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I am trying to compute

$$\int_0^\infty \frac{\ln^2 x}{1+x^2} dx $$

using a complex contour integration of $f(z) = \ln^2 z / (1+z^2)$ around a closed contour with the segments:

$$C_1 : [-R,-\rho]\\C_2: \{z: |z| = \rho, 0 \leq \arg z \leq \pi\}\\ C_3: [\rho , R] \\ C_4: \{z: |z| = R, 0 \leq \arg z \leq \pi\}.$$

This contour encloses one singular point $z = i$ where I found $Res(f,i) = i\pi^2/8.$

My difficulty is in handling the $\int_{C_1}f(z) dz$ integral on the real interval $[-R,-\rho]$ which does not vanish as $R \to \infty$ and $\rho \to 0$.

Thank you.

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  • $\begingroup$ Related: math.stackexchange.com/questions/1976510 $\endgroup$ – Watson Jan 27 '17 at 21:14
  • $\begingroup$ I'm interested in this specific contour in particular how to make it work with the integral over the segment on the negative real axis. I know there are other ways to get the value. $\endgroup$ – scobaco Jan 27 '17 at 21:19
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By the residue theorem

$$\oint_{C_1 \cup C_2 \cup C_3 \cup C_4}f(z) \, dz = 2\pi i Res(f,i) = 2\pi i (i \pi^2/8) = -\frac{\pi^3}{4}.$$

It is straightforward to show that the integrals over $C_2$ and $C_4$ vanish in the limit as $R \to \infty$ and $\rho \to 0$.

Hence,

$$\tag{1} -\frac{\pi^3}{4} = \int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx + \lim_{R \to \infty, \rho \to 0} \int_{C_1} \frac{\ln^2 z}{1 + z^2} \, dz.$$

Now we can focus on the second integral on the RHS -- the source of your difficulty.

Note that

$$\begin{align}\lim_{R \to \infty, \rho \to 0} \int_{C_1} \frac{\ln^2 z}{1 + z^2} \, dz &= \int_{-\infty}^0 \frac{ \ln^2 x}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 (-x)}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ (\ln x + i \pi)^2}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx + 2\pi i\int_{0}^\infty \frac{ \ln x }{1 + x^2} \, dx - \pi^2\int_{0}^\infty \frac{ 1 }{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx + 0 - \pi^2 \frac{\pi}{2} \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx - \frac{\pi^3}{2}\end{align}.$$

The second of the three integrals on the RHS can be shown to be $0$ by showing that integrals over $[0,1]$ and $[1,\infty)$ must cancel (using a change of variables $u = 1/x)$.

Substituting into (1) we obtain

$$-\frac{\pi^3}{4} = 2\int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx - \frac{\pi^3}{2}.$$

Thus,

$$\int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx = \frac{\pi^3}{8}.$$

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  • $\begingroup$ Ah, whoops, I set that up incorrectly. Still learning :-) $\endgroup$ – Simply Beautiful Art Jan 28 '17 at 4:30
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One notes that $I_1$ is the original integral from $-R$ to $-\rho$ with an integrand shifted by $\pi i$, or $-1$.

$$\lim_{R\to\infty}\lim_{\rho\to0}I_1=\int_{-\infty}^0\frac{\ln^2(x)}{1+x^2}\ dx$$

with the simple substitution of $x=-u$, we get...

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  • $\begingroup$ This is not correct. The value of the integral over $[0,\infty)$ should be $\pi^3/8$. Since $2\pi i Res(f,i) = -\pi^3/4$, if the integral over $C_1$ is equal to that over $C_3$ you will get the wrong answer. Also Wolfram Alpha shows the two intergrals are not equal. SInce $\ln(-x) = \ln|x| + i \pi$, I can't see how they are equal. $\endgroup$ – scobaco Jan 27 '17 at 22:28
  • $\begingroup$ @scobaco Hm, I did $u=-x$, rather than actually evaluating. I'll check this. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 22:31
  • $\begingroup$ @scobaco You need to check that. WolframAlpha says the two are equal to $\pi^3/8$. wolframalpha.com/input/… $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 22:34
  • $\begingroup$ @scobaco sorry for the previous comment for being somewhat wrong. My thoughts on this are for you to check how $I_2$ plays in, because I think it does and that my answer is perfectly right. $\endgroup$ – Simply Beautiful Art Jan 28 '17 at 3:01
  • $\begingroup$ @SimplyBeautifulArt: I think you set up the integral in WolframAlpha incorrectly. Answer below confirms that the contributions from $C_1$ and $C_3$ are different. But your general suggestion was indeed correct. $\endgroup$ – RRL Jan 28 '17 at 3:55

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