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I am struggling with this integral:

$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$

What I tried so far:

$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$ $\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{\cos^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$ $\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{-\sin^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$ $\displaystyle =\frac{1}{2}\int _{0}^{\frac{\pi }{2}}\frac{\cos 2x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$

The answer should come out to be $\dfrac{-7\pi^2}{72}$.

Any help will be appreciated.

Quanto
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Nothing special
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    For the record, wolfram alpha can't solve the indefinite integral nor can it find an exact form for the definite integral (although it agrees when you ask it if it's $\frac{-7\pi^2}{72}$) – TheBestMagician Apr 11 '22 at 01:06
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    Seriously, unless someone gives you sufficient incentive, you'd have several better things to do with your time than to worry about this. There are endless calculus questions of similar nature, with no serious mathematical significance. To exaggerate, it's like asking for the product of two 15-digit integers: sure, we know the algorithm, but... no one actually cares about the outcome. The genuine question is about the algorithm, not about the specific answer. – paul garrett Apr 11 '22 at 01:21
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    After making the substitution $u = \tan x$, one can evaluate the integral using a standard trick in contour integration: Evaluate $\int_{\Gamma_\epsilon} \frac{\log^2 u ,dt}{(1 - u + u^2) (1 + u^2)}$ over an appropriate family ${\Gamma_{\epsilon}}$ of keyhole contours then take the limit $\epsilon \searrow 0$ and split the result into real and imaginary parts. – Travis Willse Apr 11 '22 at 02:38
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    See this excellent solution for an example of applying this method to a similar integral: https://math.stackexchange.com/a/621665/155629 – Travis Willse Apr 11 '22 at 02:41

3 Answers3

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Substitute $t=\tan x$\begin{align} &\int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x \\ =&\int_0^1\frac{\ln t}{(1 - t + t^{2})(1 + t^{2})}dt + \int_1^\infty \frac{\ln t}{(1 - t + t^{2})(1 + t^{2})}\overset{t\to 1/t}{dt}\\ =& \int_0^1\frac{(1-t^2)\ln t}{(1 - t + t^{2})(1 + t^{2})}dt = \int_0^1\frac{2t\ln t}{1 + t^{2}}\>\overset{ibp}{dt} -\int_0^1\frac{(2t-1)\ln t}{1 - t + t^{2}}\>\overset{ibp}{dt}\\ =& \>-\int_0^1 \frac{\ln (1+t^2)}{t}\>\overset{t^2\to t}{dt} +\int_0^1 \frac{\ln (1+t^3)}t \>\overset{t^3\to t}{dt}-\int_0^1 \frac{\ln (1+t)}tdt\\ =& \>-\frac76 \int_0^1 \frac{\ln (1+t)}tdt= -\frac76 \cdot\frac{\pi^2}{12}=-\frac{7\pi^2}{72} \end{align} where $\int_0^1 \frac{\ln (1+t)}tdt=\frac{\pi^2}{12}$

Quanto
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Since $$ \displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x=\int_0^{\infty} \frac{\ln t}{(1-t+t^2)(1+t^2)}dt, $$ consider $$ \begin{aligned} \mathscr{I}(s)&=\int_0^{\infty} \frac{t^s}{(1-t+t^2)(1+t^2)}dt \\&=\int_0^{\infty} \frac{t^{s-1}}{1-t+t^2}dt-\int_0^{\infty} \frac{t^{s-1}}{1+t^2}dt \\&=\int_0^{\infty} \frac{t^{s-1}}{1+t^3}dt+\int_0^{\infty} \frac{t^{s}}{1+t^3}dt-\int_0^{\infty} \frac{t^{s-1}}{1+t^2}dt. \end{aligned} $$ With Beta function, we have $$ \int_{0}^{\infty}\frac{t^{s-1}}{1+t^{a}}dt=\frac{\pi \csc(\frac{\pi s}{a})}{a} $$ thus $$ \mathscr{I}(s)=\frac{\pi \csc(\frac{\pi s}{3})}{3}+\frac{\pi \csc(\frac{1}{3} \pi (s+1))}{3}-\frac{\pi \csc(\frac{\pi s}{2})}{2}. $$ In conclusion,

