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
  • 2
    $\begingroup$ @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 $I$:


  • 5
    $\begingroup$ +1 cannot believe "The world's sneakiest substitution" was staring at me in the face. $\endgroup$ – IAmNoOne Apr 26 '14 at 23:32
  • 1
    $\begingroup$ $\beta(2m+1) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}}$ 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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.