Prove some identities! (Round two)

**Drexel28** · January 3rd 2011, 06:41 PM

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 $\displaystyle \sin\left(\pi\sqrt{z}\right)=\pi\sqrt{z}\prod_{n=1 }^{\infty}\left(1-\frac{z}{n^2}\right)$ and thus $\displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)=\sum_{n=1}^{\infty}\l og\left(1-\frac{z}{n^2}\right)$ but expanding the logarithm gives $\displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=-\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{z^m}{n ^{2m}m}$ doing the quick verification that the inner sum can be interchanged this implies that $\displaystyle \log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right)\right)=\sum_{m=1}^{\i nfty}\frac{\zeta(2m)}{m}z^m$ . Integrating both sides from $0$ to $1$ etc. etc

**chiph588@** · January 3rd 2011, 06:42 PM

Originally Posted by Random Variable

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

$\displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}$ $\displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx$

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

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

**Drexel28** · January 3rd 2011, 10:13 PM

Originally Posted by chiph588@

$\displaystyle \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} \implies \log(\pi \sqrt{x}) - \log(\sin(\pi \sqrt{x})) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{n}$ $\displaystyle \implies \sum_{n=1}^\infty \frac{\zeta(2n)}{n(n+1)} = \int_0^1 \log(\pi\sqrt{x}) dx - \int_0^1 \log(\sin(\pi \sqrt{x})) dx$

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

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

Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

Spoiler:

**chiph588@** · January 4th 2011, 08:10 AM

Originally Posted by Drexel28

Gah! I made a stupid Calc I error substituting! Haha! Anways, let's finish this

Spoiler:

Wow! You made that integral look easy!

**Drexel28** · January 4th 2011, 10:30 AM

Originally Posted by chiph588@

Wow! You made that integral look easy!

I actually found a much better way.

Spoiler:

**chiph588@** · January 5th 2011, 04:56 PM

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

**Drexel28** · January 5th 2011, 05:25 PM

Originally Posted by chiph588@

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

Of course assuming $m>n+1$ . I'll assume that $b\ne 0$ since this is trivial otherwise. Similarly, I'll assume that $\frac{a}{b}>0$ .

Spoiler:

**Drexel28** · January 5th 2011, 05:30 PM

This one's not too bad.

Compute:

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

**Random Variable** · January 5th 2011, 06:22 PM

Are you sure that integral converges?

**Drexel28** · January 5th 2011, 06:27 PM

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)!

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

**simplependulum** · January 6th 2011, 01:33 AM

Originally Posted by Drexel28

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

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

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

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

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

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

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

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

Integration by parts

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

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

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

Problem :

How do you prove this

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

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

**Drexel28** · January 6th 2011, 07:44 AM

Originally Posted by simplependulum

Problem :

How do you prove this

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

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

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

**Random Variable** · January 6th 2011, 09:18 AM

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

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

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

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

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

let $t = u^{2}$

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

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

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

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

change to polar coordinates

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

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

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

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

**chiph588@** · January 6th 2011, 11:54 AM

Originally Posted by Drexel28

This one's not too bad.

Compute:

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

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

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

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.

**simplependulum** · January 6th 2011, 11:55 PM

Here is my method :

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

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

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

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

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

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

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

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

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

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

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

Thread: Prove some identities! (Round two)

LinkBack

Thread Tools

Display

Similar Math Help Forum Discussions

Can't prove these identities?

prove the following identities.

Prove Identities

Prove these identities.

help Prove identities

Search Tags