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How to show that $(1)$

$$2\int_{0}^{\pi\over 2}\ln^2(\tan^2{x})\mathrm dx=\pi^3?\tag1$$

$u=\tan^2{x}$ then $\mathrm dx={\mathrm du\over \sqrt{u}(u-1)}$

$$\int_{0}^{\infty}\ln^2{u}{\mathrm du\over \sqrt{u}(u-1)}\tag2$$

$$-\sum_{n=0}^{\infty}\int_{0}^{\infty}u^{n-1/2}\ln^2{u}\mathrm du\tag3$$

huhhh? $(3)$ seems to be diverging.

gymbvghjkgkjkhgfkl
  • 8,090
  • Let $u=\tan(x)$ which makes $$I=8\int_0^\infty \frac{ \log ^2(u)}{u^2+1},du$$ which "looks" better (please check you bounds). I suppose that the antiderivative would involve polylogarithms with complex arguments (using $1+u^2=(1+iu)(1-iu)$ and fraction decomposition. But I am stuck. – Claude Leibovici Nov 04 '17 at 08:58
  • @123: please be respectful towards other users. I am not the only one wondering what is the purpose of asking questions if one does not learn anything from their answers. This is yet another question of yours which is trivial to tackle by differentiation under the integral sign, since $\int_{0}^{+\infty}\frac{u^\alpha}{u^2+1},du$ is known through Euler's Beta function, and at this point you should know that. – Jack D'Aurizio Nov 04 '17 at 18:07
  • And by Claude Leibovici's comment, this problem is just an instance of a well-known one: https://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-values-of-etas – Jack D'Aurizio Nov 04 '17 at 18:08
  • @JackD'Aurizio: There seems to be a connection between the users of this post and this, though the one above seems to have a small typo. (It should be $\tan^2 \frac{x}2$, not $\tan^2 x$.) – Tito Piezas III Dec 20 '17 at 17:07
  • @TitoPiezasIII: it is just the same user asking almost the same question. – Jack D'Aurizio Dec 20 '17 at 17:14

1 Answers1

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Setting $I = \int^{\pi/2}_0 \ln^2 (\tan^2 x) \, dx$, after letting $u = \tan x$ the integral becomes $$I = 4 \int^\infty_0 \frac{\ln^2 u}{1 + u^2} \, du.$$ Writing this integral as $$I = 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + 4 \int^\infty_1 \frac{\ln^2 u}{1 + u^2} \, du,$$ if, in the second of these integrals we set $u = 1/x$, one has $$I = 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du = 8 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du.$$ Writing the term $1/(1 + x^2)$ that appears in the integrand as a geometric series, namely $$\frac{1}{1 + x^2} = \sum^\infty_{n = 0} (-1)^n x^{2n}, \quad |x| < 1,$$ after interchanging the summation with the integration (by Tonelli's theorem) we have $$I = 8 \sum^\infty_{n = 0} (-1)^n \int^1_0 x^{2n} \ln^2 x \, dx.$$ Integrating by parts twice, we find $$I = 16 \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^3}.$$ To evaluate the sum we write \begin{align*} I &= 16 \sum^\infty_{n = 0} \frac{(-1)^n}{(2n + 1)^3} = 2 \sum^{\infty}_{n = 0} \frac{(-1)^n}{(n + \frac{1}{2})^3}\\ &= 2 \sum^{\infty}_{n = 0,n \in \text{even}} \frac{1}{(n + \frac{1}{2})^3} - 2 \sum^{\infty}_{n = 0,n \in \text{odd}} \frac{1}{(n + \frac{1}{2})^3}. \end{align*} Reindexing, in the first sum let $n \mapsto 2n$ while in the second sum let $n \mapsto 2n + 1$. Thus \begin{align*} I &= 2 \sum^\infty_{n = 0} \frac{1}{(2n + \frac{1}{2})^3} - 2 \sum^\infty_{n = 0} \frac{1}{(2n + \frac{3}{2})^3}\\ &= \frac{1}{4} \sum^\infty_{n = 0} \frac{1}{(n + \frac{1}{4})^3} - \frac{1}{4} \sum^\infty_{n = 0} \frac{1}{(n + \frac{3}{4})^3}\\ &= -\frac{1}{4} \left [\zeta(3,\frac{3}{4}) - \zeta(3, \frac{1}{4}) \right ], \end{align*} where $\zeta (s,z)$ is the Hurwitz zeta function.

Now the polygamma function $\psi^{(m)} (z)$ is related to the Hurwitz zeta function by $$\zeta (1 + m, z) = \frac{(-1)^{m + 1}}{m!} \psi^{(m)}(z).$$ Using this result to rewrite the Hurwitz zeta functions in terms of polygamma functions we have \begin{align*} \zeta (3, \frac{1}{4}) &= \zeta (1 + 2, \frac{1}{4}) = -\frac{1}{2} \psi^{(2)} (\frac{1}{4})\\ \zeta (3, \frac{3}{4}) &= \zeta (1 + 2, \frac{3}{4}) = -\frac{1}{2} \psi^{(2)} (\frac{3}{4}) \end{align*} So our integral becomes $$I = \frac{1}{8} \left [\psi^{(2)}(\frac{3}{4}) - \psi^{(2)} (\frac{1}{4}) \right ].$$ Now making use of the reflection formula for the polygamma function, namely $$\psi^{(m)}(1 - z) + (-1)^{m + 1} \psi^{(m)}(z) = (-1)^m \pi \frac{d^m}{dz^m} \cot (\pi z),$$ setting $m = 2$ we have \begin{align*} \psi^{(2)}(\frac{3}{4}) - \psi^{(2)} (\frac{1}{4}) &= \psi^{(2)}(1 - \frac{1}{4}) - \psi^{(2)} (\frac{1}{4})\\ &= \pi \left. \frac{d^2}{dz^2} \cot (z \pi) \right |_{z = 1/4}\\ &= \pi [2 \pi^2 \cot (z\pi) \text{cosec}^2 (z \pi)]_{z = 1/4}\\ &= \pi \cdot 4\pi^2 = 4 \pi^3. \end{align*} So finally $$I = \frac{1}{8} \cdot 4 \pi^3 = \frac{\pi^3}{2},$$ as required.

omegadot
  • 11,736
  • There might be a small error somewhere in your answer. In another post, we have the very similar integral $$\int_{0}^{\pi/2}\ln^2\left(\tan^2\color{red}{x\over 2}\right)\mathrm dx=\frac{\pi^3}2$$ but in this post, the OP is asking about $$\int_{0}^{\pi/2}\ln^2\left(\tan^2\color{blue}x\right)\mathrm dx=,??$$ – Tito Piezas III Dec 20 '17 at 16:58
  • It seems both these integrals have the same answer. – omegadot Dec 20 '17 at 17:16
  • I've asked a general question here. – Tito Piezas III Dec 21 '17 at 04:16
  • Yes, I understand. The integral as given here (your blue one) does indeed have a value equal to $\pi^3/2$, which turns out to be exactly the same value as the other integral you mention (your red one). – omegadot Dec 21 '17 at 04:21