Definite integral using the method of residues

Solution 1:

Define a branch cut for $\log$ going from the origin to the lower half plane, and restrict to the branch where $\log(z) = \ln|z|$ for $z>0$ and $\log(z) = \ln|z| + \pi i$ for $z <0$. Then we have:

$$ \int_{-\infty}^\infty\frac{\log(z)}{1+z^2}dz=2\int_{0}^\infty\frac{\log(z)}{1+z^2}dz+\pi i\int_{0}^\infty\frac{1}{1+z^2}dz $$

We can evaluate the left hand size by closing the contour with a great semicircle in the upper half plane. Since $f(z)|z|\rightarrow 0$ as $|z| \rightarrow 0$, we can ignore the contribution from the singularity at the origin. Likewise since $f(z)|z|\rightarrow 0$ as $|z| \rightarrow +\infty$ we can ignore the contribution from the great semicircle in the upper half plane. The function has a one singularity in the upper half plane, at $z=i$. So by the residue theorem:

$$ \begin{equation} \int_{-\infty}^\infty\frac{\log(z)}{1+z^2}dz= 2 \pi i \ \mbox{Res}\left[ \frac{\log(z)}{1+z^2}; z = i \right]= 2 \pi i \left[ \frac{\log(z)}{i+z}\right]_{z=i}= 2 \pi i \frac{\pi i /2}{i+i} = \frac{\pi^2 i}{2} \end{equation} $$ And since $\int_{0}^\infty\frac{1}{1+z^2}=\pi/2$, the original equation becomes:

$$ \frac{\pi^2 i}{2}=2\int_{0}^\infty\frac{\log(z)}{1+z^2}dz+\pi i\left(\frac{\pi}{2}\right) $$ $$ \Rightarrow \int_{0}^\infty\frac{\log(z)}{1+z^2}dz=0 $$

Solution 2:

Residue Approach

We will compute $$\renewcommand{\Re}{\operatorname{Re}}\newcommand{\Res}{\operatorname*{Res}} \int_\gamma\frac{\log(z)^2}{1+z^2}\,\mathrm{d}z\tag1 $$ where $\gamma$ is the keyhole contour $$ [ri,R+ri]\cup\sqrt{R^2+r^2}e^{i\left[\tan^{-1}\left(\frac rR\right),2\pi-\tan^{-1}\left(\frac rR\right)\right]}\cup[R-ri,-ri]\cup re^{i\left[\frac{3\pi}2,\frac\pi2\right]}\tag2 $$ where $r\to0$ and $R\to\infty$. We will put the branch cut along the positive real axis. On the the upper side of the real axis, $\log(z)$ is $\log(\Re(z))$. On the lower side, it is $\log(\Re(z))+2\pi i$.

The integral counter-clockwise along $\gamma$ is $$ \begin{align} \int_\gamma\frac{\log(z)^2}{1+z^2}\,\mathrm{d}z &=\int_0^\infty\frac{\log(x)^2-(\log(x)+2\pi i)^2}{1+x^2}\,\mathrm{d}x\\ &=\int_0^\infty\frac{4\pi^2-4\pi i\log(x)}{1+x^2}\,\mathrm{d}x\\ &=2\pi^3-4\pi i\int_0^\infty\frac{\log(x)}{1+x^2}\,\mathrm{d}x\tag3 \end{align} $$ The integral is also equal to $2\pi i$ times the sum of the residues of $\frac{\log(z)^2}{1+z^2}$: $$ \Res_{z=i}\left(\frac{\log(z)^2}{1+z^2}\right)=\frac{\pi^2i}8\tag{4a} $$ $$ \Res_{z=-i}\left(\frac{\log(z)^2}{1+z^2}\right)=-\frac{9\pi^2i}8\tag{4b} $$ Thus, the integral is $2\pi i\left(\frac{\pi^2i}8-\frac{9\pi^2i}8\right)=2\pi^3$. This, in conjunction with $(3)$ yields $$ \int_0^\infty\frac{\log(x)}{1+x^2}\,\mathrm{d}x=0\tag5 $$

Non-Residue Approach

To get this directly, without contour integration or breaking up the interval of integration, substitute $x\mapsto\frac{1}{x}$: $$ \int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x=-\int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x\tag6 $$ and therefore the integral is $0$.

Solution 3:

In the book 'COMPLEX VARIABLES Introductions and Applications' page 247-248 Example 4.3.5, contains a complete answer to your question. He begins by evaluating $I=\int_{0}^{\infty}\frac{\log^{2}(x)}{1+x^2}dx$ and after a few steps, arrive at the following expression $2\int_{0}^{\infty}\frac{\log^{2}(x)}{1+x^{2}}dx +2\pi i\int_{0}^{\infty}\frac{\log(x)}{1+x^{2}}dx=\frac{\pi^3}{4}$. So $$\int_{0}^{\infty}\frac{\log(x)}{1+x^{2}}=0.$$ I suggest reading! Any questions post to us.