Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$

Question

I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$ First,I tried to evaluate like this: $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$

$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$

$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$ $$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$ but I can't proceed next step,help me,thanks.

Answer 1

At the price of special functions, the antiderivative could be computed $$I=\int\frac{x^2}{ \sin x}\,dx=-4 i x \text{Li}_2\left(e^{i x}\right)+i x \text{Li}_2\left(e^{2 i x}\right)+4 \text{Li}_3\left(e^{i x}\right)-\frac{1}{2} \text{Li}_3\left(e^{2 i x}\right)-2 x^2 \tanh ^{-1}\left(e^{i x}\right)$$ where appear the polylogarithm functions.

$$\lim_{x\to \frac{\pi }{2}} \, I=2 \pi C\qquad \text{and} \qquad\lim_{x\to 0} \, I=\frac{7 }{2}\zeta (3)\implies \int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=2 \pi C-\frac{7 }{2}\zeta (3)$$ as given by Wolfram Alpha. This evaluates a $\approx 1.54798$.

For a fast approximation, we could use the superb approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (have a look here). This would make $$I \approx J= - \int \left(\frac{x^2}{4}+\frac{5 \pi ^3}{16 (x-\pi )}+\frac{5 \pi ^2}{16} \right)\,dx=-\frac{x^3}{12}-\frac{5 \pi ^2 x}{16}-\frac{5}{16} \pi ^3 \log (\pi -x)+\frac{19 \pi ^3}{48}$$ $$\lim_{x\to \frac{\pi }{2}} \, J=\frac{\pi ^3}{48} \left(11-15 \log \left(\frac{\pi }{2}\right)\right)\qquad \text{and} \qquad\lim_{x\to 0} \, J =\frac{\pi ^3}{48} (19-15 \log (\pi ))$$ leading to the approximation $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx\approx \frac{\pi ^3}{48} (15 \log (2)-8)\approx 1.54851$$ which is not too bad.

Tha advantage of such approximation is that it allows a fast evaluation of $$K(t)=\int_{0}^{t}\frac{x^2}{ \sin x}dx$$ The table below compares the approximation to the exact result $$\left( \begin{array}{ccc} t & \text{approximation} & \text{exact} \\ \frac{\pi }{20} & 0.01221 & 0.01236 \\ \frac{\pi }{10} & 0.04936 & 0.04976 \\ \frac{3 \pi }{20} & 0.11258 & 0.11312 \\ \frac{\pi }{5} & 0.20358 & 0.20409 \\ \frac{\pi }{4} & 0.32475 & 0.32508 \\ \frac{3 \pi }{10} & 0.47939 & 0.47945 \\ \frac{7 \pi }{20} & 0.67196 & 0.67176 \\ \frac{2 \pi }{5} & 0.90847 & 0.90807 \\ \frac{9 \pi }{20} & 1.19701 & 1.19650 \\ \frac{\pi }{2} & 1.54851 & 1.54798 \\ \frac{11 \pi }{20} & 1.97802 & 1.97746 \\ \frac{3 \pi }{5} & 2.50657 & 2.50583 \\ \frac{13 \pi }{20} & 3.16447 & 3.16315 \\ \frac{7 \pi }{10} & 3.99696 & 3.99445 \\ \frac{3 \pi }{4} & 5.07529 & 5.07091 \\ \frac{4 \pi }{5} & 6.52008 & 6.51359 \\ \frac{17 \pi }{20} & 8.55922 & 8.55230 \\ \frac{9 \pi }{10} & 11.7067 & 11.7077 \\ \frac{19 \pi }{20} & 17.6067 & 17.6510 \end{array} \right)$$