$$ \begin{aligned} \int_0^{\frac{\pi}{2}}\frac{\ln \tan x}{1-\tan x+\tan^2 x}dx&=\lim_{s\to 0}\frac{\partial }{\partial s}\mathscr{I}(s) \\&=\lim_{s\to 0}\left(-\frac{\pi^{2} \csc(\frac{\pi s}{3}) \cot(\frac{\pi s}{3})}{9}-\frac{\pi^{2} \csc(\frac{1}{3} \pi s+\frac{1}{3} \pi) \cot(\frac{1}{3} \pi s+\frac{1}{3} \pi)}{9}+\frac{\pi^{2} \csc(\frac{\pi s}{2}) \cot(\frac{\pi s}{2})}{4}\right) \\&=-\frac{7 \pi^{2}}{72} \end{aligned} $$

metamorphy
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Wahrscheinlichkeitsrechnung
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$$\begin{align*} I &= \int_0^\tfrac\pi2 \frac{\ln(\tan x)}{1-\tan x+\tan^{2} x} \, dx \\ &= \int_0^\infty \frac{\ln t}{(1-t+t^2)(1+t^2)} \, dt \tag1 \\ &= \int_0^1 \frac{1-t^2}{(1-t+t^2)(1+t^2)} \ln t \, dt \tag2 \\ &= \int_0^1 \left(\frac{2t}{1+t^2} + \frac{1-t-2t^2}{1+t^3}\right) \ln t \, dt \tag3 \\ &= \sum_{n=0}^\infty (-1)^n \int_0^1 \left(2t^{2n+1} + t^{3n} - t^{3n+1} - 2t^{3n+2}\right) \ln t \, dt \tag4 \\ &= \sum_{n=0}^\infty (-1)^{n+1} \left(\frac2{(2n+2)^2} + \frac1{(3n+1)^2} - \frac1{(3n+2)^2} - \frac2{(3n+3)^2}\right) \tag5 \\ &= \frac5{18} \sum_{n=1}^\infty \frac{(-1)^n}{n^2} + \frac19 \sum_{n=1}^\infty (-1)^n \left(\frac1{\left(n-\frac23\right)^2} - \frac1{\left(n-\frac13\right)^2}\right) \\ &= -\frac{5\pi^2}{216} + \frac1{36} \sum_{k=1}^\infty \left(\frac1{\left(k-\frac13\right)^2} - \frac1{\left(k-\frac16\right)^2} + \frac1{\left(k-\frac23\right)^2} - \frac1{\left(k-\frac56\right)^2}\right) \tag6 \\ &= -\frac{5\pi^2}{216} + \frac1{36} \left[\psi^{(1)}\left(\frac23\right) - \psi^{(1)}\left(\frac56\right) + \psi^{(1)}\left(\frac13\right) - \psi^{(1)}\left(\frac16\right)\right] \tag7 \\ &= -\frac{5\pi^2}{216} + \frac1{36} \left(\frac{4\pi^2}3 - 4\pi^2\right) = \boxed{-\frac{7\pi^2}{72}} \tag8 \end{align*}$$

  • $(1)$ : substitute $t=\tan x$
  • $(2)$ : split the integral at $t=1$, substitute $t\mapsto\dfrac1t$ for $t\ge1$, and recombine
  • $(3)$ : partial fractions; introduce a factor of $1+t$ to get a nicer denominator in the second term
  • $(4)$ : invoke Macluarin series of $\dfrac1{1-t}$
  • $(5)$ : integrate by parts and shift index of summation
  • $(6)$ : the first sum is well-known; expand the latter summand into cases of even and odd $n$
  • $(7)$ : trigamma function
  • $(8)$ : trigamma reflection formula,

$$\psi^{(1)}\left(z\right) + \psi^{(1)}\left(1-z\right) = \pi^2 \csc^2\left(\pi z\right)$$

user170231
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