# Thread: Prove some identities! (Round two)

1. Originally Posted by Random Variable
It would appear that the answer is just a multiple of that impossible-looking integral.
I'm a little busy now, but I'll think about it. The methodology is to note that $\displaystyle \displaystyle \sin\left(\pi\sqrt{z}\right)=\pi\sqrt{z}\prod_{n=1 }^{\infty}\left(1-\frac{z}{n^2}\right)$ and thus $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)=\sum_{n=1}^{\infty}\l og\left(1-\frac{z}{n^2}\right)$ but expanding the logarithm gives $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{z^m}{n ^{2m}m}$ doing the quick verification that the inner sum can be interchanged this implies that $\displaystyle \displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=\sum_{m=1}^{\i nfty}\frac{\zeta(2m)}{m}z^m$. Integrating both sides from $\displaystyle 0$ to $\displaystyle 1$ etc. etc

2. Originally Posted by Random Variable
It would appear that the answer is just a multiple of that impossible-looking integral.
$\displaystyle \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}$ $\displaystyle \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx$

For the second integral, let $\displaystyle t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt$

Thus the second integral equals $\displaystyle \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt$.

3. Originally Posted by chiph588@
$\displaystyle \displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}$ $\displaystyle \displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx$

For the second integral, let $\displaystyle t=\pi\sqrt{x} \implies dt = \frac\pi{2\sqrt{x}}dx \implies dx = \frac2{\pi^2} t dt$

Thus the second integral equals $\displaystyle \displaystyle \frac2{\pi^2}\int_0^\pi t\log(\sin t) dt$.
Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

Spoiler:

Let $\displaystyle \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)}$ then we have shown that

$\displaystyle \displaystyle S=\log(\pi)-\frac{1}{2}-I$

Where

$\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt$

Letting $\displaystyle \pi-y=t$ in this integral gives

$\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig ht)\text{ }dt-I$

And thus

$\displaystyle \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt$

Let $\displaystyle \displaystyle t=\frac{\pi}{2}-y$ to get that

$\displaystyle \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t ext{ }dy$

But, since $\displaystyle \cos(y)$ is symmetric on $\displaystyle \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ we get that

$\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co s(y)\right)\text{ }dy$

Letting $\displaystyle \displaystyle y=\frac{\pi}{2}-u$ gives that

$\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(u)\right)\text{ }du$

This integral is well-known (see here fore example) and so we get

$\displaystyle \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)$

So, finally we may conclude that

$\displaystyle \displaystyle S=\log(2\pi)-\frac{1}{2}$

4. Originally Posted by Drexel28
Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

Spoiler:

Let $\displaystyle \displaystyle S=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)}$ then we have shown that

$\displaystyle \displaystyle S=\log(\pi)-\frac{1}{2}-I$

Where

$\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^{\pi}t\log(\sin(t))\text{ }dt$

Letting $\displaystyle \pi-y=t$ in this integral gives

$\displaystyle \displaystyle I=\frac{2}{\pi^2}\int_0^\pi\left(\pi-y\right)\log\left(\sin(t)\right)\text{ }dt=\frac{2}{\pi}\int_0^{\pi}\log\left(\sin(t)\rig ht)\text{ }dt-I$

And thus

$\displaystyle \displaystyle \pi I=\int_0^{\pi}\log\left(\sin(t)\right)\text{ }dt$

Let $\displaystyle \displaystyle t=\frac{\pi}{2}-y$ to get that

$\displaystyle \displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(y)\right)\t ext{ }dy$

But, since $\displaystyle \cos(y)$ is symmetric on $\displaystyle \displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ we get that

$\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\co s(y)\right)\text{ }dy$

Letting $\displaystyle \displaystyle y=\frac{\pi}{2}-u$ gives that

$\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(u)\right)\text{ }du$

This integral is well-known (see here fore example) and so we get

$\displaystyle \displaystyle I=\frac{\frac{-\pi\log(2)}{2}}{\frac{\pi}{2}}=-\log(2)$

So, finally we may conclude that

$\displaystyle \displaystyle S=\log(2\pi)-\frac{1}{2}$

Wow! You made that integral look easy!

5. Originally Posted by chiph588@
Wow! You made that integral look easy!
I actually found a much better way.
Spoiler:

Looking at my last solution see that we have

$\displaystyle \displaystyle \pi I=\int_0^\pi \log(\sin(t))\text{ }dt$

and

$\displaystyle \displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\si n(t)\right)\text{ }dt=\int_0^{\frac{\pi}{2}}\log\left(\cos(t)\right) \text{ }dt$

Thus, from the equality immediately preceding this sentence

\displaystyle \displaystyle \begin{aligned}\pi I &= \int_0^{\frac{\pi}{2}}\left(\log(\sin(t)\right)+\l og\left(\cos(t)\right)\right)\text{ }dt\\ &= \int_0^{\frac{\pi}{2}}\log\left(\frac{1}{2}\sin\le ft(2t\right)\right)\\ &= \frac{-\pi\log(2)}{2}+\int_0^{\frac{\pi}{2}}\log\left(\si n(2t)\right)\text{ }dt\\ &= \frac{-\pi\log(2)}{2}+\frac{1}{2}\int_0^\pi \log(\sin(w))\text{ }dw\quad\quad 2t\mapsto w\\ &= \frac{-\pi\log(2)}{2}+\frac{\pi}{2}I\end{aligned}

Solving we find that $\displaystyle \displaystyle I=-\log(2)$ as desired. Thus, we've completely circumvented the need to calculate$\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}\log\left(\sin(x)\right)\tex t{ }dx$ directly, and have even found an alternate way of computing it.