Answer 2

Observe we have \begin{align} I=\int^{\pi/2}_0 \frac{x^2}{\sin x}\ dx = \int^{\pi/2}_0 \frac{x^2}{\cos\left(\frac{\pi}{2}-x \right)}\ dx = \int^{\pi/2}_0 \frac{(\frac{\pi}{2}-u)^2}{\cos u}\ du. \end{align} Then using integration by parts, we see that \begin{align} I&=\left(\frac{\pi}{2}-u\right)^2\left\{\log\left|1 + \sin u\right|-\log|\cos u|\right\}\bigg|^{\pi/2}_0 + 2\int^{\pi/2}_0\left(\frac{\pi}{2}-u \right)\log|\sec u + \tan u|\ du\\ &= 2\pi \left(\frac{1}{2}\int_{0}^{\pi/2}\log|\sec u+\tan u|\ du \right)-\frac{7}{2}\left(\frac{4}{7}\int^{\pi/2}_0 u \log|\sec u+\tan u|\ du \right)\\ &= 2\pi G - \frac{7}{2}\zeta(3). \end{align}

Here, I have used the facts that \begin{align} G= \frac{1}{2}\int_{0}^{\pi/2}\log|\sec u+\tan u|\ du \end{align} and \begin{align} \zeta(3) = \frac{4}{7}\int^{\pi/2}_0 u \log|\sec u+\tan u|\ du. \end{align} See here for reference.

Answer 3

$$\mathcal{J}=\int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = \int_{0}^{1}\frac{\arcsin^2(x)}{x\sqrt{1-x^2}}\,dx=\sum_{n\geq 1}\frac{2^{2n-1}}{n^2\binom{2n}{n}}\int_{0}^{1}\frac{x^{2n-1}}{\sqrt{1-x^2}}\,dx \tag{1}$$ by the Maclaurin series of $\arcsin^2(x)$. Euler's Beta function then leads to $$ \mathcal{J}=\sum_{n\geq 1}\frac{16^n}{4n^3 \binom{2n}{n}^2}=\phantom{}_4 F_3\left(1,1,1,1;\tfrac{3}{2},\tfrac{3}{2},2;1\right)\tag{2} $$ where the RHS is a manageable hypergeometric function (similar objects are evaluated both here and here) and as already shown by Claude Leibovici, $\mathcal{J}=4\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du $ is simply given by a combination of a dilogarithm and a trilogarithm. Indeed $$ \int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta =-\pi G+\frac{7}{2}\zeta(3)\tag{3}$$ leading to $\mathcal{J}=2\pi G-\frac{7}{2}\zeta(3)$, has already been a key lemma in this historical thread.
An alternative way for proving this identity is just to write $\frac{x}{\sin x}$ and $|x|$ as Fourier cosine series.
The Shafer-Fink inequality leads to $$ \int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = 4 \int_{0}^{1}\frac{\arctan^2(u)}{u}\,du \approx \frac{6}{7}(3\sqrt{2}-5)+9\log\left(\frac{2\sqrt{2}+1}{3}\right)\approx 1.54.\tag{4}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x & = \left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\bracks{-\ic\ln\pars{z}}^{2} \over \pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] & = \left.2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \end{align}

$\ds{\ln}$ is the $\ds{\log}$-principal branch. Integration of $\ds{{\ln^{2}\pars{z} \over 1 - z^{2}}}$ along the path $\ds{C_{x}\cup C_{R}\cup C_{y}}$ vanishes out such that

$\ds{\int_{\large C_{R}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z = -\int_{\large C_{y}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z - \int_{\large C_{x}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z}$

Then, \begin{align} \int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x &\ =\ \overbrace{-2\,\Re\int_{1}^{0}{\bracks{\ln\pars{y} + \pi\ic/2}^{\, 2} \over 1 + y^{2}}\,\ic\,\dd y}^{\ds{\mbox{along}\ C_{y}}}\ -\ \overbrace{2\,\Re\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x} ^{\ds{\mbox{along}\ C_{x}}} \\[5mm] & = -2\pi\,\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y - 2\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x \end{align}

