# Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$

What starting point would you recommend me for the one below?

$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$$

EDIT

Thanks to Felix Marin, we know the integral evaluates to

$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$

• I assume you have already taken $\log\sin x=\displaystyle\int\cot x~dx$ into consideration. Nov 14, 2014 at 13:45
• @Lucian thank you, but unfortunately I didn't think of it. Nov 14, 2014 at 13:50
• $\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$. Nov 26, 2014 at 3:26
• Do you care for an answer with all steps or the result is enough? Nov 26, 2014 at 14:01

\begin{align} \mathcal{I}=&\int^1_0x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\\ =&\left(\int^{1/2}_0+\int_{1/2}^1\right)x\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag1\\ =&\int^{1/2}_0(2x-1)\tan(\pi x)\ln(\sin(\pi x))\ {\rm d}x\tag2\\ =&-\frac{2}{\pi^2}\int^{\pi/2}_0x\cot{x}\ln(\cos{x})\ {\rm d}x\tag3\\ =&-\frac{1}{\pi}\int^{\pi/2}\tan{x}\ln(\sin{x})\ {\rm d}x+\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag4\\ =&\frac{2}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{2}{\pi^2}\int^{\pi/2}_0x\tan{x}\ln(\sin{x})\ {\rm d}x\tag5\\ =&\frac{1}{\pi^2}\int^{\pi/2}_0\ln(\sin{x})\ln(\cos{x})\ {\rm d}x-\frac{1}{2\pi}\int^{\pi/2}_0\tan{x}\ln(\sin{x})\ {\rm d}x\tag6\\ =&\frac{1}{8\pi^2}\frac{\partial^2{\rm B}}{\partial a\partial b}\left(\frac{1}{2},\frac{1}{2}\right)-\frac{1}{8\pi}\frac{\partial{\rm B}}{\partial b}\left(0^+,1\right)\tag7\\ =&\frac{1}{8\pi^2}\left(4\pi\ln^2{2}-\frac{\pi^3}{6}\right)-\frac{1}{8\pi}\left(-\frac{\pi^2}{6}\right)=\boxed{\Large{\color{red}{\dfrac{\ln^2{2}}{2\pi}}}}\\ \end{align}

Explanation:

$(1)$: Split the integral at $\displaystyle\frac12$.

$(2)$: Substituted $\displaystyle x\mapsto1-x$ in the second integral.

$(3)$: Substituted $\displaystyle x\mapsto\frac{1}{2}-\frac{x}{\pi}$.

$(4)$: Substituted $\displaystyle x\mapsto\frac{\pi}{2}-x$.

$(5)$: Integrated by parts.

$(6)$: Took the average of $(4)$ and $(5)$.

$(7)$: $\displaystyle {\rm B}(a,b)=2\int^{\pi/2}_0\sin^{2a-1}{x}\cos^{2b-1}{x}\ {\rm d}x$

• A very nice job (+1) Nov 26, 2014 at 16:30
• @Chris's sis Thank you. Nov 26, 2014 at 16:34
• +1. It's a very clever answer. I was struggling with it but the annoying $\large x$ factor defeated me. Nov 30, 2014 at 2:06
• @Jack D'Aurizio & Felix Marin, Thank you for your kind words. Nov 30, 2014 at 7:48

Integrating by parts, we get \begin{align} &\int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\\ &=\int_0^{1/2}(2x-1)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1a} \\ &=-\frac1\pi\int_0^{1/2}(2x-1)\log(\sin(\pi x))\,\mathrm{d}\log(\cos(\pi x))\tag{1b}\\ &=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x +\int_0^{1/2}(2x-1)\cot(\pi x)\log(\cos(\pi x))\,\mathrm{d}x\tag{1c}\\ &=\small\frac2\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x +\int_0^{1/2}(-2x)\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1d}\\ &=\small\frac1\pi\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x -\frac12\int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x\tag{1e}\\ \end{align} Explanation:
$\text{(1a)}$: Subtract $\frac12\int_0^1\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x=0$ and use the symmetry of $x\mapsto1-x$
$\text{(1b)}$: Apply $\mathrm{d}\log(\cos(\pi x))=\pi\tan(\pi x)\,\mathrm{d}x$
$\text{(1c)}$: Integrate by Parts
$\text{(1d)}$: Substitute $x\mapsto\frac12-x$
$\text{(1e)}$: Average $\text{(1a)}$ and $\text{(1d)}$