6. This isn't too hard to calculate, but I think it's a good integral to know. Compute $\displaystyle \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx$.

7. Originally Posted by chiph588@
This isn't too hard to calculate, but I think it's a good integral to know. Compute $\displaystyle \displaystyle \int_0^\infty \frac{x^n}{(ax+b)^m} dx$.
Of course assuming $\displaystyle m>n+1$. I'll assume that $\displaystyle b\ne 0$ since this is trivial otherwise. Similarly, I'll assume that $\displaystyle \frac{a}{b}>0$.

Spoiler:

Let

$\displaystyle \displaystyle I=\int_0^{\infty}\frac{x^n}{(ax+b)^m}\text{ }dx$

Then,

$\displaystyle \displaystyle b^mI=\int_0^{\infty}\frac{x^n}{(1+cx)^m}\text{ }dx$

where $\displaystyle \displaystyle c=\frac{a}{b}$. Let $\displaystyle 1+cx=z$ so that

$\displaystyle \displaystyle c^{n+1}b^mI=\int_1^{\infty}\frac{(z-1)^n}{z^m}\text{ }dz$

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

\displaystyle \displaystyle \begin{aligned}c^{n+1}b^m I &=\int_0^1 u^{m-2}\left(\frac{1}{u}-1\right)^n\text{ }du\\ &=\int_0^1 u^{m-2-n}(1-u)^n\text{ }du\\ &=B\left(m-1-n,n+1\right)\end{aligned}

and thus

$\displaystyle \displaystyle I=\frac{b^{m-n-1}}{a^{n+1}}B\left(m-1-n,m+1\right)$

8. This one's not too bad.

Compute:

$\displaystyle \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx \right)\text{ }dx$

9. Are you sure that integral converges?

10. Originally Posted by Random Variable
Are you sure that integral converges?
Assuming I didn't miscalculate something? Anyways, try this integral. It's much more surprising (albeit not too hard)!

$\displaystyle \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}$

11. Originally Posted by Drexel28

$\displaystyle \displaystyle \int_0^1\log\left(\Gamma(x)\right)\cos\left(2\pi n x\right)\text{ }dx,\quad n\in\mathbb{N}$

$\displaystyle \displaystyle I= \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx$

$\displaystyle \displaystyle = \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx ]$

Sub. $\displaystyle x = 1-t$ on the second integral and change the dummy variable , we have

$\displaystyle \displaystyle I= \frac{1}{2} [ \int_0^1 \ln[\Gamma(x)]\cos(2\pi nx)~dx + \int_0^1 \ln[\Gamma(1-x)]\cos(2\pi nx)~dx$

$\displaystyle \displaystyle = \frac{1}{2} \int_0^1 \ln[ \Gamma(x) \Gamma(1-x) ] \cos(2\pi nx ) ~dx$

$\displaystyle \displaystyle = \frac{1}{2} \int_0^1 [ \ln(\pi) - \ln[\sin(\pi x)] ] \cos(2\pi nx ) ~dx$

$\displaystyle \displaystyle = - \frac{1}{2\pi} \int_0^{\pi} \ln[\sin(x)] \cos(2nx)~dx$

Integration by parts

$\displaystyle \displaystyle = - \frac{1}{4n\pi} \left[ \ln[\sin(x)] \sin(2nx)\right]_0^\pi + \frac{1}{4n\pi} \int_0^{\pi} \frac{\sin(2nx)\cos(x)}{\sin(x)}~dx$

$\displaystyle \displaystyle = \frac{1}{8n\pi} \int_0^{\pi} \frac{\sin(2n+1)x + \sin(2n-1)x}{\sin(x)}~dx$

$\displaystyle \displaystyle = \frac{1}{8n\pi} (2\pi) = \frac{1}{4n}$

Problem :

How do you prove this

$\displaystyle \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt$

My solution is to make a substitution , followed by changing the order of a double integral .

