Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$

Question

Can anyone help me to find $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Any help would be appreciated. Thanks in advance.

Answer 1

Rewrite $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\ dx. \end{align} $$ Let $\,u=\cos x$, then $\,du=-\sin x\,dx$. For $\,0 < x < \frac{\pi}{2}$, we have $\,0 < u < 1$. Now, the integral turns out to be $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\,dx&=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sqrt{1-\cos^2 x})\,dx\\ &=-\frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du.\tag1\\ \end{align} $$ Next, use Maclaurin series for natural logarithm: $$ \ln(1-u^2)=-\sum_{n=1}^\infty \frac{u^{2n}}{n}.\tag2\\ $$ Substitute $\,(2)$ to $\,(1)$, yield $$ \begin{align} \frac{1}{2}\int_0^1\frac{\ln(1-u^2)}{u}\,du&=-\frac{1}{2}\int_0^1\sum_{n=1}^\infty \frac{u^{2n}}{un}\,du\\ &=-\frac{1}{2}\sum_{n=1}^\infty\int_0^1 \frac{u^{2n-1}}{n}\,du\\ &=-\frac{1}{4}\sum_{n=1}^\infty \left.\frac{u^{2n}}{n^2}\right|_{u=0}^1\\ &=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}.\tag3 \end{align} $$ The infinite series in $(3)$ is defined as Riemann zeta function $\,\zeta (2)=\dfrac{\pi^2}{6}$. Thus, $$ \begin{align} \int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx&=-\frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2}\\ &=-\frac{1}{4}\cdot \frac{\pi^2}{6}\\ &=\large\color{blue}{-\frac{\pi^2}{24}}.\\ \end{align} $$

Answer 2

$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)log(\sin(x))}{\cos(x)}dx$$

Let $t=\cos(x), \;\ \sin(x)=\sqrt{1-t^{2}}, \;\ dx=\frac{-1}{\sqrt{1-t^{2}}}dt$

$$1/2\int_{0}^{1}\frac{log(1-t^{2})}{t}dt$$

This integral can be done using the series for $$log(1-t^{2})=-\sum_{k=1}^{\infty}\frac{t^{2k}}{k}$$, and is rather famous.

But, it evaluates to $$\frac{-\pi^{2}}{24}$$

Answer 3

Subst. $t=\sin{x}$, then the given integral becomes \begin{align*} \int_0^1 \, \frac{t\, \log{t}}{1-t^2} dt &= \int_{0}^{1} \, \log{t} \, \sum_{k\ge 0} t^{2k+1} \, dt\\ &= \sum_{k\ge 0} \int_{0}^{1} \, \left(\log{t}\right)\, t^{2k+1}\, dt\\ &= \sum_{k\ge 0} -\frac{1}{4 \, {\left(k^{2} + 2 \, k + 1\right)}}\\ &= -\frac{1}{4} \zeta{(2)}\\ &= -\frac{\pi^2}{24} \end{align*}

In general, we can have \begin{align*} \int_{0}^{\pi/2} \, \left(\log{\sin{x}}\right)^n\, \tan{x}\, dx = (-1)^n\, \frac{ n!\, \zeta(n + 1)}{2^{n + 1}} \end{align*}

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\overbrace{\color{#00f}{\large\int_{0}^{\pi/2}\tan\pars{x}\ln\pars{\sin\pars{x}}\,\dd x}} ^{\ds{t\ \equiv \sin\pars{x}}}\ =\ \overbrace{\int_{0}^{1}{t\ln\pars{t} \over 1 - t^{2}}\,\dd t} ^{\ds{t\ \equiv \expo{-\xi}}}\ =\ -\int_{0}^{\infty}{\xi\expo{-2\xi} \over 1 - \expo{-2\xi}}\,\dd\xi \\[3mm]&= -\,{1 \over 4}\int_{0}^{\infty}{\xi\expo{-\xi} \over 1 - \expo{-\xi}}\,\dd\xi =\,{1 \over 4}\int_{0}^{\infty}\ln\pars{1 - \expo{-\xi}}\,\dd\xi ={1 \over 4}\int_{0}^{\infty} \pars{-\sum_{n = 1}^{\infty}{\expo{-n\xi} \over n}}\,\dd\xi \\[3mm]&=-\,{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n}\ \underbrace{\int_{0}^{\infty}\expo{-n\xi}\,\dd\xi}_{\ds{=\ {1 \over n}}} =-\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{=\ {\pi^{2} \over 6}}} = \color{#00f}{\Large -\,{\pi^{2} \over 24}} \end{align}

Answer 5

Here is one line proof using beta function

$$\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx=\lim_{a\rightarrow 1,b\rightarrow 1^+}\frac{1}{2}\frac{\partial B}{\partial a}\left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)=-\frac{\pi^2}{24}$$

Q.E.D.

Answer 6

I just finished this answer and noticed that equation $(6)$ there answers this question. I did a small amount of customization for this question.

Let $t=\sin(x)$ and $e^{-u/2}=t$ $$ \begin{align} \int_0^{\pi/2}\tan(x)\log(\sin(x))\,\mathrm{d}x &=\int_0^1\frac{t}{1-t^2}\log(t)\,\mathrm{d}t\\ &=-\frac14\int_0^\infty\frac{e^{-u}}{1-e^{-u}}u\,\mathrm{d}u\\ &=-\frac14\int_0^\infty\sum_{k=1}^\infty ue^{-ku}\,\mathrm{d}u\\ &=-\frac14\sum_{k=1}^\infty\frac1{k^2}\\ &=-\frac{\pi^2}{24} \end{align} $$ This is only slightly different from Felix Marin's answer, using the series for $\dfrac{x}{1-x}$ instead of $\log(1-x)$.

Answer 7

Let $$ I(a,b)=\int_0^{\pi/2}(\sin x)^{1-a}(\cos x)^{1-b}dx.$$ Then $$ I(a,b)=\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right)}{2 \Gamma \left(-\frac{a}{2}-\frac{b}{2}+2\right)} $$ and hece \begin{eqnarray} \frac{\partial I(a,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(1-\frac{b}{2}\right) \left(\psi ^{(0)}\left(-\frac{a}{2}-\frac{b}{2}+2\right)-\psi ^{(0)}\left(1-\frac{a}{2}\right)\right)}{4 \Gamma \left(-\frac{a}{2}-\frac{b}{2}+2\right)}. \end{eqnarray} Thus \begin{eqnarray} \frac{\partial I(0,b)}{\partial a}&=&\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi ^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma \left(2-\frac{b}{2}\right)}. \end{eqnarray} Noting $$ \Gamma(x)\approx\frac{1}{x} \text{ near }x=0 $$ we have \begin{eqnarray} \lim_{b\to2}\frac{\partial I(0,b)}{\partial a}&=&\lim_{b\to2}\frac{\Gamma \left(1-\frac{b}{2}\right) \left(\psi ^{(0)}\left(2-\frac{b}{2}\right)+\gamma \right)}{4 \Gamma \left(2-\frac{b}{2}\right)}\\ &=&\lim_{b\to2}\frac14 \frac{\psi^{(0)}(2-\frac{b}{2})+\gamma}{1-\frac{b}{2}}\\ &=&\lim_{b\to2}\frac14 \psi^{(1)}(2-\frac{b}{2})\\ &=&\frac{\pi^2}{24}. \end{eqnarray} So $$ \int_0^{\pi/2}\tan x\ln(\sin x)dx=-\frac{\pi^2}{24}. $$