Using $$\log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{2}$$ $$\log(\cos(x))=-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\tag{3}$$ we get \begin{align} &\int_0^{1/2}\log(\sin(\pi x))\log(\cos(\pi x))\,\mathrm{d}x\\ &=\color{#C00000}{\int_0^{1/2}(-\log(2))^2\,\mathrm{d}x}\\ &-\color{#00A000}{\log(2)\int_0^{1/2}\sum_{k=1}^\infty\left[-1-(-1)^k\right]\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\ &+\color{#0000FF}{\int_0^{1/2}\sum_{j=1}^\infty\sum_{k=1}^\infty(-1)^k\frac{\cos(2\pi jx)}{j}\frac{\cos(2\pi kx)}{k}\,\mathrm{d}x}\\ &=\color{#C00000}{\frac{\log(2)^2}2}-\color{#00AA00}{0}+\color{#0000FF}{\frac14\sum_{k=1}^\infty\frac{(-1)^k}{k^2}}\\ &=\frac{\log(2)^2}2-\frac{\pi^2}{48}\tag{4} \end{align} since $\int_0^{1/2}\cos(2\pi jx)\cos(2\pi kx)\,\mathrm{d}x=0$ when $j\ne k$.

Substituting $t=\sin(\pi x)$ and $e^{-u/2}=t$, we get \begin{align} \int_0^{1/2}\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x &=\frac1\pi\int_0^1\frac{t}{1-t^2}\log(t)\,\mathrm{d}t\\ &=-\frac1{4\pi}\int_0^\infty\frac{e^{-u}}{1-e^{-u}}u\,\mathrm{d}u\\ &=-\frac1{4\pi}\int_0^\infty\sum_{k=1}^\infty ue^{-ku}\,\mathrm{d}u\\ &=-\frac1{4\pi}\sum_{k=1}^\infty\frac1{k^2}\\ &=-\frac{\pi}{24}\tag{6} \end{align} Combining $(1)$, $(4)$, and $(6)$ yields $$\int_0^1x\tan(\pi x)\log(\sin(\pi x))\,\mathrm{d}x =\frac{\log(2)^2}{2\pi}\tag{7}$$

• Could you elaborate $(4)$? It hasn't seemed clear yet to me. Thanks. Nov 27, 2014 at 17:14
• @Venus It can be shown using the calculus of residues Nov 27, 2014 at 17:17
• @ZubinMukerjee: it is simpler than that. It only needs $(2)$ and $(3)$. I will update the answer to show this.
– robjohn
Nov 27, 2014 at 17:18
• @Venus: I have expanded $(4)$. I hope that helps.
– robjohn
Nov 27, 2014 at 19:56
• @JackD'Aurizio: Thanks! Formulas $(2)$ and $(3)$ have proven very useful. I didn't even think of them as Fourier Series since I usually derive them using $\log\left(\frac{e^{ix}+e^{-ix}}2\right)$ and $\log\left(\frac{e^{ix}-e^{-ix}}{2i}\right)$; but as you point out, they are Fourier Series.
– robjohn
Nov 30, 2014 at 20:15

