Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$

Question

Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$

My attempt. Taking $\sin x \sin y=t$, $\iint\sin x \ dx \ dy$. I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.

I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?

Is this the correct way. Is there any easier way to go about it?

Answer 1

Here is an approach using power series expansions.

It is known that for $t\in (-1,1)$ $$\frac{1}{\sqrt{1-t}} =(1-t)^{-1/2}=\sum_{n=0}^\infty \binom{-1/2}{n} (-t)^{n}= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} t^{n},$$ and for $t\in [-1,1]$, $$\arcsin(t) =\int_0^t\frac{ds}{\sqrt{1-s^2}}= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ t^{2n+1}}{2n+1}.$$ Hence, the given integral is equal to \begin{align*}I&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\int_0^{\pi/2}\int_0^{\pi/2} \sin x (\sin x \sin y)^{2n+1} \ dx \ dy\\ &=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}W_{2n+2} W_{2n+1} \\ &=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\cdot \frac{\pi}{4(n+1)}\\ &=\frac{\pi}{4}\left(2\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}-\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{n+1}\right)\\ &=\frac{\pi}{4}\left(2\arcsin(1)-\int_0^1\frac{dt}{\sqrt{1-t}}\right)\\ &=\frac{\pi(\pi-2)}{4}. \end{align*} where $W_r=\int_{0}^{\pi/2}\sin^r(t)dt $ denotes the Wallis integral of order $r$.

Answer 2

First consider the following identity: $$\int_0^{\pi/2} \sin^n(x)\,\mathrm{d}x = \frac{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{2\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}$$ Now consider the Taylor expansion of $a\sin^{-1}(ab)$ around $b = 0$: $$a\sin^{-1}(ab) = \sum_{k=0}^{\infty} \frac{(1/2)_k^2a^{2k+2}b^{2k+1}}{(3/2)_kk!}$$ where $(x)_k$ is the rising factorial (this is used in the theory of hypergeometric functions). Using the above identity, $$\int_0^{\pi/2}\int_0^{\pi/2} \sin^{2k+2}(x)\sin^{2k+1}(y)\,\mathrm{d}x\,\mathrm{d}y = \frac{\sqrt{\pi}\operatorname{\Gamma}\left(k+\frac{3}{2}\right)}{2\operatorname{\Gamma}(k+2)}\cdot \frac{\sqrt{\pi}\operatorname{\Gamma}(k+1)}{2\operatorname{\Gamma}\left(k+\frac{3}{2}\right)} = \frac{\pi}{4(k+1)}$$ Therefore, by dominated convergence, \begin{align*} \int_0^{\pi/2}\int_0^{\pi/2} \sin(x)\sin^{-1}(\sin(x)\sin(y))\,\mathrm{d}x\,\mathrm{d}y &= \int_0^{\pi/2}\int_0^{\pi/2} \sum_{k=0}^{\infty} \frac{(1/2)_k^2\sin^{2k+2}(x)\sin^{2k+1}(y)}{(3/2)_kk!} \\ &= \frac{\pi}{4}\sum_{k=0}^{\infty} \frac{(1/2)_k^2}{(3/2)_k(k+1)!} \end{align*} Now, we note that $$\frac{(1/2)_k^2}{(3/2)_k(k+1)!} = \frac{2(1/2)_k^2}{(3/2)_kk!}-\frac{2k+1}{(k+1)!}\frac{(1/2)_k^2}{(3/2)_k} = \frac{2(1/2)_k^2}{(3/2)_kk!}-\frac{(1/2)_k}{(k+1)!}$$ We know that $$\sum_{k=0}^{\infty} \frac{2(1/2)_k^2x^{2k+1}}{(3/2)_kk!} = 2\sin^{-1}(x)$$ so $$\sum_{k=0}^{\infty} \frac{2(1/2)_k^2}{(3/2)_kk!} = 2\sin^{-1}(1) = \pi$$ and we can figure out that $$\sum_{k=0}^{\infty} \frac{(1/2)_kx^k}{(k+1)!} = \frac{2}{1+\sqrt{1-x}}$$ by checking this, so $$\sum_{k=0}^{\infty} \frac{(1/2)_k}{(k+1)!} = 2$$ Therefore, the value of the integral is $$\int_0^{\pi/2}\int_0^{\pi/2} \sin(x)\sin^{-1}(\sin(x)\sin(y))\,\mathrm{d}x\,\mathrm{d}y = \frac{\pi(\pi-2)}{4}$$

