I need help to evaluate the integral \begin{align*} \int_{0}^{\infty}\frac{\ln(x)}{(x+1)^{3}}dx \end{align*} using the Residue theorem. I think that I could consider the contour integral \begin{align*} \oint_{C}\frac{\ln(z)}{(z+1)^{3}}dz, \end{align*} where $C$ is the quarter circle in the upper-right quadrant. But I'm not sure how to continue.

Could someone help me?

  • $\begingroup$ I suggest to use the logarithm with $[0,+\infty[$ as the branch cut and then use a pacman contour. $\endgroup$ – C. Dubussy Mar 5 '17 at 10:07
  • $\begingroup$ the integral should be $$-\frac{1}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 5 '17 at 10:07
  • $\begingroup$ Could you please explain further. How do I set up the pacman contour? $\endgroup$ – Math Mar 5 '17 at 10:24

Take the function $$f\left(z\right)=\frac{\log^{2}\left(z\right)}{\left(1-z\right)^{3}}$$ and the branch of the logarithm corresponding to $-\pi<\arg\left(z\right)\leq\pi$. Take the keyhole contour and define $\Gamma$ and $\gamma$ respectively the large circumference of radius $R$ and the small circumference of radius $\rho$ (see here). It is not difficult to prove that the integrals over the circumferences vanish so $$2\pi i\underset{z=1}{\textrm{Res}}\left(f\left(z\right)\right)=\int_{0}^{\infty}\frac{\log^{2}\left(-x+i\epsilon\right)}{\left(x-i\epsilon+1\right)^{3}}dx-\int_{0}^{\infty}\frac{\log^{2}\left(-x-i\epsilon\right)}{\left(x+i\epsilon+1\right)^{3}}dx$$ $$\stackrel{\epsilon\rightarrow0}{\rightarrow}\int_{0}^{\infty}\frac{\left(\log\left(x\right)+i\pi\right)^{2}}{\left(x+1\right)^{3}}dx-\int_{0}^{\infty}\frac{\left(\log\left(x\right)-i\pi\right)^{2}}{\left(x+1\right)^{3}}dx$$ $$=4\pi i\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+1\right)^{3}}dx$$ and since $$\underset{z=1}{\textrm{Res}}\left(f\left(z\right)\right)=-1$$ we get $$\int_{0}^{\infty}\frac{\log\left(x\right)}{\left(x+1\right)^{3}}dx=\color{red}{-\frac{1}{2}}.$$

We can also use real methods. Note that $$I\left(a\right)=\int_{0}^{\infty}\frac{x^{a}}{\left(x+1\right)^{3}}dx=B\left(a+1,-a+2\right),\,-1<a<2$$ where $B(x,y)$ is the beta function so $$I'\left(a\right)=\int_{0}^{\infty}\frac{\log\left(x\right)x^{a}}{\left(x+1\right)^{3}}dx=\left(\psi\left(a+1\right)-\psi\left(2-a\right)\right)B\left(a+1,-a+2\right)$$ where $\psi\left(x\right)$ is the digamma function hence $$I=\lim_{a\rightarrow0}\left[\left(\psi\left(a+1\right)-\psi\left(2-a\right)\right)B\left(a+1,-a+2\right)\right]=\color{blue}{-\frac{1}{2}}.$$

  • $\begingroup$ So you mean that we get by the residue theorem that $2\pi i Res(f,1)=\int_{\Gamma}f(z)dz+\int_{\gamma}f(z)dz+\int_{0}^{\infty}f(x)dx+\int_{\infty}^{0}f(x)dx$, then $\int_{\Gamma}f(z)dz$ and $\int_{\gamma}f(z)dz$ vanish? $\endgroup$ – Math Mar 6 '17 at 10:15
  • $\begingroup$ @Math The limits $R \rightarrow \infty$ and $\rho \rightarrow 0$ give $0$ as result. If you want I can write the calculations. $\endgroup$ – Marco Cantarini Mar 6 '17 at 11:38
  • $\begingroup$ Do you mean like this: On $\Gamma$ we have $\int_{\Gamma}f(z)dz=\frac{\ln^{2}(R)}{(1-R)^{3}}2\pi R \to 0$ when $R\to \infty$. And simiarly when we look at $\gamma$ but we take $\rho \to 0$? $\endgroup$ – Math Mar 6 '17 at 15:05
  • $\begingroup$ @Math No. On $\Gamma$ you have to put $z=R e^{i \theta}$ and on $\gamma$ you have to put $z= \rho e^{i \theta}$. Then take the absolute value of the integrals. It is possible to find an upper bound that does not depend on $\theta$ but only on $R$ and $\rho$ such that the limits go to $0$. See also en.wikipedia.org/wiki/Jordan's_lemma $\endgroup$ – Marco Cantarini Mar 6 '17 at 15:33
  • $\begingroup$ Okey thank you for your help! $\endgroup$ – Math Mar 6 '17 at 16:58

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