I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.
Here is a version of a rational log integral rarely encountered.
$\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$
I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.
My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?. Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?.
Here is my favorite method:
Consider a parameter $a$ dependent integral:
$$I(a)=\int_{0}^{1}\frac{\ln(ax^{2}+1)}{x+1}dx$$
After differentiating with respect to $a$ we get:
$$\frac{dI}{da}=\int_{0}^{1}\frac{x^2dx}{(1+x)(1+ax^2)}=$$
$$=\frac{1}{2}\frac{\ln(1+a)}{a(1+a)}+\frac{\ln 2}{1+a}-\frac{\arctan\sqrt{a}}{\sqrt{a}(1+a)}$$
Integration of $I'(a)$ yields $$I(a)=\ln2\ln(1+a)-\frac{1}{4}\ln^2(1+a)-\arctan^2\sqrt{a}+\frac{1}{2}\int_{1}^{a}\frac{\ln(1+z)}{z}dz+\text{const}$$
In order to find out the constant of integration , we note that $I(0)=0$, implying that
$$\text{const}=\frac{1}{2}\int_{0}^{1}\frac{\ln(1+z)}{z}dz=\frac{1}{2}\frac{\pi^2}{12}=\frac{\pi^2}{24}$$
The last integral can be easily calculated by expanding the log into Taylor series and integrating term by term. Finally the original integral $I$ yields $I=I(1)$