Calculate: $$\int_{0}^{\infty}\frac{\ln x}{(x+1)^{3}}\mathrm{d}x$$
My try:
Keyhole integration:
$\displaystyle \frac{\pi i\ln R\cdot e}{(Re^{\theta i}+1)^{3}}\rightarrow 0$ (we take $r$ as large as we want)
and here is the confusion :
around the circle the residue is $0$:
as $\displaystyle \frac{x\ln x}{(1+x)^{3}}\rightarrow0$ when we approach to $0$. Therefore, the residue is $0$, and the whole integration of the keyhole is $0$. which leads that the result is $0$.
but if we take the pole in the keyhole, $x=-1$ this is $3$rd order pole, and its residue is $\displaystyle \left. -\frac{1}{x^{2}}\right|_{x=-1}=-1$ meaning that the whole integral is $-2\pi i$ which means that that the result should be $-\pi i$.
Both of the results are incorrect. Can you spot my mistakes?
Best Answer
$$J = \int_0^\infty \frac{ \log x \, dx} {(1+x)^3}.$$
Consider $$\oint_C \frac{(\log z)^2 \, dz}{(1+z)^3}$$ around a suitable keyhole contour $C$ that starts at $\epsilon$ goes to $R$, a large (almost) circle of radius $R$, back (below the branch cut) to $\epsilon$ and then clockwise around the origin.
There is a third order pole inside at $z_0 = -1$. The residue there is $$\text{Residue}_{z=-1} \left[\frac{ (\log z)^2}{(1+z)^3}\right] = 1-i\pi.$$
$$\begin{aligned} \oint_C \frac{(\log z)^2 \, dz}{(1+z)^3} &= \int_\epsilon^\infty \frac{(\log x)^2 \, dx}{(1+x)^3} -\int_\epsilon^\infty \frac{(\log x+2i\pi)^2 \, dx}{(1+x)^3}+\int_0^{2\pi} \frac{(\log (Re^{i\theta}))^2 \, Rie^{i\theta} }{(1+Re^{i\theta})^3}\, d\theta -~\int_0^{2\pi} \frac{(\log (\epsilon e^{i\theta}))^2 \, \epsilon i \, e^{i\theta} }{(1+\epsilon e^{i\theta})^3}\, d\theta \end{aligned} $$
Let $R\to\infty$ and $\epsilon\to 0$. The integrals along the "circles" go to zero.
Also, $$\displaystyle \int_0^\infty \frac{dx}{(1+x)^3}=\int_1^\infty \frac{dp}{p^3} = \left. -\frac{p^{-2}}{2} \right|_1^\infty = \frac{1}{2}.$$
So, we have $$-4i\pi J + 4\pi^2 \left( \frac{1}{2}\right) = 2\pi i (1-i\pi).$$
$$J=-\frac{1}{2}$$