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Compute the following integration without harmonic series or Fourier series :

$I=\displaystyle\int_0^{\frac{π}{4}}x\ln(\tan x)dx$

Wolfram alpha give $I=\frac{7\zeta(3)-4πC}{16}$ Where $C$ : Catalan's constant

My try :

put : $y=\tan x$ then $dx=\frac{dy}{1+y^2}$

Then :

$I=\displaystyle\int_0^{1}\frac{\arctan x\ln x}{1+x^2}dx$ Then define : $I(a,b)=\displaystyle \int_0^{1}\frac{\arctan (ax)\ln x}{1+x^2}dx$

Then :

$\frac{dI(a,b)}{da}=\displaystyle\int_0^{1}\frac{x\ln x}{(1+a^{2}x^{2})(1+x^2)}dx$

Use partial fraction

$\frac{dI(a,b)}{da}$ $=\displaystyle\int_0^{1}\frac{a^{2}x\ln x}{(1+a^{2}x^{2})(a^{2}-1)}dx$

$-\displaystyle\int_0^{1}\frac{x\ln x}{(1+a^{2}x^{2})(a^{2}-1)}dx$

But I don't know how I complete

Please give me ideas to approach it .

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  • $\begingroup$ You can find some solutions here: math.stackexchange.com/q/2714146/515527 after a integration by parts. $$I=-\frac18\int_0^\frac{\pi}{2} \frac{x^2}{\sin x}dx$$ $\endgroup$ – カカロット Apr 29 at 19:05
  • $\begingroup$ See my answer: math.stackexchange.com/a/3200545/186817 and search for $I-J$. $\endgroup$ – FDP Apr 29 at 19:07
  • $\begingroup$ @FDP you can make that an answer! Also you have solved this integral in the link I gave which makes two solutions from you already. Although I am not sure if this counts as a duplicate. $\endgroup$ – カカロット Apr 29 at 19:11
  • $\begingroup$ Thank you very much @Zacky & FDP $\endgroup$ – user668815 Apr 29 at 22:52
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The Clausen Function $\operatorname{Cl}_2(z)$ can be defined as a Fourier-type Series. On the other hand it is afterall a very nice special function with a convenient relation to the Dilogarithm and capable of providing closed-form anti-derivatives for logarithmo-trigonometirc integrals such as the given one.

All we need are two identities involving the Clausen Function. To be precise

\begin{align*} \int_0^z\log(\tan t)\mathrm dt~&=~-\frac12\operatorname{Cl}_2(2z)-\frac12\operatorname{Cl}_2(\pi-2z)\tag1\\ \int_0^z\operatorname{Cl}_2(t)\mathrm dt~&=~\zeta(3)-\operatorname{Cl}_3(z)\tag2 \end{align*}

Using Integration By Parts and firstly $(1)$ and then $(2)$ we obtain

\begin{align*} \int_0^\frac\pi4x\log(\tan x)\mathrm dx&=\left[x\left(-\frac12\operatorname{Cl}_2(2x)-\frac12\operatorname{Cl}_2(\pi-2x)\right)\right]_0^\frac\pi4+\frac12\int_0^\frac\pi4\operatorname{Cl}_2(2x)+\operatorname{Cl}_2(\pi-2x)\mathrm dx\\ &=-\frac\pi4\operatorname{Cl}_2\left(\frac\pi2\right)+\frac14\int_0^\frac\pi2\operatorname{Cl}_2(x)\mathrm dx+\frac14\int_0^\frac\pi2\operatorname{Cl}_2(\pi-x)\mathrm dx\\ &=-\frac\pi4G+\frac14\int_0^\pi\operatorname{Cl}_2(x)\mathrm dx\\ &=-\frac\pi4G+\frac14\left[\zeta(3)-\operatorname{Cl}_3(\pi)\right]\\ &=-\frac\pi4G+\frac14\left[\zeta(3)+\eta(3)\right]\\ &=-\frac\pi4G+\frac7{16}\zeta(3) \end{align*}

$$\therefore~\int_0^\frac\pi4x\log(\tan x)\mathrm dx~=~-\frac\pi4G+\frac7{16}\zeta(3)$$

All here used identities can be rather easy deduced from the integral and series representation of the Clausen Function. In my opinion it gives an elegant way of solving such and similiar integrals.


