Complex analysis contour integral: $\int_0^{\infty}\frac{\log(x)}{x^2-1}$

Question

I am working on the integral $\displaystyle\int_0^{\infty}\frac{\log(x)}{x^2-1}$. I see it done here $\int_0^\infty\frac{\log x dx}{x^2-1}$ with a hint. but I am wondering if it is possible to integrate on a different contour, the keyhole. I know that we could set the contour integral equal to the residue at $z = i$ which is $\dfrac{\pi}{2}$ since it is a simple pole. I also know that the integral around the large circle will be zero.

Answer 1

I'm not sure the semicircle is exactly what you want. The standard keyhole contour works quite well here, as I will show below.

Note that the singularity at $x=1$ is removable in this integral. We evaluate this integral by appealing to the residue theorem; consider

$$\oint_{C} dz \frac{\log^2{z}}{z^2-1}$$

where $C$ is a keyhole contour with respect to the positive real axis. By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C$ increases without bound, we get

$$\oint_{C} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$

This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C$. As the only pole is at $z=-1$, we see that

$$\begin{align}\oint_{C} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi \frac{\pi^2}{2}\end{align}$$

Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$. The Cauchy principal value is zero:

$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon} \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$

The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:

$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$

so that

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$

Therefore

$$\int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{4}$$