# Prove some identities! (Round two)

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• Jan 7th 2011, 02:33 PM
Random Variable
$\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x} \ dx \ \text{for} \ 0<a<1$
• Jan 7th 2011, 03:09 PM
Drexel28
Quote:

Originally Posted by Random Variable
$\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x} \ dx \ \text{for} \ 0<a<1$

Spoiler:

Let

$\displaystyle \displaystyle I(a)=\int_0^{\infty}\frac{x^{a-1}}{1+x}\text{ }dx\quad a\in(0,1)$

Let $\displaystyle x^{a-1}=u$ then $\displaystyle x=u^{\frac{1}{a-1}}$ and so $\displaystyle \displaystyle dx=\frac{1}{(a-1)}u^{\frac{1}{a-1}-1} du$ and thus

$\displaystyle \displaystyle I(a)=\int_0^{\infty}\frac{u}{1+u^{\frac{1}{a-1}}} \frac{1}{(a-1)}u^{1-\frac{1}{1-a}}\text{ }du=\frac{1}{a-1}\int_0^{\infty}\frac{du}{1+u^{\frac{1}{1-a}}}$

but by an earlier post this is equal to $\displaystyle \displaystyle I(a)==\frac{\pi}{\sin(\pi a)}$. Thus,

$\displaystyle \displaystyle -\pi^2\cot(\pi a)\csc(\pi a)=I'(a)=\int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x}\text{ }dx$

• Jan 7th 2011, 03:36 PM
Random Variable
I used contour integration and the Pac-Man contour to find I(a), unaware that by substitution I could make it look like that integral I asked about.

You and simplependulum are too good. I give up. (Clapping)
• Jan 7th 2011, 06:33 PM
Random Variable
Doesn't $\displaystyle \displaystyle \frac{1}{a-1} \int^{\infty}_{0} \frac{1}{1+ u^{\frac{1}{1-a}}} \ du = \frac{1}{a-1} \frac{\pi (1-a)}{\sin \pi(1-a)} = -\frac{\pi}{\sin \pi a}$ ?
• Jan 7th 2011, 06:51 PM
Drexel28
Quote:

Originally Posted by Random Variable
Doesn't $\displaystyle \displaystyle \frac{1}{a-1} \int^{\infty}_{0} \frac{1}{1+ u^{\frac{1}{1-a}}} \ du = \frac{1}{a-1} \frac{\pi (1-a)}{\sin \pi(1-a)} = -\frac{\pi}{\sin \pi a}$ ?

See! we all make stupid mistakes sometimes (especially me :) ). Find mine! .Meanwhile, here is a correct proof:

Spoiler:

Let

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

Let $\displaystyle x^a=u$ and thus $\displaystyle x^{a-1}=u^{1-\frac{1}{a}}$ and $\displaystyle \displaystyle dx=\frac{1}{a}u^{\frac{1}{a}-1}$. Thus,

$\displaystyle \displaystyle I(a)=\frac{1}{a}\int_0^{\infty}\frac{u^{1-\frac{1}{a}}}{1+u^{\frac{1}{a}}}u^{\frac{1}{a}-1}\text{ }dx=\frac{1}{a}\int_0^{\infty}\frac{du}{1+u^{\frac {1}{a}}}$

So, by the same previous problem $\displaystyle \displaystyle I(a)=\frac{\pi}{\sin(\pi a)}$ and thus

$\displaystyle \displaystyle -\pi^2\cot(\pi a)\csc(\pi a)=I'(a)=\int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x}\text{ }dx$

• Jan 7th 2011, 07:14 PM
Random Variable
The only other thing I see wrong in your first solution is what you substituted back in for dx. But even if you did that correctly, you would still get the negative of what you should get. So what else is wrong?
• Jan 7th 2011, 07:17 PM
Drexel28
Quote:

Originally Posted by Random Variable
The only other thing I see wrong in your first solution is what you substituted back in for dx. But even if you did that correctly, you would still get the negative of what you should get. So what else is wrong?

I let $\displaystyle u=x^{a-1}$ and since $\displaystyle a<1$ we have that $\displaystyle x^{a-1}\to 0$ as $\displaystyle x\to\infty$, not $\displaystyle \infty$!
• Jan 7th 2011, 07:28 PM
Random Variable
So that's where the negative was hiding. Sneaky bastard.

EDIT: I was referring to the negative; not you.
• Jan 8th 2011, 10:52 AM
Random Variable
How about $\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x^{2}} \ dx \ , \ \text{for} \ 0<a<2$

This one I worked out without using contour integration.
• Jan 8th 2011, 11:10 AM
Drexel28
Quote:

Originally Posted by Random Variable
How about $\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1} \ln x}{1+x^{2}} \ dx \ , \ \text{for} \ 0<a<2$

This one I worked out without using contour integration.