However, $\ds{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y = -G}$ where $\ds{G}$ is the Catalan Constant such that

\begin{align} \int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x & = 2\pi G - 2\sum_{n = 0}^{\infty}\ \overbrace{\int_{0}^{1}\ln^{2}\pars{x}x^{2n}\,\dd x} ^{\ds{2 \over \pars{2n + 1}^{3}}} \\[5mm] & = 2\pi G - 4\bracks{\sum_{n = 1}^{\infty}{1 \over n^{3}} - \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}} = 2\pi G - {7 \over 2}\sum_{n = 1}^{\infty}{1 \over n^{3}} \\[5mm] & = \bbx{2\pi G - {7 \over 2}\,\zeta\pars{3}} \approx 1.5480 \\ & \end{align}

Answer 5

$\begin{align} J&=\int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx& \end{align}$

Perform the change of variable,

$\displaystyle y=\tan\left(\frac{x}{2}\right)$,

$\begin{align} J&=4\int_0^1 \frac{\arctan^2 x}{x}\,dx\\ &=4\Big[\ln x\arctan ^2 x\Big]_0^1-8\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ &=-8\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx\\ \end{align}$

For $x\in [0;1]$, define $F$,

$\begin{align} F(x)&=\int_0^x \frac{\ln t}{1+t^2}\,dt\\ &=\int_0^1 \frac{x\ln(xt)}{1+x^2t^2}\,dt \end{align}$

Observe that,

$\displaystyle F(0)=0$ and, $\displaystyle F(1)=-\text{G}$.

$\text{G}$ is the Catalan constant.

$\begin{align}J&=-8\Big[F(x)\arctan x\Big]_0^1+8\int_0^1 \int_0^1 \frac{x\ln(tx)}{(1+t^2x^2)(1+x^2)}\,dt\,dx\\ &=2G\pi+8\int_0^1 \int_0^1 \frac{x\ln x}{(1+t^2x^2)(1+x^2)}\,dt\,dx+8\int_0^1 \int_0^1 \frac{x\ln t}{(1+t^2x^2)(1+x^2)}\,dt\,dx\\ &=2G\pi+8\int_0^1 \Big[\frac{\arctan(tx)\ln x}{1+x^2}\Big]_{t=0}^{t=1}\,dx+4\int_0^1 \Big[\frac{(\ln(1+t^2x^2)-\ln(1+x^2))\ln t}{t^2-1}\Big]_{x=0}^{x=1}\,dt\\ &=2G\pi+8\int_0^1 \frac{\arctan x\ln x}{1+x^2}\,dx+4\int_0^1 \frac{(\ln(1+t^2)-\ln 2)\ln t }{t^2-1}\,dt\\ &=2G\pi-J+4\int_0^1 \frac{(\ln 2-\ln(1+t^2))\ln t }{1-t^2}\,dt\\ \end{align}$

Therefore,

$\displaystyle J=\text{G}\pi+2\int_0^1 \frac{(\ln 2-\ln(1+x^2))\ln x }{1-x^2}\,dx$

For $x\in[0;1]$, define,

$\begin{align}H(x)&=\int_0^x \frac{\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt\\ \end{align}$

Observe that,

$\displaystyle H(0)=0$ and $\displaystyle H(1)=-\frac{\pi^2}{8}$.

$\begin{align}J&=\text{G}\pi+2\Big[(\ln 2-\ln(1+x^2))H(x)\Big]_0^1+4\int_0^1\int_0^1\frac{x^2\ln(tx)}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=\text{G}\pi+4\int_0^1\int_0^1\frac{x^2\ln(tx)}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=\text{G}\pi+4\int_0^1\int_0^1\frac{x^2\ln t}{(1+x^2)(1-t^2x^2)}\,dt\,dx+4\int_0^1\int_0^1\frac{x^2\ln x}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=\text{G}\pi+4\int_0^1\Big[\frac{\ln t}{1+t^2}\left(\frac{\ln(1+tx)}{2t}-\frac{\ln(1-tx)}{2t}-\arctan x\right)\Big]_{x=0}^{x=1}\,dt+\\ &2\int_0^1 \Big[\frac{x\ln x}{1+x^2}\ln\left(\frac{1+tx}{1-tx}\right)\Big]_{t=0}^{t=1}\,dx\\ &=\text{G}\pi+2\int_0^1 \frac{\ln t}{t(1+t^2)}\ln\left(\frac{1+t}{1-t}\right)\,dt-\pi\int_0^1 \frac{\ln t}{1+t^2}\,dt+2\int_0^1 \frac{x\ln x}{1+x^2}\ln\left(\frac{1+x}{1-x}\right)\,dx\\ &=2\text{G}\pi+2\int_0^1 \frac{\ln x}{x}\ln\left(\frac{1+x}{1-x}\right)\,dx\\ \end{align}$

