I wish to compute the integral $\int_1^\infty \frac{(\log x)}{x^2(1 + x^\alpha)} \mathrm{d} x$ for some $\alpha$ using the residue method. Here is my attempt. Would appreciate if anyone could help me out with what is going wrong. Thanks!
Consider the complex integral
$\oint_{C_R} \frac{(\log z)^2}{z^2(1 + z^\alpha)} \mathrm{d}z$ (note that $\log z$ is squared) where $C$ is the contour given by,$[1,R] \cup \{ Re^{i\theta} : \theta \in [0 , \epsilon] \} \cup \{ x e^{i \epsilon} : x \in [R,1] \} \cup \{ e^{i\theta} : \theta \in [0 , \epsilon] \}$ as shown in the figure. Consider the branch corresponding to $\log z \in [-\pi,\pi]$.
The integrand is holomorphic inside the contour so the complex integral comes out to be $0$ by the residue theorem.
Considering its individual components, the integral over the large circle vanishes as $R \to \infty$. On the other hand, the integral over the small radius $1$ circle is of the order of $-\epsilon^3$. And the integration on the sloped line comes out to be,
$$-\int_{1}^\infty \frac{(\log x + i \epsilon)^2}{1+x^\alpha e^{i \epsilon \alpha}} \frac{e^{-2i \epsilon}}{x^2}\mathrm{d} x$$
Adding together the contour integral over the real line gives,
$$\int_{1}^\infty \frac{(\log x)^2}{1+x^\alpha} \frac{1}{x^2} – \frac{(\log x + i \epsilon)^2}{1+x^\alpha e^{i \epsilon \alpha}} \frac{e^{-2i \epsilon}}{x^2}\mathrm{d} x$$
In the limit $\epsilon \to 0$,
$$ \int_1^\infty \epsilon \frac{\log x}{1 + x^\alpha} – O (\epsilon^3)= 0$$
gives that the original integral itself goes to $0$, which is obviously incorrect. Where am I going wrong?
I think that choosing a different contour, say one that picks $\epsilon = \frac{2 \pi}{\alpha}$ as opposed to $\epsilon \to 0$ might work, but that is a different question and I am interested in finding out the mistake in my understanding.
Best Answer
$$\int_{1}^{+\infty}\frac{\log(x)}{x^2(1+x^\alpha)}\,dx=\int_{0}^{+\infty}\frac{z\,dz}{e^z(1+e^{\alpha z})}=\sum_{n\geq 0}(-1)^n \int_{0}^{+\infty}\frac{z}{e^{(1+\alpha)z}}e^{-n\alpha z}\,dz=\sum_{n\geq 0}\frac{(-1)^n}{(\alpha(n+1)+1)^2} $$ is related to the Hurwitz $\zeta$ function, so i highly doubt that the LHS can be computed through the residue theorem if $\alpha\not\in\mathbb{Q}$.