$\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have $$I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy$$ Perform integration by parts by taking $u=y$, we have \begin{align} I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\right|_0^\pi+\frac{1}{2\pi^2}\int_0^\pi\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\,dy\\ &=\left.\frac{y}{4\pi^2}\Li\left(\cos^2y\right)\,\right|_0^\pi-\frac{1}{4\pi^2}\int_0^\pi\Li\left(\cos^2y\right)\,dy\\ &=\frac{\Li\left(1\right)}{4\pi}-\frac{1}{4\pi^2}\int_0^{\pi}\sum_{k=1}^\infty\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{use series representation of dilogarithm}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\int_0^{\pi}\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{justified by Fubini-Tonelli theorem}}\\ &=\frac{\pi}{24}-\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^{\pi/2}\cos^{2k}y\,\,dy\quad\Rightarrow\quad\color{red}{\mbox{by symmetry argument}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{k^2\,\Gamma\left(k+1\right)}\quad\Rightarrow\quad\color{red}{\mbox{Wallis' integrals}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi}\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k^2\,(k!)^2}\tag1 \end{align} $\def\arctanh{{\rm{\,arctanh}\,}}$Here is the tedious part (and also the hardest part). I use Mathematica to help me out to find generating function of the following series. Let us start with $$\sum_{k=0}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)x^k}{\Gamma\left(k+1\right)}=\pi\sum_{k=0}^\infty\frac{(2k)!\,\,x^k}{4^k\,(k!)^2}=\frac{\pi}{\sqrt{1-x}}$$ Divide by $x$ and then integrate it, we have $$\sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k\,(k!)^2}=\int\left[\frac{1}{x\,\sqrt{1-x}}-\frac{1}{x}\right]\,dx=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+C_1$$ Taking the limit as $x\to0$, we obtain $$C_1=\lim_{x\to0}\left(2\arctanh\left(\sqrt{1-x}\,\right)+\ln x\right)=\ln4$$ Hence $$\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k\,(k!)^2}x^k=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+\ln4$$ Repeat the process once more, we obtain \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}=&\,-2\int\frac{\arctanh\left(\sqrt{1-x}\,\right)}{x}\,dx-\int\frac{\ln x}{x}\,dx+\ln4\int\frac{dx}{x}\\ =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x+C_2\\ \end{align} Taking the limit as $x\to0$, we obtain $$C_2=-2\ln^22$$ Hence \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2} =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x-2\ln^22\tag2 \end{align} Thus, by putting $x=1$ to $(2)$ then $(1)$ becomes $$I=\frac{\pi}{24}-\frac{1}{4\pi}\left(\frac{\pi^2}{6}-2\ln^2 2\right)=\frac{\ln^22}{2\pi}$$ and we are done.

• @Chris'ssis Thanks. I'm still thinking for the evaluation of series. Nov 26, 2014 at 20:20
• +1 You could start with $$\frac{1}{\sqrt{1-x^2}}=\sum^\infty_{k=0}\frac{1}{2^{2k}}\binom{2k}{k}x^{2k}$$ Divide by $x$ and integrate and repeat the process again. Nov 27, 2014 at 3:18
• @M.N.C.E. It doesn't work. I have tried it at school although without your advice I've thought about it. Thanks anyway. Or maybe I don't see how to use it, feel free to use my answer. I don't mine. $\ddot\smile$ Nov 27, 2014 at 16:38
• @Anastasiya-Romanova I wish I'll see you soon! Dec 2, 2014 at 12:52
• Credit where credit is due! Good speed! :-) Dec 2, 2014 at 13:31

Let $I = \int_0^1 x \tan(\pi x) \log(\sin(\pi x)) dx$. We have $$I= \int_0^1 \dfrac{x \tan(\pi x)}2 \log(1-\cos^2(\pi x))dx$$ Hence, $$2I = -\sum_{k=1}^{\infty}\dfrac1k\int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx \,\,\,\,\,\,\,\, \spadesuit$$ \begin{align} \int_0^1 x \tan(\pi x) \cos^{2k}(\pi x)dx & = \int_0^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\ & = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx + \int_{1/2}^1 x\sin(\pi x) \cos^{2k-1}(\pi x)dx\\ & = \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2} (1-x)\sin(\pi x) \cos^{2k-1}(\pi x)dx\\ & = 2 \int_0^{1/2} x\sin(\pi x) \cos^{2k-1}(\pi x)dx - \int_0^{1/2}\sin(\pi x) \cos^{2k-1}(\pi x)dx\\ & = \frac2{\pi^2}\int_0^1 t^{2k-1} \arccos(t) dt - \frac1{\pi}\int_0^1 t^{2k-1} dt\\ & = \frac1{2\pi^{3/2}} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} - \frac1{2 \pi k} \end{align} Use this in $\spadesuit$ and the approrpriate Taylor series will give you the answer. I am away now since I have a class to teach. Will add more details if needed sometime later.

