1. ## Lebesgue integration questions

I have a couple of integrals that I need to evaluate - they are in the context of Lebesgue integrals. They seem similar to questions regarding contour integration that I've done before, but I'm not sure if that applies here.

I need to show the following:

1) $\int_{0}^{\infty}\frac{\sin(\alpha x)}{e^x-1}dx=\sum_{n=1}^{\infty}\frac{\alpha}{\alpha^2 + n^2}$

2) $\int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$

Thanks for any help!

2. Originally Posted by angela87
I have a couple of integrals that I need to evaluate - they are in the context of Lebesgue integrals. They seem similar to questions regarding contour integration that I've done before, but I'm not sure if that applies here.

I need to show the following:

1) $\int_{0}^{\infty}\frac{\sin(\alpha x)}{e^x-1}dx=\sum_{n=1}^{\infty}\frac{\alpha}{\alpha^2 + n^2}$

2) $\int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$

Thanks for any help!
For the first one multiply the numerator and denomiantor of the integrand by $e^{-x}$

This gives

$\int_{0}^{\infty}\frac{e^{-x}\sin(\alpha x)}{1-e^{-x}}dx$

Note that $e^{-x} < 1$ for all $x \in (0,\infty)$

Expand the denomiantor in a geometric series to get

$\int_{0}^{\infty}e^{-x}\sin(\alpha x) \sum_{n=0}^{\infty}e^{-nx}dx=\sum_{n=0}^{\infty}\sin(\alpha x)e^{-(n+1)x}dx$

Integrate by parts to get

$\sum_{n=0}^{\infty}\frac{\alpha}{(n+1)^2+\alpha}$

Just reindex the sereis and your done Note that some of the above steps still need formal justification.

For the 2nd one it is a similar trick

write $\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$

Then use integration by parts again.

Good luck

3. good show...i had just started looking into bessel functions to get the first one. i guess the wisdom of that would depend on knowing whether this problem set came from an undergrad course or not. anyway, nice moves.

4. Originally Posted by angela87
2) $\int_{0}^{1}\frac{\log x}{1+x}dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$
we have $\int_{0}^{1}{x^{n}\ln x\,dx}=\int_{0}^{1}{\int_{1}^{x}{\frac{x^{n}}{t}\, dt}\,dx}=-\int_{0}^{1}{\int_{0}^{t}{\frac{x^{n}}{t}\,dx}\,dt }=-\frac{1}{(n+1)^{2}},$ so $\int_{0}^{1}{\frac{\ln x}{1+x}\,dx}=\sum\limits_{n=0}^{\infty }{(-1)^{n}\int_{0}^{1}{x^{n}\ln x\,dx}}=\sum\limits_{n=0}^{\infty }{\frac{(-1)^{n+1}}{(n+1)^{2}}}=\sum\limits_{n=1}^{\infty }{\frac{(-1)^{n}}{n^{2}}},$ as required.

but you're gonna have to justify why we can swap sum and integral, so there's your job.