But, for $0\leq x<1$,

$\displaystyle \frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)=2\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}$

Therefore,

$\begin{align}\int_0^1 \frac{\ln x}{x}\ln\left(\frac{1+x}{1-x}\right)\,dx&=2\int_0^1 \left(\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}\right)\ln x\,dx\\ &=2 \sum_{n=0}^{\infty}\int_0^1 \frac{x^{2n}\ln x}{2n+1}\,dx\\ &=-2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^3}\\ &=-2\left(\sum_{n=1}^{\infty} \frac{1}{n^3}-\sum_{n=1}^{\infty} \frac{1}{(2n)^3}\right)\\ &=-2\left(\zeta(3)-\frac{1}{8}\zeta(3)\right)\\ &=-\frac{7}{4}\zeta(3)\\ \end{align}$

Therefore,

$ \boxed{J=2\text{G}\pi-\frac{7}{2}\zeta(3)}$

Answer 6

As pointed out within the other answers we want to prove that

$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=2\pi G-\frac72\zeta(3)$$

As the OP showed $\mathfrak{I}$ can be reduced to a linear combination of $x$ and the function $\log(\cot x)$

$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=8\int_0^{\pi/4}x\log(\cot x)~dx$$

By applying the definition of the cotangent function followed up by the usage of the well-known Fourier series expansions of $\log(\cos x)$ and $\log(\sin x)$ this can be further simplified. Therefore we get

$$\small\begin{align} \mathfrak{I}=8\int_0^{\pi/4}x\log(\cot x)~dx&=8\left[\int_0^{\pi/4}x\log(\cos x)~dx-\int_0^{\pi/4}x\log(\sin x)~dx\right]\\ &=8\left[\int_0^{\pi/4}x\left(-\log(2)-\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}\right)~dx-\int_0^{\pi/4}x\left(-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right)~dx\right]\\ &=8\left[-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\underbrace{\int_0^{\pi/4}x\cos(2nx)~dx}_I+\sum_{n=1}^{\infty}\frac{1}{n}\underbrace{\int_0^{\pi/4}x\cos(2nx)~dx}_I\right]\\ \end{align}$$

The inner integral $I$ can be easily evaluated using IBP which leads to

$$I=\int_0^{\pi/4}x\cos(2nx)~dx=\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}$$

Plugging this into our original formula and followed by a little bit of algebraic manipulation we get

$$\small\begin{align} \mathfrak{I}&=8\left[-\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left(\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}\right)+\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{\pi}{8n}\sin\left(n\frac{\pi}2\right)-\frac1{(2n)^2}\right)\right]\\ &=2\left[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}-\sum_{n=1}^{\infty}\frac{1}{n^3}\right]+\pi\left[\sum_{n=1}^{\infty}\frac{1}{n^2}\sin\left(n\frac{\pi}2\right)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\sin\left(n\frac{\pi}2\right)\right] \end{align}$$

