Thread: Prove some identities! (Round two)

1. Originally Posted by simplependulum
$\int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1 }~dx dy$

Sub. $xy = u$ ( $y$ being constant ) we have

$\int_0^1 \int_0^1 \frac{\ln^s(xy)}{xy - 1 }~dx~dy$

$= \int_0^1 \int_0^y \frac{\ln^s(u)}{u-1} \frac{du}{y} ~dy$

$= \int_0^1 \int_u^1 \frac{dy}{y}~ \frac{\ln^s(u)}{u-1}~du$

$= - \int_0^1 \frac{\ln^{s+1}(u)}{u-1} ~du$

Sub. $u = e^{-t}$

$= - \int_0^{\infty} \frac{(-1)^{s+1} t^{s+1} }{ 1 - e^t }~dt$

$= (-1)^{s+1} \int_0^{\infty} \frac{t^{s+1}}{e^t - 1 } ~dt$

$= (-1)^{s+1} \Gamma(s+2) \zeta(s+2)$
Right, it's not that much harder.

2. I've come across the following identity multiple times, but I've never been able to prove it.

Show $\displaystyle \int_{0}^{\infty} \frac{1}{1+x^{a}} \ dx = \frac{\pi}{a} \csc \Big(\frac{\pi}{a}}\Big) \ \text{for} \ a>1$

3. Originally Posted by Random Variable
I've come across the following identity multiple times, but I've never been able to proof it.

Show $\displaystyle \int_{0}^{\infty} \frac{1}{1+x^{a}} \ dx = \frac{\pi}{a} \csc \Big(\frac{\pi}{a}}\Big) \ \text{for} \ a>0$
It should be $a>1$.

Here is one method. There is another I know using contour integration if you're interested.

Method 1:

Spoiler:

Let

$\displaystyle I(a)=\int_0^{\infty}\frac{dx}{1+x^a}$

Let $1+x^a=u$ and so $du=a x^{a-1}$ or $dx=\frac{1}{a}\left(u-1\right)^{\frac{1}{a}-1}$. Thus,

$\displaystyle I(a)=\frac{1}{a}\int_1^{\infty}\frac{(u-1)^{\frac{1}{a}-1}}{u}\text{ }du$

Let $z=\frac{1}{u}$ to get that

\displaystyle \begin{aligned}I(a) &=\frac{1}{a}\int_0^1 z^{\frac{-1}{a}}\left(1-z\right)^{\frac{1}{a}-1}\\ &= \frac{1}{a}B\left(1-\frac{1}{a},\frac{1}{a}\right)\\ &= \frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Ga mma\left(1-\frac{1}{a}\right)}{\Gamma(1)}\\ &=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\ left(1-\frac{1}{a}\right)\\ &= \frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\end{al igned}

4. Well, that was awfully simple.

And yeah, the integral obviously diverges for $a \le 1$

simplependulum posted some problems on the bottom of the last page that you'll most likely find much more interesting.

Originally Posted by Drexel28
It should be $a>1$.

Here is one method. There is another I know using contour integration if you're interested.

Method 1:

Spoiler:

Let

$\displaystyle I(a)=\int_0^{\infty}\frac{dx}{1+x^a}$

Let $1+x^a=u$ and so $du=a x^{a-1}$ or $dx=\frac{1}{a}\left(u-1\right)^{\frac{1}{a}-1}$. Thus,

$\displaystyle I(a)=\frac{1}{a}\int_1^{\infty}\frac{(u-1)^{\frac{1}{a}-1}}{u}\text{ }du$

Let $z=\frac{1}{u}$ to get that

\displaystyle \begin{aligned}I(a) &=\frac{1}{a}\int_0^1 z^{\frac{-1}{a}}\left(1-z\right)^{\frac{1}{a}-1}\\ &= \frac{1}{a}B\left(1-\frac{1}{a},\frac{1}{a}\right)\\ &= \frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Ga mma\left(1-\frac{1}{a}\right)}{\Gamma(1)}\\ &=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\ left(1-\frac{1}{a}\right)\\ &= \frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\end{al igned}

5. Originally Posted by Random Variable
Well, that was awfully simple.

And yeah, the integral obviously diverges for $a \le 1$

simplependulum posted some problems on the bottom of the last page that you'll most likely find much more interesting.
The one with the AGM is well known but a pain to verify, so I won't do that one. The other, I haven't been able to find a nice answer for it yet :S

6. Originally Posted by Random Variable
What's required to switch the order of summation? Absolute convergence of the inner sum?
Yes, this can be seen (for example) on page 202 of Apostol.

