I'm trying to compute this integral, $$\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx \hbox{ where } 0<a<1$$
I drew a typical Pacman contour with branch cut at positive real axis. Then, we have $$\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz=\left(\int_{L_{1}}+\int_{L_{2}}+\int_{C_{R}}+\int_{C_{\epsilon}}\right)\frac{\ln z}{z^{a}(z+1)}dz$$
The function has a pole at $z=-1$. Then, by residue theorem, the whole integral becomes
\begin{align*}
\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz&=2\pi i\text{res}(\frac{\ln z}{z^{a}(z+1)};-1)\\&=2\pi i\lim_{z\rightarrow-1}(z+1)\frac{\ln z}{z^{a}(z+1)}\\&=2\pi i\lim_{z\rightarrow-1}\frac{\ln z}{z^{a}}\\&=2\pi i\frac{\ln e^{i\pi}}{e^{i\pi a}}\\&=2\pi i\frac{\ln1+i\pi}{e^{i\pi a}}\\&=-\frac{2\pi^{2}}{(e^{i\pi})^{a}}
\end{align*}
On $L_1$, $z=xe^{i\epsilon}$ where $\epsilon\leq x\leq R$
On $L_{2}$, $z=xe^{i(2\pi-\epsilon)}$ where $\epsilon\leq x\leq R$
On $C_{R}$, $z=Re^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$
Lastly on On $C_{\epsilon}$, $z=\epsilon e^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$
So I tried to compute them separately and I anticipate the curves will both go to $0$ by ML inequality, and two linear integrals will consist of the term $\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$ and hence simply solve for it. But I'm having trouble physically solving it out. I think the log is throwing me off. Can anybody please help me?
Best Answer
The integral can be evaluated to $$\pi^2\cot\pi a~\csc\pi a$$
I will provide a proof later.
Let $$J(a)=\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$$ $$I(a)=\int_0^\infty\frac{x^{-a}}{x+1}dx$$
Clearly, $J(a)=-I'(a)$.
Evaluation of $I(a)$ is easier.
Let $$f(z)=\frac{z^{-a}}{z+1}=\frac{\exp(-a(\ln|z|+i\arg z))}{z+1}\hbox{ where }\arg z\in[0,2\pi).$$
Let $C$ be the keyhole contour centered at the origin, avoiding the branch cut of $z^{-a}$.
By residue theorem, $$\oint_C f(z)dz=2\pi i\operatorname*{Res}_{z=-1}f(z)=2\pi i(-1)^{-a}=2\pi ie^{-\pi i a}\qquad{(1)}$$
Also, $$\oint_C =\int_{\text{large circle}}+\int_{\text{small circle}}+\int_{\text{upper real axis}}+\int_{\text{lower real axis}}$$
You can easily prove that the first two integrals tend to zero.
Moreover, $$\int_{\text{upper real axis}}=\int^\infty_0 f(te^{i0})dt=I(a)$$ $$\begin{align} \int_{\text{lower real axis}} &=\int_\infty^0 f(te^{i2\pi})dt \\ &=\int_\infty^0\frac{(te^{2\pi i})^{-a}}{t+1}dt \\ &=-e^{-2\pi i a}\int^\infty_0\frac{t^{-a}}{t+1}dt \\ &=-e^{-2\pi i a}I(a) \\ \end{align} $$
Back to $(1)$, $$I(a)-e^{-2\pi i a}I(a)=2\pi ie^{-\pi i a}$$ $$(e^{\pi i a}-e^{-\pi i a})I(a)=2\pi i$$ $$\frac{e^{\pi i a}-e^{-\pi i a}}{2i}I(a)=\pi$$ $$(\sin{\pi a})I(a)=\pi$$ $$I(a)=\pi\csc{\pi a}$$
Therefore, $J(a)=-I'(a)=\pi^2\cot\pi a~\csc\pi a$. $$\color{red}{\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx=\pi^2\cot\pi a~\csc\pi a}$$
Some special values are $$J(1/6)=2\sqrt3\pi^2, ~J(1/4)=\sqrt2\pi^2, ~J(1/3)=2\pi^2/3, ~J(1/2)=0$$
$J(a)$ satisfies the functional equation $$J(a)=-J(1-a).$$