The first terms can be evaluated in terms of the Riemann Zeta Function $\zeta(s)$ and the Dirichlet Eta Function $\eta(s)$ whereas for the second term we have to consider some basic properties of the Sine function. For $n\in\mathbb{N}>0$ the function $\sin\left(n\frac{\pi}2\right)$ will be zero for all even $n$ and $-1$ and $1$ respectively for odd $n$ starting with $\sin\left(\frac{\pi}2\right)=1$ for $n=1$. Therefore all even terms vanish while the odd ones will remain with a oscillating negative sign. This leads to

$$\small\begin{align} \mathfrak{I}&=2\left[\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}-\sum_{n=1}^{\infty}\frac{1}{n^3}\right]+\pi\left[\sum_{n=1}^{\infty}\frac{1}{n^2}\sin\left(n\frac{\pi}2\right)-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\sin\left(n\frac{\pi}2\right)\right]\\ &=2[-\eta(3)-\zeta(3)]+\pi\left[\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}(-1)^n-\sum_{n=0}^{\infty}\frac{(-1)^{2n+1}}{(2n+1)^2}(-1)^n\right]\\ &=-2[(1-2^{-2})\zeta(3)+\zeta(3)]+2\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\\ \Leftrightarrow\mathfrak{I}&=-\frac72\zeta(3)+2\pi G \end{align}$$

where within the last step the functional relation between the Riemann Zeta Function and the Dirichlet Eta Function aswell as the series defintions of Catalan's Constant $G$ where used.

Answer 7

We can adapt the formula derived in $(2)$ of this answer: $$ \log(2\cos(x/2))=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(kx)\tag{1a} $$ Substituting $x\mapsto\pi-x$ in $\text{(1a)}$, we get $$ \log(2\sin(x/2))=\sum_{k=1}^\infty\frac{-1}k\cos(kx)\tag{1b} $$ Subtracting $\text{(1a)}$ from $\text{(1b)}$, the even terms cancel and we get $$ \bbox[5px,border:2px solid #C0A000]{\log(\tan(x/2))=\sum_{k=0}^\infty\frac{-2}{2k+1}\cos((2k+1)x)}\tag2 $$

---

Therefore, $$ \begin{align} \int_0^{\pi/2}\frac{x^2}{\sin(x)}\,\mathrm{d}x &=\int_0^{\pi/2}x^2\,\mathrm{d}\log(\tan(x/2))\tag3\\ &=-2\int_0^{\pi/2}x\log(\tan(x/2))\,\mathrm{d}x\tag4\\ &=\sum_{k=0}^\infty\frac4{2k+1}\int_0^{\pi/2}x\cos((2k+1)x)\,\mathrm{d}x\tag5\\ &=\sum_{k=0}^\infty\frac4{(2k+1)^2}\int_0^{\pi/2}x\,\mathrm{d}\sin((2k+1)x)\tag6\\ &=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[x\sin((2k+1)x)+\frac{\cos((2k+1)x)}{2k+1}\right]_0^{\pi/2}\tag7\\ &=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[\frac\pi2(-1)^k-\frac1{2k+1}\right]\tag8\\ &=\bbox[5px,border:2px solid #C0A000]{2\pi\mathrm{G}-\frac72\zeta(3)}\tag9 \end{align} $$ Explanation:
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: apply $(2)$
$(6)$: prepare to integrate by parts
$(7)$: integrate by parts
$(8)$: apply the limits of integration
$(9)$: evaluate, where $\mathrm{G}$ is Catalan's Constant

Answer 8

I will present an evaluation that makes use of the following two Euler sums: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3) \qquad \text{and} \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} = \frac{23}{16} \zeta (3) - \pi \mathbf{G}.$$ Here $\mathbf{G}$ is Catalan's constant. For a proof of the first, see either here or Eq. (646) in this link. For a proof of the second, see Eq. (659) of this link.

Begin by enforcing a substitution of $x \mapsto 2 \arctan x$. Then $$\int_0^{\frac{\pi}{2}} \frac{x^2}{\sin x} \, dx = 4 \int_0^1 \frac{\arctan^2 x}{x} \, dx.\tag1$$

Since $\displaystyle{\arctan x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1}}$ for $|x| < 1$, applying the Cauchy product to the product between the two inverse tangent functions one obtains the following Maclaurin series expansion for $\arctan^2 x$ of: $$\arctan^2 x = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) x^{2n},$$ where $H_n$ is the $n$th Harmonic number.