Spoiler:

Let

$\displaystyle \displaystyle I(a,m)=\int_0^{\infty}\frac{x^{a-1}}{1+x^m},\quad m>1,\;\; 0<a<m$

Then, let $\displaystyle x^m=u$ to get that

$\displaystyle \displaystyle I(a,m)=\int_0^{\infty}\frac{u^{\frac{a-1}{m}}}{1+u}\frac{du}{m u^{1-\frac{1}{m}}}=\frac{1}{m}\int_0^{\infty}\frac{u^{\ frac{a}{m}-1}}{1+u}\text{ }du$

Note though that since $\displaystyle \displaystyle 0<\frac{a}{m}<1$ we may apply the methods in the previous problem to conclude that

$\displaystyle \displaystyle I(a,m)=\frac{1}{m}\frac{\pi}{\sin\left(\frac{\pi a}{m}}\right)}$

Thus,

$\displaystyle \displaystyle \int_0^{\infty}\frac{x^{a-1}\log(x)}{1+x^m}\text{ }dx=\frac{\partial}{\partial a}I(a,m)=-\left(\frac{\pi}{m}\right)^2\cot\left(\frac{\pi a}{m}\right)\csc\left(\frac{\pi a}{m}\right)$

• Jan 8th 2011, 11:43 AM
Random Variable
I can generalize my solution, which doesn't rely on the previous problem.

$\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1}}{1+x^{m}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} x^{a-1} e^{-(1+x^m)t} \ dt \ dx$

$\displaystyle \displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} e^{-t} x^{a-1} e^{-tx^{m}} \ dt \ dx$

reverse the order the integration and let $\displaystyle \displaystyle x = \Big(\frac{u}{t}\Big)^{1/m}$

$\displaystyle \displaystyle = \frac{1}{m}\int^{\infty}_{0} \int^{\infty}_{0} e^{-t} \Big(\frac{u}{t}\Big)^{\frac{1}{m}(a-1)} e^{-u} \Big(\frac{u}{t}\Big)^{\frac{1}{m} -1} \ \frac{1}{t} \ dt \ du$

$\displaystyle \displaystyle = \frac{1}{m} \int^{\infty}_{0} t^{-\frac{a}{m}} \ e^{-t} \ dt \int^{\infty}_{0} u^{\frac{a}{m} - 1} \ e^{-u} \ du$

$\displaystyle = \frac{1}{m} \Gamma(1- \frac{a}{m}) \Gamma(\frac{a}{m}) = \frac{1}{m} \pi \csc (\frac{\pi a}{m})$
• Jan 8th 2011, 11:45 AM
Drexel28
Quote:

Originally Posted by Random Variable
I can generalize my solution, which doesn't rely on the previous problem.

$\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a-1}}{1+x^{m}} \ dx = \int^{\infty}_{0} \int^{\infty}_{0} x^{a-1} e^{-(1+x^m)t} \ dt \ dx$

$\displaystyle \displaystyle = \int^{\infty}_{0} \int^{\infty}_{0} e^{-t} x^{a-1} e^{-tx^{m}} \ dt \ dx$

reverse the order the integration and let $\displaystyle \displaystyle x = \Big(\frac{u}{t}\Big)^{1/m}$

$\displaystyle \displaystyle = \frac{1}{m}\int^{\infty}_{0} \int^{\infty}_{0} e^{-t} \Big(\frac{u}{t}\Big)^{\frac{1}{m}(a-1)} e^{-u} \Big(\frac{u}{t}\Big)^{\frac{1}{m} -1} \ \frac{1}{t} \ dt \ du$

$\displaystyle \displaystyle = \frac{1}{m} \int^{\infty}_{0} t^{-\frac{a}{m}} \ e^{-t} \ dt \int^{\infty}_{0} u^{\frac{a}{m} - 1} \ e^{-u} \ du$

$\displaystyle = \frac{1}{m} \Gamma(1- \frac{a}{m}) \Gamma(\frac{a}{m}) = \frac{1}{m} \pi \csc (\frac{\pi a}{m})$

Nice!
• Jan 12th 2011, 10:52 AM
Random Variable
EDIT: On second thought, it's really not that interesting. Nevermind.
• Jan 12th 2011, 05:45 PM
Drexel28
Quote:

Originally Posted by Random Variable
EDIT: On second thought, it's really not that interesting. Nevermind.

Way to get me excited only to yank it away!
• Jan 13th 2011, 12:16 AM
Random Variable
Quote:

Originally Posted by Drexel28
Way to get me excited only to yank it away!

Sorry. I have another one.

$\displaystyle \displaystyle \int^{\infty}_{0} \frac{x^{a}}{(1+e^{x})^{2}} \ dx \ , \ a>0$
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