I am trying to show $$\int_0^{\infty} \frac{\log (x)}{(x^2+1)^2}$$. We can use the integrand $g(z) = \frac{\log(z)}{(x^2+1)^2}$ defining log as $\log(\rho e^{i\theta}) = \log(\rho) + i\theta$ and letting $\theta \in [0,2\pi]$
Now, I know that the singularities of the integrand occur at $i,-i$ one being in the upper half plane, the other in the lower half plane.
Typically the counter used for integrands which require choosing a branch cut of log is a half donut shape on the upper plane. Since there is a singularity no on the lower half plane, do I ignore it or include the residue of it in the calculation?
Also, I believe the singularities to be poles of order 3 and I do not know how to calculate residues for them. Is there an explicit formula like there is for simple poles?
Best Answer
You are trying to show... what? $$\begin{eqnarray*}\int_{0}^{+\infty}\frac{\log(x)}{(1+x^2)^2}\,dx &=& \int_{0}^{1}\frac{\log x}{(1+x^2)^2}\,dx+\underbrace{\int_{0}^{1}\frac{-x^2\log(x)}{(1+x^2)^2}\,dx}_{x\mapsto 1/x}\\&=&\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}\log(x)\,dx\end{eqnarray*} $$ and since $$\forall x\in(0,1),\qquad \frac{1-x^2}{(1+x^2)^2}=\sum_{n\geq 0}(-1)^n (2n+1) x^{2n}, $$ $$ \int_{0}^{1}x^{2n}\log(x)\,dx = -\frac{1}{(2n+1)^2}$$ we have: $$ \int_{0}^{+\infty}\frac{\log x}{(1+x^2)^2}\,dx = -\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=-\arctan(1)=\color{red}{-\frac{\pi}{4}}.$$ We do not strictly need Complex Analysis.