7. Yes , AGM is one of the most famous identities , I have a proof of it but of course it is not due to me , I was inspired to give a complete proof by this problem from American Mathematical Monthly (AMM 6672 ) :

Show that $\displaystyle I(a,b) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{a^2\cos^2{\theta} + b^2 \sin^2{\theta} }} = \int_0^{\pi/2} \frac{d\phi}{ \sqrt{(a\cos^2{\phi} + b\sin^2{\phi})(a\sin^2{\phi} + b\cos^2{\phi}) } }$ . Then use this result to show that :

$\displaystyle I(a,b) = I\left( \sqrt{ab} , \frac{a+b}{2} \right)$

Oh , I have made a very big mistake on the first problem , sorry about that

The integrand should be squared :

$\displaystyle \int_0^1 \int_0^1 \int_0^1 \frac{dxdydz}{(1 + x^2 + y^2 + z^2)^2 }$

8. Show that $\displaystyle \int^{\infty}_{0} \frac{J_{1}(x)}{\sqrt{a^{2}+x^{2}}} \ dx = \frac{1- e^{-a}}{a}$

where $\displaystyle J_{1}(x)$ is the first order Bessel function of the first kind.

The proof is fairly long, but simple and pretty cool at the same time.

9. Originally Posted by Random Variable
Show that $\displaystyle \int^{\infty}_{0} \frac{J_{1}(x)}{\sqrt{a^{2}+x^{2}}} \ dx = \frac{1- e^{-a}}{a}$

where $\displaystyle J_{1}(x)$ is the first order Bessel function of the first kind.

The proof is fairly long, but simple and pretty cool at the same time.
Not having heard about Bessel function , I thought I would not be able to solve this until I found this formula on a website :

$\displaystyle J_1(x) = \frac{1}{\pi} \int_0^{\pi} \cos[t - x\sin(t)]~dt$

$\displaystyle = \frac{1}{\pi} \int_0^{\pi} \left( \cos(t)\cos[x\sin(t)] + \sin(t)\sin[x\sin(t)] \right) ~dt$

$\displaystyle = \frac{1}{\pi} \int_0^{\pi} \sin(t)\sin[x\sin(t)] ~dt$

$\displaystyle I = \frac{1}{\pi} \int_0^{\infty} \int_0^{\pi} \frac{ \sin(t)\sin[x\sin(t)] }{\sqrt{a^2 + x^2} }~dtdx$

My method is to switch the order of the integral , followed by substitution , then switch back its order :

Sub. $\displaystyle x\sin(t) = y$ we have

$\displaystyle I = \frac{1}{\pi} \int_0^{\pi} \int_0^{\infty} \frac{\sin(y)}{\sqrt{a^2 + y^2/\sin^2(t) }}~dydt$

$\displaystyle = \frac{1}{\pi} \int_0^{\pi} \int_0^{\infty} \frac{\sin(t)\sin(y)}{\sqrt{a^2 \sin^2(t) + y^2 }}~dydt$

Then switching back !

$\displaystyle = \frac{1}{\pi }\int_0^{\infty} \int_0^{\pi} \frac{\sin(t)\sin(y)}{\sqrt{ a^2 + y^2 - a^2 \cos^2(t) }}~dtdy$

$\displaystyle = \frac{2}{a \pi} \int_0^{\infty} \sin(y) \sin^{-1}{\left( \frac{a}{\sqrt{a^2 + y^2 }}\right)}~dy$

$\displaystyle = \frac{2}{a \pi} \int_0^{\infty} \sin(y) \tan^{-1}(a/y) ~ dy$

Integration by parts ,

$\displaystyle I = \frac{2}{a \pi} \left[- \cos(y) \tan^{-1}(a/y) \right]_0^{\infty} - \frac{2}{\pi} \int_0^{\infty} \frac{\cos(y)}{y^2 + a^2 }~dy$

$\displaystyle = \frac{1}{a} - \frac{2}{\pi} ~ \frac{\pi}{2a} e^{-a}$

$\displaystyle = \frac{1}{a} ( 1 - e^{-a} )$

10. Changing the order of integration solely for the purpose of making a substitution? Where you do come up with these ideas?

