$int_0^infty frac{1}{((x-cot x)^2 +1)^2}dx=sqrt 2 pi$$
Of course one can do the same thing when there is $x^4$ in the denominator, but that's quite evilish $smile$.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=frac12int_0^infty x^{r-1} frac{1}{(x^2+2ax+1)^r}left(sqrt x+frac{1}{sqrt x}right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem. And maybe I'll add another time other ideas.
answered Jan 29 at 18:35
ZackyZacky
7,77511062
$endgroup$
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
add a comment |
$begingroup$
First I will like to give some steps and maybe some more insight of this integral.$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx+int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$$
With $xrightarrow frac{1}{x}$ in the second one we get:
$$int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx=int_0^1 left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2left(frac{1}{x^s}+1right)}mathrm dx$$
Now if we add with first part of the integral that was splited using: $displaystyle{frac{1}{x^s+1}+frac{1}{frac{1}{x^s}+1}=1},$ that is way the $s$ doesn't affect our integral. Anyway we get indeed by adding :
$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
Again via $xrightarrow frac{1}{x}$ we get:
$$I=int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
$$Rightarrow I=frac12int_0^infty left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx=frac12 int_{-infty}^infty left(frac{1}{x^2+frac{1}{x^2}+2a}right)^rleft(1+frac{1}{x^2}right)dx$$
And now by writting $displaystyle{x^2+frac{1}{x^2}=left(x-frac{1}{x}right)^2+2}$ and do a $x-frac{1}{x}=t$ we get: $$I=frac12 int_{-infty}^infty frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=xsqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=frac12 int_{-infty}^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx= frac12 int_{-infty}^infty left(frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}right)^r frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $xrightarrow x-frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $int_{-infty}^infty f(x)dx=int_{-infty}^infty f(x-cot x)dx$ to get :$$I=frac12 int_{-infty}^infty frac{1}{(x^2+2(a+1))^r}dx=frac12 int_{-infty}^infty frac{1}{((x-cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=frac12$ and $r=2$ to get: $$int_0^infty frac{1}{((x-cot x)^2 +1)^2}dx=sqrt 2 pi$$
Of course one can do the same thing when there is $x^4$ in the denominator, but that's quite evilish $smile$.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=frac12int_0^infty x^{r-1} frac{1}{(x^2+2ax+1)^r}left(sqrt x+frac{1}{sqrt x}right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem. And maybe I'll add another time other ideas.
answered Jan 29 at 18:35
ZackyZacky
7,77511062
$endgroup$
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
add a comment |
$begingroup$
First I will like to give some steps and maybe some more insight of this integral.$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx+int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$$
With $xrightarrow frac{1}{x}$ in the second one we get:
$$int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx=int_0^1 left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2left(frac{1}{x^s}+1right)}mathrm dx$$
Now if we add with first part of the integral that was splited using: $displaystyle{frac{1}{x^s+1}+frac{1}{frac{1}{x^s}+1}=1},$ that is way the $s$ doesn't affect our integral. Anyway we get indeed by adding :
$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
Again via $xrightarrow frac{1}{x}$ we get:
$$I=int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
$$Rightarrow I=frac12int_0^infty left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx=frac12 int_{-infty}^infty left(frac{1}{x^2+frac{1}{x^2}+2a}right)^rleft(1+frac{1}{x^2}right)dx$$
And now by writting $displaystyle{x^2+frac{1}{x^2}=left(x-frac{1}{x}right)^2+2}$ and do a $x-frac{1}{x}=t$ we get: $$I=frac12 int_{-infty}^infty frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=xsqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=frac12 int_{-infty}^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx= frac12 int_{-infty}^infty left(frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}right)^r frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $xrightarrow x-frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $int_{-infty}^infty f(x)dx=int_{-infty}^infty f(x-cot x)dx$ to get :$$I=frac12 int_{-infty}^infty frac{1}{(x^2+2(a+1))^r}dx=frac12 int_{-infty}^infty frac{1}{((x-cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=frac12$ and $r=2$ to get: $$int_0^infty frac{1}{((x-cot x)^2 +1)^2}dx=sqrt 2 pi$$
Of course one can do the same thing when there is $x^4$ in the denominator, but that's quite evilish $smile$.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=frac12int_0^infty x^{r-1} frac{1}{(x^2+2ax+1)^r}left(sqrt x+frac{1}{sqrt x}right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem. And maybe I'll add another time other ideas.
answered Jan 29 at 18:35
ZackyZacky
7,77511062
$endgroup$
First I will like to give some steps and maybe some more insight of this integral.$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx+int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx$$
With $xrightarrow frac{1}{x}$ in the second one we get:
$$int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx=int_0^1 left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2left(frac{1}{x^s}+1right)}mathrm dx$$
Now if we add with first part of the integral that was splited using: $displaystyle{frac{1}{x^s+1}+frac{1}{frac{1}{x^s}+1}=1},$ that is way the $s$ doesn't affect our integral. Anyway we get indeed by adding :
$$I=int_0^1left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
Again via $xrightarrow frac{1}{x}$ we get:
$$I=int_1^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx$$
$$Rightarrow I=frac12int_0^infty left(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2}dx=frac12 int_{-infty}^infty left(frac{1}{x^2+frac{1}{x^2}+2a}right)^rleft(1+frac{1}{x^2}right)dx$$
And now by writting $displaystyle{x^2+frac{1}{x^2}=left(x-frac{1}{x}right)^2+2}$ and do a $x-frac{1}{x}=t$ we get: $$I=frac12 int_{-infty}^infty frac{1}{(t^2+2(a+1))^r}dx$$ By letting $t=xsqrt{2(a+1)}$ we get rather easy using beta function the result.