Answer 3

Robert Z' approach of expanding $\arcsin z$ as a Taylor series, then exploiting $\int_{0}^{\pi/2}\sin(x)^{2k}\,dx\int_{0}^{\pi/2}\sin(x)^{2k+1}\,dx=\frac{\pi}{4k+2}$ is very slick, but there is an interesting alternative approach. The original integral clearly equals

$$ \mathcal{I}=\int_{0}^{1}\int_{0}^{1}\frac{x\arcsin(xy)}{\sqrt{(1-x^2)(1-y^2)}}\,dx\,dy \tag{1}$$ and $$ \int_{0}^{1}\frac{\arcsin(xy)}{\sqrt{1-y^2}}\,dy = \frac{4}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\int_{0}^{1}\frac{T_{2n-1}(xy)}{\sqrt{1-y^2}}\,dy\tag{2}$$ by the Fourier-Chebyshev series for the arcsin function (page 33 here). This leads to: $$\begin{eqnarray*} \mathcal{I}&=& \frac{4}{\pi}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}\int_{0}^{x}\frac{T_{2n-1}(y)}{\sqrt{x^2-y^2}}\,dy\,dx\\&=&\frac{4}{\pi}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}T_{2n-1}(y)\int_{y}^{1}\frac{x}{\sqrt{(1-x^2)(x^2-y^2)}}\,dx\,dy\\ &=&\frac{2}{\color{blue}{\pi}}\sum_{n\geq 1}\frac{1}{(2n-1)^2}\int_{0}^{1}T_{2n-1}(y)\color{blue}{\int_{y^2}^{1}\frac{1}{\sqrt{(1-x)(x-y^2)}}\,dx}\,dy\\ &=&2\sum_{n\geq 1}\frac{1}{(2n-1)^2} \int_{0}^{1}T_{2n-1}(y)\,dy \\&=&2\Big[\sum_{\substack{n\geq 1 \\ n\text{ odd}}}\frac{1}{(2n)(2n-1)^2}-\sum_{\substack{n\geq 1 \\ n\text{ even}}}\frac{1}{(2n-2)(2n-1)^2}\Big]\tag{3}\end{eqnarray*} $$

by the wonderful cancellation of the blue terms. The last series are straightforward to compute in terms of $\frac{\pi}{4},\zeta(2),\log(2)$ and Catalan's constant by partial fraction decomposition.

Answer 4

Perform integration by parts,

$\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\sin x \sin y) \ dx \ dy\\ &=\int_0^{\frac{\pi}{2}}\left(\Big[-\cos x\arcsin(\sin x \sin y)\Big]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx\right)\ dy\\ &=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\frac{\cos^2x\sin y}{\sqrt{1-\sin^2 x\sin^2 y }}\ dx \ dy\\ &=\int_0^{\frac{\pi}{2}}\left(\Big[-\frac{\cos x\text { arcsinh} \left(\tan x\cos y\right)}{\tan x}\Big]_{y=0}^{y=\frac{\pi}{2}}\right)\ dx\\ &=\int_0^{\frac{\pi}{2}} \frac{\cos x\text{ arcsinh}(\tan x)}{\tan x}\ dx \end{align}$

Perform the change of variable $u=\tan x$,

$\begin{align}J&=\int_0^{\infty}\frac{\text{ arcsinh } x}{x\left(1+x^2\right)^{\frac{3}{2}}}\ dx \end{align}$

Perform integration by parts,

$\begin{align}J&=\left[\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\text{arcsinh }x\right]_0^{\infty}-\int_0^{\infty}\left(\frac{1}{\sqrt{1+x^2}}-\text{arcsinh}\left(\frac{1}{x}\right)\right)\frac{1}{\sqrt{1+x^2}}\ dx\\ &=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\int_0^{\infty}\frac{1}{1+x^2}\ dx\\ &=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx-\frac{\pi}{2} \end{align}$