12. Originally Posted by simplependulum

Problem :

How do you prove this

$\displaystyle \displaystyle \frac{\Gamma(x) \Gamma(y) }{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1} ~dt$

My solution is to make a substitution , followed by changing the order of a double integral .
Let $\displaystyle \displaystyle B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\text{ }dt$. It's easy to show that $\displaystyle \displaystyle B(x,y)=\frac{x+y}{y}B(x,y+1)$ and more so by induction that $\displaystyle \displaystyle B(x,y)=\frac{(x+y)_n}{n! n^{x+y-1}}\frac{n! n^{y-1}}{(y)_n}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{n+y-1}\text{ }dt$. But, by definition (I guess it really depends upon your definition of the Gamma function, but this is a common one) it follows that $\displaystyle \displaystyle B(x,y)=\frac{\Gamma(y)}{\Gamma(x+y)}\lim_{n\to\inf ty}\int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^{y+n-1}\text{ }dt$ but one can verify (using the LDC theorem for example) that this integral tends to $\displaystyle \displaystyle \int_0^{\infty}t^{x-1}e^{-t}\text{ }dt$ from where the conclusion follows.

13. $\displaystyle \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt$

let $\displaystyle t = \sin^{2} \theta$

$\displaystyle = 2 \int^{\pi /2}_{0} (\sin^{2} \theta)^{r-1} (1-\sin^{2} \theta)^{s-1} \sin \theta \cos \theta \ d \theta$

$\displaystyle = 2 \int^{\pi /2}_{0} (\sin \theta)^{2r-1} (\cos \theta)^{2s-1} \ d \theta$

$\displaystyle \Gamma(x) = \int^{\infty}_{0} t^{x-1} e^{-t} \ dt$

let $\displaystyle t = u^{2}$

$\displaystyle \Gamma (x) = \int^{\infty}_{0} u^{2x-2} e^{-u^{2}} (2 u \ du)$

$\displaystyle = 2 \int^{\infty}_{0} u^{2x-1} e^{-u^{2}} \ du$

then $\displaystyle \Gamma(x) \Gamma(y) = 4 \int^{\infty}_{0} u^{2x-1} e^{-u^{2}} \ du \int^{\infty}_{0} v^{2y-1} e^{-v^{2}} \ dv$

$\displaystyle = 4 \int^{\infty}_{0} \int^{\infty}_{0} u^{2x-1} v^{2y-1} e^{-(u^{2}+v^{2})} \ du \ dv$

change to polar coordinates

$\displaystyle \Gamma(x) \Gamma (y) = 4 \int^{\pi / 2}_{0} \int^{\infty}_{0} (r \cos \theta)^{2x-1} \ (r \sin \theta)^{2y-1} e^{-r^{2}} r \ dr \ d \theta$

$\displaystyle = 4 \int^{\pi /2}_{0} \int^{\infty}_{0} (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ r^{2(x+y)-1} e^{-r^{2}} \ dr \ d \theta$

$\displaystyle = 2 \Gamma(x+y) \int^{\pi /2}_{0} (cos \theta)^{2x-1} \ (\sin \theta)^{2y-1} \ d \theta$

$\displaystyle = \Gamma (x+y) \int^{1}_{0} t^{x-1} (1-t)^{y-1} \ dt$

14. Originally Posted by Drexel28

Compute:

$\displaystyle \displaystyle \int_0^{\infty}x\sin\left(ax^2\right)\sin\left(2bx \right)\text{ }dx$
Here's a thought: $\displaystyle \displaystyle x\sin(ax^2)\sin(2bx) = \frac{x}2 (\cos(a x^2-2 b x)-\cos(a x^2+2 b x))$

$\displaystyle = \frac{x}2\left(\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)-\cos\left(\frac{(a x+b)^2}a-\frac{b^2}a\right)\right)$

Now split the cosines up into its exponentials.

I could be wrong, and won't have time to work it out... I'll be away from a computer for a week.

15. Here is my method :

$\displaystyle \displaystyle I = \int_0^1 t^{x-1} (1-t)^{y-1}~dt$

Sub. $\displaystyle \displaystyle t = \frac{1}{u+1}$

$\displaystyle \displaystyle I = \int_0^{\infty} \frac{u^{y-1}}{(u+1)^{x+y}}~du$

Consider $\displaystyle \displaystyle \Gamma(x+y) = \int_0^{\infty} e^{-z} z^{x+y-1}~dz$

Sub. $\displaystyle z = (u+1)s$ , $\displaystyle u$ as parameter , we have

$\displaystyle \displaystyle \frac{\Gamma(x+y)}{(u+1)^{x+y}} = \int_0^{\infty} e^{-(u+1)s} s^{x+y-1}~ds$

Let's come back to the original problem , we thus have

$\displaystyle \displaystyle I = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \int_0^{\infty} e^{-(u+1)s} u^{y-1} s^{x+y-1}~dsdu$

$\displaystyle \displaystyle = \frac{1}{\Gamma(x+y)}~ \int_0^{\infty} \left[ \frac{\Gamma(y)}{s^y}~ e^{-s} s^{x+y-1} \right] ~ds$

$\displaystyle \displaystyle = \frac{\Gamma(y)}{\Gamma(x+y)}~\Gamma(x)$

$\displaystyle \displaystyle = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

Page 3 of 7 First 1234567 Last