TITLE: A $\log$ integral with a parameter
QUESTION [11 upvotes]: Prove that: 
$$\int_0^\infty \frac{\ln x}{x^a+1}\;\text{d}x=-\left( \frac{\pi }{a} \right)\cot \left( \frac{\pi }{a} \right)\csc \left( \frac{\pi }{a} \right),\ \ a>1$$
For this one I consider to have $\displaystyle\int_0^\infty \ln x\int_0^\infty \text{e}^{-y(1+x^a)} \, \text{d}y \, \text{d}x$. 
And use $\displaystyle{\Gamma}'(s)=\int_0^\infty \ln(x) x^{s-1} \text{e}^{-x} \, \text{d}x$, but then I just stuck...

REPLY [6 votes]: Very similar to Sasha's elegant answer, just without the double integral!
$$I(\lambda)=\int_0^{\infty}\frac{x^{\lambda}}{x^a+1}dx$$
Let $t=\dfrac{1}{x^a+1}:$
$$\begin{align*}I(\lambda)&=\frac{1}{a}\int_0^1\left(\frac{1}{t}-1\right)^{\frac{\lambda+1-a}{a}}\frac{1}{t}dt\\&=\frac{1}{a}\int_0^1\left(1-t\right)^{\frac{\lambda+1}{a}-1}t^{\frac{-\lambda-1}{a}}dt\\&=\frac{1}{a}\text{B}\left(\frac{\lambda+1}{a},1-\frac{\lambda+1}{a}\right)\end{align*}$$
Using the Beta-Gamma relation:
$$I(\lambda)=\frac{1}{a}\Gamma\left(\frac{\lambda+1}{a}\right)\Gamma\left(1-\frac{\lambda+1}{a}\right)$$
By the lovely reflection formula:
$$I(\lambda)=\frac{\pi}{a}\csc\left(\pi\cdot\frac{\lambda+1}{a}\right)$$
Thus:
$$I'(\lambda)=-\frac{\pi^2}{a^2}\csc\left(\pi\cdot\frac{\lambda+1}{a}\right)\cot\left(\pi\cdot\frac{\lambda+1}{a}\right)$$
$$I'(0)=-\frac{\pi^2}{a^2}\csc\left(\frac{\pi}{a}\right)\cot\left(\frac{\pi}{a}\right)$$