Definite integral in closed form?

Hi,

I'd like to compute the following integral in closed form:

$\displaystyle \int_0^\infty \frac{\log(1+x) e^{-x}}{1+x} dx$

I assume this is possible via complex analysis, but I can't figure out how to extend this to a complex integration problem. Any ideas?

Re: Definite integral in closed form?

Quote:

Originally Posted by

**jens** Hi,

I'd like to compute the following integral in closed form:

$\displaystyle \int_0^\infty \frac{\log(1+x) e^{-x}}{1+x} dx$

I assume this is possible via complex analysis, but I can't figure out how to extend this to a complex integration problem. Any ideas?

Maybe try integration by parts with $\displaystyle \displaystyle \begin{align*} u = e^{-x} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} dv = \frac{\log{(1 + x)}}{1 + x} \end{align*}$...

Re: Definite integral in closed form?

Quote:

Originally Posted by

**Prove It** Maybe try integration by parts with $\displaystyle \displaystyle \begin{align*} u = e^{-x} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} dv = \frac{\log{(1 + x)}}{1 + x} \end{align*}$...

I've tried that already, but it's not conclusive. Whatever I try with integration by parts, I only manage to express the above integral in terms of a similar integral, but with the logarithm squared, or the denominator squared. So it's making things only more difficult instead of simpler.

By the way, I happened to find the exact value of this integral by playing around with WolframAlpha and doing some extra algebra. It reads as follows:

$\displaystyle e \left[ \frac{1}{12} (\pi^2+6 \gamma^2) - \sideset{_3}{_3}F (1,1,1;2,2,2|-1)\right]$

$\displaystyle = e \left[ \frac{1}{12} (\pi^2+6 \gamma^2) - \sum_{k=0}^\infty \frac{(-1)^k}{k!\;k^3(k+1)^3}\right]$

and is in agreement with the numerical computation.

But that doesn't tell me the steps to get to this result…