Hi I am trying to prove the relation $$ I:=\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}. $$ I tried expanding the log argument by using $\sin x/ \cos x=\tan x,$ and than used $\log(a/b)=\log a-\log b$, I get $$ I=\int_0^\pi \left( \log \sin \frac{x}{4}-\log\cos \frac{x}{4}\right)^2dx. $$ We can distribute this out $$ \int_0^\pi \log^2 \sin \frac{x}{4}dx +\int_0^\pi \log^2\cos \frac{x}{4}dx-2\int_0^\pi\log \sin \frac{x}{4}\log \cos \frac{x}{4}dx. $$ Now I am stuck at how to solve these. Thanks.


$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{\pi}\ln^{2}\pars{\tan\pars{x \over 4}}\,\dd x ={\pi^{3} \over 4\phantom{^{3}}}:\ {\large ?}}$

\begin{align} I&=4\int_{0}^{\pi/4}\ln^{2}\pars{\tan\pars{x}}\,\dd x =4\int_{0}^{1}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x =2\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x \\[3mm]&=2\lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{0}^{\infty}{x^{\mu} \over x^{2} + 1}\,\dd x =\lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{0}^{\infty}{x^{\pars{\mu - 1}/2} \over x + 1}\,\dd x \end{align}

With $\ds{t \equiv {1 \over x + 1}\quad\imp\quad x = {1 \over t} - 1}$: \begin{align} I&=\lim_{\mu \to 0}\partiald[2]{}{\mu} \int_{1}^{0}t\pars{1 - t}^{\pars{\mu - 1}/2}t^{\pars{1 - \mu}/2}\, \pars{-\,{\dd t \over t^{2}}} \\[3mm]&=\lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{1}t^{-\pars{1 + \mu}/2} \pars{1 - t}^{\pars{\mu - 1}/2}\,\dd t =\lim_{\mu \to 0}\partiald[2]{{\rm B}\pars{1/2 - \mu/2,1/2 + \mu/2}}{\mu} \\[3mm]&=\lim_{\mu \to 0}\partiald[2]{}{\mu} \bracks{\Gamma\pars{1/2 - \mu/2}\Gamma\pars{1/2 + \mu/2} \over \Gamma\pars{1}} =\lim_{\mu \to 0}\partiald[2]{}{\mu} \braces{\pi \over \sin\pars{\pi\bracks{1/2 + \mu/2}}} \\[3mm]&=\pi\lim_{\mu \to 0}\partiald[2]{\sec\pars{\pi\mu/2}}{\mu} =\pi\lim_{\mu \to 0}\bracks{% {1 \over 4}\,\pi^{2}\sec^{3}\pars{\pi\mu \over 2} + {1 \over 4}\,\pi^{2}\sec\pars{\pi\mu \over 2}\tan^{2}\pars{\pi\mu \over 2}} \\[3mm]&=\pi\pars{\pi^{2} \over 4} \end{align}

$\ds{{\rm B}\pars{x,y}}$ and $\ds{\Gamma\pars{z}}$ are the Beta and Gamma Functions, respectively, and we used well known properties of them.

$$ \int_{0}^{\pi}\ln^{2}\pars{\tan\pars{x \over 4}}\,\dd x = {\pi^{3} \over 4\phantom{^{3}}} $$

  • $\begingroup$ Thanks for the solution. I appreciate the use of special functions by you in this proof and in others, very helpful! I also am glad to see the usual $\partial_\mu$ you like to use! I would check this as the answer, but have already checked the one above. Thanks +1 $\endgroup$ – Jeff Faraci Apr 27 '14 at 3:25
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    $\begingroup$ @Integrals $\partial_{\mu}$ is my ghost. Thanks. $\endgroup$ – Felix Marin Apr 27 '14 at 6:45

Rescaling domain by a factor of $2$, the integral becomes: $$I=2\int_{0}^{\frac{\pi}{2}}\log^2\left(\tan{\frac{x}{2}}\right)dx.$$

If it didn't appear so beforehand, the new form of the integral should be screaming "tangent half-angle substitution" to you now.

Let $t=\tan{\frac{x}{2}}$. The integral then becomes,


Now let $u=-\log{t}$. Then, $t=e^{-u}$, $dt=-e^{-u}du$, and


The denominator can be expanded into an alternating geometric series of exponentials. Interchanging the order of summation and integration then integrating term by term should yield a recognizable series:

$$\frac{1}{1+e^{-2u}}=\sum_{n=0}^{\infty}(-1)^ne^{-2nu}\\ \implies \frac{u^2e^{-u}}{1+e^{-2u}}=\sum_{n=0}^{\infty}(-1)^n u^2 e^{-(2n+1)u}\\ \implies \int_{0}^{\infty}\frac{u^2e^{-u}}{1+e^{-2u}}du=\sum_{n=0}^{\infty}(-1)^n \int_{0}^{\infty} u^2 e^{-(2n+1)u}du=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}$$

Hence, we have the series representation of the integral $I$:


  • 4
    $\begingroup$ +1 cannot believe "The world's sneakiest substitution" was staring at me in the face. $\endgroup$ – IAmNoOne Apr 26 '14 at 23:32
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    $\begingroup$ $\beta(2m+1) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}}$ has a nice closed form in terms of the Euler numbers. math.stackexchange.com/questions/762813/… $\endgroup$ – Random Variable Apr 27 '14 at 0:00
  • $\begingroup$ @David H. This is a very nice solution and I didn't see the substitution at first at all! Thank you for this! $\endgroup$ – Jeff Faraci Apr 27 '14 at 0:07
  • $\begingroup$ @RandomVariable Your comment completes the proof of this problem, so thanks. $\endgroup$ – Jeff Faraci Apr 27 '14 at 0:14

Let $u=\tan{\dfrac{x}{4}}$, then $du=\dfrac{1}{4}\sec^2{\dfrac{x}{4}}dx=\dfrac{1}{4}(1+u^2)dx$. The integral becomes \begin{align} \int^{\pi}_0\ln^2\left(\tan\frac{x}{4}\right)dx &=4\int^{1}_0\frac{\ln^2{u}}{1+u^2}du\\ &=4\sum_{n \ge 0}(-1)^n\int^1_0x^{2n}\ln^2{x}dx\\ &=4\sum_{n \ge 0}(-1)^n\lim_{a \to 2n}\frac{d^2}{da^2}\int^1_0x^adx\\ &=4\sum_{n \ge 0}(-1)^n\lim_{a \to 2n}\frac{d}{da}\left(-\frac{1}{(a+1)^2}\right)\\ &=8\sum_{n \ge 0}\frac{(-1)^n}{(2n+1)^3}\\ &=8\beta(3)\\ &=\frac{\pi^3}{4} \end{align}


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