But the substitution $x-frac{1}{x}$ reminds us of Glasser's Master theorem. Of course in order to milk it we can take the original integral and apply this theorem how many times we want.
$$I=frac12 int_{-infty}^inftyleft(frac{x^2}{x^4+2ax^2+1}right)^rfrac{x^2+1}{x^2(x^s+1)}mathrm dx= frac12 int_{-infty}^infty left(frac{x^6-2x^4+x^2}{x^8+2ax^6-4x^6-4ax^4+7x^4+2ax^2-4x^2+1}right)^r frac{x^4-x^2+1}{(x^2-1)^2}dx$$
Where I have used $xrightarrow x-frac{1}{x}$ and $s=0$. Of course we can be mean and in the simplest form use $int_{-infty}^infty f(x)dx=int_{-infty}^infty f(x-cot x)dx$ to get :$$I=frac12 int_{-infty}^infty frac{1}{(x^2+2(a+1))^r}dx=frac12 int_{-infty}^infty frac{1}{((x-cot x)^2+2(a+1))^r}dx$$ And by setting $a+1=frac12$ and $r=2$ to get: $$int_0^infty frac{1}{((x-cot x)^2 +1)^2}dx=sqrt 2 pi$$
Of course one can do the same thing when there is $x^4$ in the denominator, but that's quite evilish $smile$.
Or another thing would be to let $x^2=t$ in order to get some Mellin transforms, for example: $$I=frac12int_0^infty x^{r-1} frac{1}{(x^2+2ax+1)^r}left(sqrt x+frac{1}{sqrt x}right)dx$$
Also those two theorems might also do some milk with this integral: Laplace transform property and Plancherel theorem. And maybe I'll add another time other ideas.
answered Jan 29 at 18:35
ZackyZacky
7,77511062
answered Jan 29 at 18:35
ZackyZacky
7,77511062
answered Jan 29 at 18:35
ZackyZacky
7,77511062
answered Jan 29 at 18:35
ZackyZacky
7,77511062
7,77511062
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
add a comment |
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
1
1
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
$begingroup$
This is truly an integral milking (+1). Nice answer
$endgroup$
– clathratus
Jan 29 at 19:53
add a comment |
$begingroup$
This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}int_n...int (a+1)^{1-r} , da^n=frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}left( frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}},left(frac{d}{da}right)^n (a+1)^{1-r} =frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}} left( -(1-r-1) (1-r-2)...(1-r-n) ,(a+1)^{1-n-r} right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
$endgroup$
add a comment |
$begingroup$
This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}int_n...int (a+1)^{1-r} , da^n=frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}left( frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}},left(frac{d}{da}right)^n (a+1)^{1-r} =frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}} left( -(1-r-1) (1-r-2)...(1-r-n) ,(a+1)^{1-n-r} right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
$endgroup$
add a comment |
$begingroup$
This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}int_n...int (a+1)^{1-r} , da^n=frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}left( frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}},left(frac{d}{da}right)^n (a+1)^{1-r} =frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}} left( -(1-r-1) (1-r-2)...(1-r-n) ,(a+1)^{1-n-r} right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
$endgroup$
This is not an answer just a pause for thought on your question:
Math. Operation A, Integrate w.r.t $a$, $n$ times consecutively
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}int_n...int (a+1)^{1-r} , da^n=frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}}left( frac{ (a+1)^{1+n-r} }{(1-r+1)(2-r+1)...(n-r+1)}right)$$
Math. Operation B, Differentiate w.r.t $a$, $n$ times
$$frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}},left(frac{d}{da}right)^n (a+1)^{1-r} =frac{ Gamma left(r-frac{1}{2}right) } {2^r, Gamma (r)} sqrt{frac{pi }{2}} left( -(1-r-1) (1-r-2)...(1-r-n) ,(a+1)^{1-n-r} right)$$
Math. Operation C, Differentiate w.r.t $r$, $n$ times as you have indicated in your example
Integrating with respect to $r$ doesn't seem to be possible (at least for me).
So I have couple of thoughts for you while your creative well refills: Starting with $n=1$, are all basic permutations of repeated elementary operations 1,2 and 3 possible and does it matter in what order the desired mathematical operations are applied? Are there any other possibilities?
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
answered Jan 28 at 23:45
James ArathoonJames Arathoon
1,608423
1,608423
add a comment |
add a comment |
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What exactly is the question?
$endgroup$
– mrtaurho
Jan 28 at 17:19
$begingroup$
@mrtaurho see the edit
$endgroup$
– clathratus
Jan 28 at 17:22
2
$begingroup$
Ah, okay. I see now; to be honest it was really not clear but I guess like this it is more reasonable.
$endgroup$
– mrtaurho
Jan 28 at 17:23