Yep, I thought contours would make that a pushover. Try a real method if you wish.

I calculated the residues and saw it was coming together.

What exactly is it you want to prove regarding the Dirichlet Beta?.

The derivative of DB is related to the derivative of Zeta...as I kind of mention below. Here is a link. Is this what you mean?.

http://engineeringandmathematics.blogsp ... ation.htmlAdamchik wrote a paper dedicated to integrals of the type $log(log(u))$ and so on.

He even derived a general form:

$$\int_{0}^{1}\frac{u^{p-1}}{1+u^{n}}\log\log(1/u)du=\frac{\gamma+\log(2n)}{2n}\left[\psi(\frac{p}{2n})-\psi(\frac{n+p}{2n})\right]+\frac{1}{2n}\left[\zeta'(1,\frac{p}{2n})-\zeta'(1,\frac{n+p}{2n})\right]$$

for our integral, let $p=1, \;\ n=1$

note that $$\zeta'(1,1/4)-\zeta'(1,3/4)=\pi\left[\gamma+4\log(2)+3\log(\pi)-4\log\Gamma(1/4)\right]$$

and $$\psi(1/4)-\psi(3/4)=-\pi$$

Also, $$\beta(s)=\frac{1}{4^{s}}\left[\zeta(s,1/4)-\zeta(s,3/4)\right]$$

$$\Gamma(s)\beta(s)=\int_{0}^{\infty}\frac{x^{s-1}e^{-x}}{1+e^{-2x}}dx$$

You know this of course, but I list them to show the relation.