Updated

We have $$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}2 \dfrac1{(1-x)^{3/2}}$$ Integrate the above once and divide by $x$ to obtain $$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} + \frac{c_1}x$$ To obtain $c_1$, take $\lim_{x \to 0}$, the left hand side is $\Gamma(1/2)$. For the right hand side limit to even exist, we need $c_1 = -\sqrt{\pi}$. Hence, $$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k\Gamma(k)} x^{k-1} = \dfrac{\sqrt{\pi}}{x(1-x)^{1/2}} - \frac{\sqrt{\pi}}x$$ Integrate again and divide by $x$ to obtain $$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \sqrt{\pi} \left(\dfrac{\log\left(1-\sqrt{1-x} \right) - \log\left(1+\sqrt{1-x} \right)}x \right) - \sqrt{\pi}\frac{\log(x)}x + \frac{c_2}x$$ To obtain $c_2$, take limit $x \to 0$. For the right hand side limit to even exist, we need $c_2 = 2\sqrt{\pi} \log(2)$. Hence, $$\sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^2\Gamma(k)} x^{k-1} = \dfrac{2\sqrt{\pi}\log2-2\sqrt{\pi} \log\left(1+\sqrt{1-x}\right)}{x}$$ Integrate again to obtain \begin{align} \sum_{k=1}^{\infty} \dfrac{\Gamma(k+1/2)}{k^3\Gamma(k)} x^{k} = \frac{\sqrt{\pi}}{2} \left(2\text{Li}_2(1/2(\sqrt{1-x}-1)) - 2\text{Li}_2(1/2(\sqrt{1-x}+1))\\ + \left(\log(1-\sqrt{1-x}) - \log(1+\sqrt{1-x}) \right)\left(\log(1-\sqrt{1-x}) + \log(1+\sqrt{1-x}) -2\log2\right)\right) - \sqrt{\pi} \frac{\log^2(x)}2 + 2\sqrt{\pi} \log2 \log(x) + c_3 \end{align} Now plug in $x=1$ in the above to get the sum of the first series and the sum of the second series is $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\pi^2}6$$

I shall add more details if needed.

• @Adhvaitha Why don't you post the whole solution, showing the simplicity of evaluating $$\sum_{k=1}^{\infty} (-1)^k \frac{\Gamma(k+1/2)}{k^3 \Gamma(k)}$$? Then I guess there will be crowds trying to upvote the answer. Nov 26, 2014 at 22:07
• @Chris'ssis Here's a one line proof: "the integral is trivially equal to $\frac{\sqrt{\pi}}{6}(\pi^2-12\log^2(2))$." I think I won.
– Pedro
Nov 26, 2014 at 23:32
• @PedroTamaroff young Tamarof, you may also note in an easier way that by letting $e^{-x}\mapsto x$, we get that $$\int_0^{\infty}\frac{x^2 e^{-x}}{(1-e^{-x})^{3/2}}\ dx=\int_0^1 \frac{\log^2(x)}{(1-x)^{3/2}} \ dx$$ and that clearly shows the last integral can be tackled by beta function. Basically it can be approached in lots of ways. Nov 26, 2014 at 23:59
• @Chris'ssis Yes, it is trivial that is trivial.
– Pedro
Nov 27, 2014 at 0:02
• Let an empty space stand for "trivial". Then the following is the shortest proof I have been able to produce: ${}{}{}{}$
– Pedro
Nov 27, 2014 at 0:06

We can get the serials Ana calculated.

• Dec 1, 2014 at 10:44
• @姚改成 Good job (+1) Dec 1, 2014 at 10:50
• Ok,I will learn it,thank you.@Shaun Dec 1, 2014 at 12:19
• Nice answer! +1 Dec 2, 2014 at 6:28