The solution I had in mind doesn't use the integral representation of the Bessel function.

let $\displaystyle I(b) = \int^{\infty}_{0} \frac{b J_{1}(bx)}{\sqrt{a^{2}+x^{2}}} \ dx \ , b \ge 0$

then using $\displaystyle \frac{d}{dx} [x^{p} J_{p}(x)] = x^{p}J_{p-1}(x)$

$\displaystyle I'(t) = \int^{\infty }_{0} \frac{bx J_{0}(bx)}{\sqrt{a^{2}+x^{2}}} \ dx =\int^{\infty}_{0} \frac{t J_{0}(t)}{\sqrt{c^{2}+t^{2}}} \ dt$ where $\displaystyle t=bx$ and $\displaystyle c=ab$

By defintion $\displaystyle J_{0}(t)$ satisfies $\displaystyle tf''(t) + f'(t) + tf(t) = 0$

Now here comes the cool part.

let $\displaystyle u(c) = \int^{\infty}_{0} \frac{t J_{0}(t)}
{\sqrt{a^{2}+x^{2}}} \ dt$

$\displaystyle v(c) = \int^{\infty}_{0} \frac{J^{'}_{0}(t)}
{\sqrt{a^{2}+x^{2}}} \ dt$

and $\displaystyle w(c) = \int^{\infty}_{0} \frac{t J^{''}_{0}(t)}
{\sqrt{a^{2}+x^{2}}} \ dt$

This directly implies that $\displaystyle u + v + w = 0$.

Then using the facts that $\displaystyle J_{0}(0) = 1, \ J_{0}(\infty) = 0, \ \frac{d}{dx} [x^{-p} J_{p}(x)] = -x^{-p} J_{p+1}(x), \ J_{1}(0) = 0$ and $\displaystyle J_{1} (\infty) = 0$, you'll find that $\displaystyle u'(c) = -1 - cv(c)$ and $\displaystyle v'(c) = \frac{w(c)}{c}$

Solve the system of equations simultaneously to get homogenous second order linear ODE $\displaystyle u''(c) = u$ which has the general solution $\displaystyle u(c) = Ae^{c} + Be^{-c}$

$A$ must be zero because $u(c)$ is a decreasing function, and $B$ must be $1$ because Bessel functions are normalized.

$\displaystyle u(c) = e^{-c} = e^{-ab}$

but $\displaystyle I'(b) = u(c)$

so $I(b) = \frac{1-e^{-ab}}{a}$ ( since $I(0) = 0$)

The integral we want is $I(1) = \frac{1-e^{-a}}{a}$

11. So that this thread remains active, try the following fairly easy problems:

$\displaystyle \text{Show that} \int_{0}^{\frac{\pi}{2}} \sin^{2n} x \ dx = \frac{\pi (2n)!}{2^{2n+1} (n!)^{2}} \ \text{for} \ n = 0,1,2,...$

$\displaystyle \text{and} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1} x \ dx = \frac{2^{2n} (n!)^{2}}{(2n+1)!} \ \text{for} \ n = 0,1,2,... \ .$

12. Originally Posted by Random Variable
So that this thread remains active, try the following fairly easy problems:

$\displaystyle \text{Show that} \int_{0}^{\frac{\pi}{2}} \sin^{2n} x \ dx = \frac{\pi (2n)!}{2^{2n+1} (n!)^{2}} \ \text{for} \ n = 0,1,2,...$

$\displaystyle \text{and} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1} x \ dx = \frac{2^{2n} (n!)^{2}}{(2n+1)!} \ \text{for} \ n = 0,1,2,... \ .$
Two words: Beta Function (by the way, what a powerful tool, huh?)

Here's a nice one: compute $\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)}$

13. Originally Posted by chiph588@
Two words: Beta Function (by the way, what a powerful tool, huh?)

Here's a nice one: compute $\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)}$
Is this really what you meant to put? I feel like this might not have a nice closed form considering that it's equivalent to integrating $\displaystyle \int_0^1\frac{\log(\sin(\pi x))}{\sqrt{x}}\text{ }dx$.

14. Originally Posted by Drexel28
Is this really what you meant to put? I feel like this might not have a nice closed form considering that it's equivalent to integrating $\displaystyle \int_0^1\frac{\log(\sin(\pi x))}{\sqrt{x}}\text{ }dx$.
Yes I did mean to put that. The integral you posted is a bit off though. You should get something similar and yes it's a hard one to compute.

15. It would appear that the answer is just a multiple of that impossible-looking integral.

Originally Posted by chiph588@
Yes I did mean to put that. The integral you posted is a bit off though. You should get something similar and yes it's a hard one to compute.

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