Substituting the Maclaurin series expansion for $\arctan^2 x$ into (1), after changing the order of the summation with the integration one obtains \begin{align} \int_0^{\frac{\pi}{2}} \frac{x^2}{\sin x} \, dx &= 4 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n - 1} \, dx\\ &= -2 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} + \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2}\\ &= -2 \left (\frac{23}{16} \zeta (3) - \pi \mathbf{G} \right ) - \frac{5}{8} \zeta (3)\\ &= 2 \pi \mathbf{G} - \frac{7}{2} \zeta (3), \end{align} as required.

Answer 9

$$since\ I=\int_{0}^{\frac{\pi }{2}}\frac{x^2}{sinx}dx=-2\int_{0}^{\frac{\pi }{2}}xln(tan(\frac{x}{2}))dx=-8\int_{0}^{\frac{\pi }{2}}t \ln(tan(t))dt\\ \\ =-8\int_{0}^{1}\frac{lnt\arctan(t)}{1+t^2}dt=-4\pi \int_{0}^{1}\frac{ln(t)}{1+t^2}dt+8\int_{0}^{1}\frac{ln(t)arctan(t)}{1+t^2}dt\\ \\ =4\pi G-8\int_{1}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\\\ \therefore 2I=4\pi G-8\int_{0}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\ , let\ Q=\int_{0}^{\infty }\frac{ln(t)arctan(t)}{1+t^2}dt\\ \\ \therefore Q(a)=\int_{0}^{\infty }\frac{ln(t)arctan(at)}{1+t^2}dt\ \therefore Q'(a)=\frac{1}{4}\int_{0}^{\infty }\frac{ln(t)dt}{(1+t)(1+a^2t)}=\frac{1}{4(1-a^2)}\int_{0}^{\infty }[\frac{1}{1+t}-\frac{a^2}{1+a^2t}]lntdt\\ \\ \therefore Q'(a)=-\frac{\pi }{4(1-a^2)}\frac{d}{da}(\frac{1-a^-2a}{sin(\pi a)})_{a}^{=0}=\frac{\pi }{8(1-a^2)}lim[\frac{-2a.a^{-2a}ln^{2}a}{a^2}]=\frac{lna^{2}}{2(1-a^2)}\ \therefore Q=\frac{1}{4}\int_{0}^{1}\frac{(lna)^{2}}{1-a}da+\int_{0}^{1}\frac{(lna)^{2}}{1+a}da=\frac{7}{8}\zeta 3\\ \\ \therefore I=2\pi G-\frac{7}{2}\zeta(3)\\ \\ ahmed hegazi$$

Answer 10

First of all, we express the integrand as a power series of exponential functions. $$ \begin{aligned} I & =2 i \int_0^{\frac{\pi}{2}} \frac{x^2}{e^{x i}-e^{-x i}} d x \\ & =2 i \int_0^{\frac{\pi}{2}} x^2 e^{-x i} \sum_{k=0}^{\infty} e^{-2 k x i} d x \\ & =2 i \sum_{k=0}^{\infty} \int_0^{\frac{\pi}{2}} x^2 e^{-(2 k+1) x i} d x \\ & =2 i \sum_{k=0}^{\infty}\left[\int_0^{\frac{\pi}{2}} x^2 \cos (2 k+1) x d x-i \int_0^{\frac{\pi}{2}} x^2 \sin (2 k+1) xdx\right] \end{aligned} $$ Comparing their real parts yields $$ I=2 \underbrace{ \sum_{k=0}^{\infty} \int_0^{\frac{\pi}{2}} x^2 \sin (2 k+1) x d x}_{J_k} $$

Using integration by parts twice, we get $$ \begin{aligned} J_k & =-\frac{1}{2k+1}\left[x^2 \cos (2 k+1) x\right]_0^{\frac{\pi}{2}}+\frac{n}{2 k+1} \int_0^{\frac{\pi}{2}} x \cos (2 k+1) x d x \\ & =\frac{2}{(2 k+1)^2}\left[x\sin (2 k+1) x\right]_0^{\frac{\pi}{2}}-\frac{2}{(2 k+1)^3 }\int_0^{\frac{\pi}{2}} \sin (2 k+1) xdx \\ & = \frac{(-1)^k \pi}{(2 k+1)^2}-\frac{2(-1)^k }{(2 k+1)^3}\end{aligned} $$ Plugging back the summation yields $$ \begin{aligned} I & =\pi \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2}-2 \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^3} \\ & =\pi G-2\left(1-\frac{1}{8}\right)\zeta(3)\\&= \pi G-\frac{7}{4}\zeta(3) \end{aligned} $$ where $G$ is the Catalan’s constant.

