Say we want to evaluate $\int^\infty_0 \frac{(\log(x))^2}{x^2+1}\,dx $ via integrating over a complex contour.

My lecturer said to use the function $f(z) = \frac{(\log(z) - \frac{i \pi}{2})^2}{z^2+1}\ $ with the branch cut along the negative imaginary axis over the following contour:

enter image description here

I understand the contour choice: we include the $i$ pole to take the residue of, and we avoid the branch cut. However, I don't understand the intuition behind why they've change the function from $\log(z)$ to $\log(z) - \frac{i \pi}{2}$. Why is that of benefit to us?


Note that with the branch cut along the negative real axis, we see that along the positive real axis


while along the negative real axis


It is easy to show that $\int_{-\infty}^0 \frac{\log^2(|x|)}{1+x^2}\,dx=\int_0^{\infty} \frac{\log^2(|x|)}{1+x^2}\,dx$.

Hence, $\int_{-\infty}^\infty \frac{(\log(x)-i\pi/2)^2}{x^2+1}\,dx=2\int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx-\frac12 \pi^2\int_0^\infty \frac{1}{x^2+1}\,dx$. So, the cross term involving $\log(x)$ is no involved.

Note that had we analyzed $\oint_C \frac{\log^2(z)}{z^2+1}\,dz$, we need to evaluate the integral $\int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx$. However, upon enforcing the substitution $x\to 1/x$, we would readily find that this integral is $0$.

Therefore, we have

$$\begin{align} \oint_C \frac{(\log(z)-i\pi/2)^2}{z^2+1}\,dz&=2\int_\epsilon^R \frac{\log^2(x)}{x^2+1}\,dx-\frac12\pi^2\underbrace{\int_\epsilon^R \frac{1}{x^2+1}\,dx}_{\to -\pi^3/4\,\text{as}\,\epsilon\to 0\,\text{and}\,R\to\infty}\\\\ &+\underbrace{\int_\pi^0 \frac{\log^2(\epsilon e^{i\phi})}{(\epsilon e^{i\phi})^2+1}\,i\epsilon e^{i\phi}\,d\phi}_{\to 0\,\text{as}\,\epsilon\to 0}+\underbrace{\int_0^\pi \frac{\log^2(R e^{i\phi})}{(R e^{i\phi})^2+1}\,i R e^{i\phi}\,d\phi}_{\to 0\,\text{as}\,R\to \infty}\\\\ &=2\pi i \text{Res}\left(\frac{(\log(z)-i\pi/2)^2}{z^2+1}, z=i\right)\\\\ &=0 \end{align}$$

Note that had we analyzed the integral $\oint_C \frac{\log^2(z)}{z^2+1}\,dz$, the reside would not be zero. However, this poses no significant challenge or complication.

Therefore, as $\epsilon\to 0$ and $R\to \infty$ we see that

$$\int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx=\frac{\pi^3}{8}$$

Although analysis of the integral $\oint_C \frac{(\log(z)-i\pi/2)^2}{z^2+1}\,dz$ facilitates analysis in that we forgo the need to evaluate the integral $\int_0^\infty \frac{\log^2(x)}{x^2+1}\,dx=0$, this does not seem to add any significant benefit of efficiency over analysis of the integral $\oint_C \frac{\log^2(z)}{z^2+1}\,dz$.

  • $\begingroup$ Thanks so much for your response. So, there is no particular reason why my lecturer chose this integrand, and I could have just as validly chosen $\oint_C \frac{\log^2(z)}{z^2+1}\,dz$ instead? $\endgroup$ – hhattiecc May 20 '17 at 21:05
  • $\begingroup$ You're welcome. My pleasure. Perhaps the instructor wanted to avoid having to evaluate $\int_0^\infty \frac{\log(x)}{x^2+1}\,dx=0$ and also wanted a residue of $0$. $\endgroup$ – Mark Viola May 20 '17 at 23:03

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