In the following integral perform the change of variable $u=\frac{1}{x}$,

$\begin{align} K&=\int_0^{\infty}\frac{\text{arcsinh}\left(\frac{1}{x}\right)}{\sqrt{1+x^2}}\ dx\\ &=\int_0^{\infty}\frac{\text{arcsinh }x}{x\sqrt{1+x^2}}\ dx\\ \end{align}$

Perform the change of variable $u=\text{arcsinh }x$,

$\begin{align} K&=\int_0^{\infty}\frac{x}{\sinh x}\ dx \end{align}$

and according to Contour integral of $\int_{0}^{\infty}\frac{x}{\sinh x}\operatorname{dx}$ ,

$\displaystyle K=\frac{\pi^2}{4}$

Therefore,

$\boxed{\displaystyle J=\frac{\pi^2}{4}-\frac{\pi}{2}}$

Addenda:

1)Another way to compute $K$.

Performing integration by parts,

$\begin{align}K&=\left[x\ln\left(\tanh\left(\frac{x}{2}\right)\right)\right]_0^{\infty}-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\ &=-\int_0^{\infty}\ln\left(\tanh\left(\frac{x}{2}\right)\right)\ dx\\ \end{align}$

Perform the change of variable $y=\tanh\left(\frac{x}{2}\right)$,

$\begin{align}K&=2\int_0^1 \frac{\ln x}{x^2-1}\ dx\\ &=\int_0^1 \frac{\ln x}{x-1}\ dx-\int_0^1 \frac{\ln x}{x+1}\ dx \end{align}$

$\begin{align}L&=\int_0^1 \frac{\ln x}{x-1}\ dx+\int_0^1 \frac{\ln x}{x+1}\ dx\\ &=\frac{1}{2}\int_0^1 \frac{2x\ln\left(x^2\right)}{x^2-1}\ dx \end{align}$

Perform the change of variable $u=x^2$,

$\begin{align}L&=\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx \end{align}$

Therefore,

$\begin{align}\int_0^1 \frac{\ln x}{x+1}\ dx=-\frac{1}{2}\int_0^1 \frac{\ln x}{x-1}\ dx \end{align}$

Therefore,

$\begin{align}K&=\frac{3}{2}\int_0^1 \frac{\ln x}{x-1}\ dx \end{align}$

Knowing that,

$\begin{align}\int_0^1 \frac{\ln x}{x-1}\ dx=\dfrac{\pi^2}{6}\end{align}$

Therefore,

$\boxed{\displaystyle K=\dfrac{\pi^2}{4}}$

2)Playing around with lindep, intnum, functions of GP-PARI, i suspect that the following formulae are true:

$\begin{align} \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arcsin(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2-\ln 2\\ \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\sin x \sin y) \ dx \ dy&=\frac{1}{2}\pi\\ \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arccos(\cos x \cos y) \ dx \ dy&=\frac{1}{8}\pi^2+\ln 2\\ \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan(\sin x \sin y) \ dx \ dy&=\frac{1}{8}\pi^2-\frac{1}{4}\pi\ln 2\\ \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \sin x \arctan^2(\sin x \sin y) \ dx \ dy&=\frac{1}{32}\pi^3+\frac{1}{8}\pi^2\ln 2-\frac{1}{2}\text{G}\pi\\ \end{align}$

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

Answer 5

I understand I am a bit late to the party, but there is a more rudimentary method to solve this. $$I = \int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) dx dy$$ Substituting $\sin x\sin y=\sin\theta$, we get $$I = \int_0^{\pi/2}\int_0^{x} \sin x \frac{\theta \cos \theta}{\sin x\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} d\theta dx$$ We change order of integration to get $$I = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta}{\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} dx d\theta = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta \sin x}{\sqrt{\cos^2 \theta- \cos^2 x}} dx d\theta$$ This is now integrable: $$I = \int_0^{\pi/2} \theta \cos \theta [-\sin^{-1}(\frac{\cos x}{\cos \theta})]_\theta ^{\pi /2} d\theta = \frac \pi 2 \int_0^{\pi/2}\theta \cos \theta d\theta = \frac \pi 2 (\frac \pi 2 -1)$$