Answer 11

Another contour integration approach:

Let's integrate the function $$f(z) = \frac{z^{2}}{\sin z}$$ around a tall rectangular contour with vertices at $z=0$, $z=\frac{\pi}{2}$, $z= \frac{\pi}{2}+ i R$, and $z= i R$.

There are no singularities inside contour, and the singularity at $z=0$ is removable.

Also, since the magnitude of $\frac{1}{\sin z}$ decays exponentially to zero as $\Im(z) \to \infty$, the integral vanishes on the top of the contour as $R \to \infty$.

Therefore, integrating around the contour and then letting $R \to \infty$, we get $$\int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx + \int_{0}^{\infty} \frac{(\pi /2 + it)^{2}}{\cosh t} \, i \, \mathrm dt + \int_{\infty}^{0} \frac{(it)^{2}}{i \sinh t} \, i \, \mathrm dt = 0. $$

And equating the real parts on both sides the equation, we get $$\int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx = \pi \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt - \int_{0}^{\infty} \frac{t^{2}}{\sinh t} \, \mathrm dt. $$

An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{2 \Gamma(s)}\int_{0}^{\infty} \frac{t^{s-1}}{\cosh t} \, \mathrm dt, \quad \Re(s) >0.$$

And an integral representation of the Dirichlet lambda function is $$\lambda(s) = \frac{1}{2 \Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1}}{\sinh t} \, \mathrm dt, \quad \Re(s) >1.$$

Therefore, $$\begin{align} \int_{0}^{\pi/2} \frac{x^{2}}{\sin x} \, \mathrm dx &= 2 \pi \Gamma(2) \beta(2) - 2 \Gamma(3) \lambda(3) \\ &= 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} -4 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{3}} \\ &= 2 \pi G - 4 \left(1-2^{-3} \right) \zeta(3) \\ &= 2 \pi G - \frac{7}{2} \zeta(3). \end{align}$$

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The same approach also shows that $$ \int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm dx = \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt =2 \Gamma(2) \beta(2) =2G $$ and

$$\begin{align} \int_{0}^{\pi/2} \frac{x^{3}}{\sin x} \, \mathrm dx &= \frac{3 \pi^{2}}{4} \int_{0}^{\infty} \frac{t}{\cosh t} \, \mathrm dt - \int_{0}^{\infty} \frac{t^{3}}{\cosh t} \, \mathrm dt \\ &= \frac{3 \pi^{2}}{2} \Gamma(2) \beta(2) - 2 \Gamma(4) \beta(4) \\ &= \frac{3 \pi^{2}G}{2} - 12 \beta(4). \end{align}$$

Answer 12

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{x^{2}}{\sin x} d x =& \int_{0}^{\frac{\pi}{2}} x^{2} d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\ =& {\left[x^{2} \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} x \ln \left(\tan \frac{x}{2}\right) d x } \\ =&-8 \int_{0}^{\frac{\pi}{4}} y \ln (\tan y) d y \textrm{, where }y=2x\\ =&-8\left[-\frac{\pi}{4} G+\frac{7}{16} \zeta(3)\right] \\ =& 2 \pi G-\frac{7}{2} \zeta(3) \end{aligned} $$

where $ \int_{0}^{\frac{\pi}{4}} x \ln (\tan x) d x= -\frac{\pi}{4} G+\frac{7}{16} \zeta(3) $ from my post .

Can we go further with the integral $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{x^{n}}{\sin x} d x ?\tag*{} $