The integral you obtained after the substitution $\tan x\mapsto x$ can be solved using the Inverse Tangent Integral $\operatorname{Ti}_2(z)$, another auxiliary function with roots within the theory of Polylogarithms.

Applying Integration By Parts twice yields to

\begin{align*} I=\int_0^1\frac{\arctan x}{1+x^2}\log x~\mathrm dx&=\underbrace{\left[\frac12\arctan^2x\log x\right]_0^1}_{\to0}-\frac12\int_0^1\arctan x\frac{\arctan x}x\mathrm dx\\ &=-\frac12\left[\operatorname{Ti}_2(x)\arctan x\right]_0^1+\frac12\int_0^1\frac{\operatorname{Ti}_2(x)}{1+x^2}\mathrm dx\\ &=-\frac\pi8G+\frac12\int_0^\frac\pi4\operatorname{Ti}_2(\tan x)\mathrm dx\\ &=-\frac\pi8G+\frac12\int_0^\frac\pi4\sum_{n\ge1}\frac{\sin[(4n-2)x]}{(2n-1)^2}+x\log(\tan x)\mathrm dx\tag{$\star$}\\ &=\frac12I-\frac\pi8G+\frac12\sum_{n\ge1}\frac1{(2n-1)^2}\int_0^\frac\pi4\sin[(4n-2)x]\mathrm dx\\ &=\frac12I-\frac\pi8G+\frac12\sum_{n\ge1}\frac1{(2n-1)^2}\left[\frac{\cos[(4n-2)x]}{4n-2}\right]_0^\frac\pi4\\ &=\frac12I-\frac\pi8G+\frac14\sum_{n\ge0}\frac1{(2n+1)^3}\\ &=\frac12I-\frac\pi8G+\frac14\lambda(3)\\ &=\frac12I-\frac\pi8G+\frac7{32}\zeta(3) \end{align*}

$$\therefore~I~=~\int_0^1\frac{\arctan x}{1+x^2}\log x~\mathrm dx~=~-\frac\pi4G+\frac7{16}\zeta(3)$$

Here $\lambda(s)$ is the Dirichlet Lambda Function. The result used at $(\star)$ is due to Ramanuajan and in his spirit I will omit a proof here. However, there are some steps within this second approach which are, indeed, in need of a more careful argumentation but the purpose of showing this approach is to illustrate the possibilities which may be used in order to evaluate this integral

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  • $\begingroup$ Nice (+1). How does one prove $(2)$? $\endgroup$ – clathratus Apr 29 at 20:49
  • $\begingroup$ @clathratus Utilizing the series representation of the Clausen Function, switching the order of integration and summation and you are basically done. All that remains is to recognise the so gained expression in terms of the Zeta Function and the generalised Clausen Function $/operatorname{Cl}_3(z)$. $\endgroup$ – mrtaurho Apr 29 at 20:54
  • $\begingroup$ Right. Also $$\mathrm{Cl}_{2n+1}'(x)=-\mathrm{Cl}_{2n}(x)$$ and $$\mathrm{Cl}_{2n+1}(0)=\sum_{k\geq1}\frac{\cos0}{k^{2n+1}}=\zeta(2n+1)$$ $\endgroup$ – clathratus Apr 29 at 21:02
  • $\begingroup$ @clathratus Precisely! Those are the more general results. Since there were not needed in this form here I decided to concentrate on the case where $n=2$. $\endgroup$ – mrtaurho Apr 29 at 21:08
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\begin{align}\int_0^{\frac{\pi}{4}}x\ln(\tan x)\,dx=\int_0^{\frac{\pi}{4}}x\ln(\sin x)\,dx-\int_0^{\frac{\pi}{4}}x\ln(\cos x)\,dx\end{align}

and see: https://math.stackexchange.com/a/3200545/186817

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  • $\begingroup$ In the linked answer you compute the RHS (denoted there as $I-J$) exactly as integral $\displaystyle\int_0^{1}\frac{\arctan x\ln x}{1+x^2}dx$, which OP is trying to evaluate. $\endgroup$ – user Apr 29 at 20:08
  • $\begingroup$ Yes I think so. $\endgroup$ – FDP Apr 29 at 20:31
  • $\begingroup$ Does it then make a sense to transform the LHS into RHS? $\endgroup$ – user Apr 29 at 20:37
  • $\begingroup$ Yes it does if the OP reads only first lines and last lines of my answer. $\endgroup$ – FDP Apr 29 at 20:45

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