Prove some identities! (Round two)

**Random Variable** · January 7th 2011, 02:33 PM

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

**Drexel28** · January 7th 2011, 03:09 PM

Originally Posted by Random Variable

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

Spoiler:

**Random Variable** · January 7th 2011, 03:36 PM

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.

**Random Variable** · January 7th 2011, 06:33 PM

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

**Drexel28** · January 7th 2011, 06:51 PM

Originally Posted by Random Variable

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

See! we all make stupid mistakes sometimes (especially me

). Find mine! .Meanwhile, here is a correct proof:

Spoiler:

**Random Variable** · January 7th 2011, 07:14 PM

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?

**Drexel28** · January 7th 2011, 07:17 PM

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 $u=x^{a-1}$ and since $a<1$ we have that $x^{a-1}\to 0$ as $x\to\infty$ , not $\infty$ !

**Random Variable** · January 7th 2011, 07:28 PM

So that's where the negative was hiding. Sneaky bastard.

EDIT: I was referring to the negative; not you.

**Random Variable** · January 8th 2011, 10:52 AM

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

This one I worked out without using contour integration.

**Drexel28** · January 8th 2011, 11:10 AM

Originally Posted by Random Variable

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

This one I worked out without using contour integration.

How about more generally:

Spoiler:

**Random Variable** · January 8th 2011, 11:43 AM

I can generalize my solution, which doesn't rely on the previous problem.

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

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

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

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

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

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

**Drexel28** · January 8th 2011, 11:45 AM

Originally Posted by Random Variable

I can generalize my solution, which doesn't rely on the previous problem.

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

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

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

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

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

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

Nice!

**Random Variable** · January 12th 2011, 10:52 AM

EDIT: On second thought, it's really not that interesting. Nevermind.

**Drexel28** · January 12th 2011, 05:45 PM

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!

**Random Variable** · January 13th 2011, 12:16 AM

Originally Posted by Drexel28

Way to get me excited only to yank it away!

Sorry. I have another one.

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

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