A Sine and Inverse Sine integral

A demonstration of methods

While reviewing an old text book an integral containing sines and sine inverse was encountered, namely, $$\int_{0}^{\pi/2} \int_{0}^{\pi/2} \sin(x) \, \sin^{-1}(\sin(x) \, \sin(y)) \, dx \, dy = \frac{\pi^{2}}{4} - \frac{\pi}{2}.$$

One can express $\sin^{-1}(z)$ as a series and integrate, but are there other methods nearly as efficient?

Solution 1:

Define the real function $\mathcal{I}:[-1,1]\rightarrow\mathbb{R}$ via the parametric integral

$$\mathcal{I}{\left(a\right)}:=\iint_{[0,\frac{\pi}{2}]^{2}}\mathrm{d}\theta\,\mathrm{d}\varphi\,\sin{\left(\varphi\right)}\arcsin{\left(a\sin{\left(\varphi\right)}\sin{\left(\theta\right)}\right)}.$$

If we perform the integration over $\varphi$ first, we can reduce the integral to an algebraic one by integrating by parts:

$$\begin{align} f{\left(a;\theta\right)} &:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sin{\left(\varphi\right)}\arcsin{\left(a\sin{\left(\varphi\right)}\sin{\left(\theta\right)}\right)}\\ &=\left[-\cos{\left(\varphi\right)}\arcsin{\left(a\sin{\left(\varphi\right)}\sin{\left(\theta\right)}\right)}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left(-1\right)a\cos^{2}{\left(\varphi\right)}\sin{\left(\theta\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}\sin^{2}{\left(\theta\right)}}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{a\cos^{2}{\left(\varphi\right)}\sin{\left(\theta\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}\sin^{2}{\left(\theta\right)}}}.\\ \end{align}$$

Now we attack $\mathcal{I}{\left(a\right)}$ by changing the order of integration. Given $a\in[-1,1]$, we have

$$\begin{align} \mathcal{I}{\left(a\right)} &=\iint_{[0,\frac{\pi}{2}]^{2}}\mathrm{d}\theta\,\mathrm{d}\varphi\,\sin{\left(\varphi\right)}\arcsin{\left(a\sin{\left(\varphi\right)}\sin{\left(\theta\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\mathrm{d}\varphi\,\sin{\left(\varphi\right)}\arcsin{\left(a\sin{\left(\varphi\right)}\sin{\left(\theta\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,f{\left(a;\theta\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{a\cos^{2}{\left(\varphi\right)}\sin{\left(\theta\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}\sin^{2}{\left(\theta\right)}}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{a\cos^{2}{\left(\varphi\right)}\sin{\left(\theta\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}\sin^{2}{\left(\theta\right)}}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,a\cos^{2}{\left(\varphi\right)}\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}\left(1-y^{2}\right)}};~~~\small{\left[\cos{\left(\theta\right)}=y\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,a\cos^{2}{\left(\varphi\right)}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}+a^{2}y^{2}\sin^{2}{\left(\varphi\right)}}}.\\ \end{align}$$

Recall that the inverse hyperbolic sine may be defined via

$$\operatorname{arsinh}{\left(z\right)}:=\int_{0}^{z}\frac{\mathrm{d}x}{\sqrt{1+x^{2}}}=z\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1+z^{2}t^{2}}};~~~\small{z\in\mathbb{R}}.$$

Then,

$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,a\cos^{2}{\left(\varphi\right)}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}+a^{2}y^{2}\sin^{2}{\left(\varphi\right)}}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos^{2}{\left(\varphi\right)}}{\sin{\left(\varphi\right)}}\cdot\frac{a\sin{\left(\varphi\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}}}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{1+\frac{a^{2}\sin^{2}{\left(\varphi\right)}}{1-a^{2}\sin^{2}{\left(\varphi\right)}}y^{2}}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos{\left(\varphi\right)}\cot{\left(\varphi\right)}\operatorname{arsinh}{\left(\frac{a\sin{\left(\varphi\right)}}{\sqrt{1-a^{2}\sin^{2}{\left(\varphi\right)}}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1-x^{2}}}{x}\operatorname{arsinh}{\left(\frac{ax}{\sqrt{1-a^{2}x^{2}}}\right)};~~~\small{\left[\sin{\left(\varphi\right)}=x\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1-x^{2}}}{x}\operatorname{artanh}{\left(ax\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1-x^{2}}}{x}\int_{0}^{1}\mathrm{d}t\,\frac{ax}{1-a^{2}x^{2}t^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,\frac{a\sqrt{1-x^{2}}}{\left(1-a^{2}x^{2}t^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{a\left(1-x^{2}\right)}{\left(1-a^{2}t^{2}x^{2}\right)\sqrt{1-x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{at^{2}\sqrt{1-x^{2}}}-\frac{\left(1-a^{2}t^{2}\right)}{at^{2}\left(1-a^{2}t^{2}x^{2}\right)\sqrt{1-x^{2}}}\right]\\ &=\int_{0}^{1}\mathrm{d}t\,\left[\int_{0}^{1}\mathrm{d}x\,\frac{1}{at^{2}\sqrt{1-x^{2}}}-\int_{0}^{1}\mathrm{d}x\,\frac{\left(1-a^{2}t^{2}\right)}{at^{2}\left(1-a^{2}t^{2}x^{2}\right)\sqrt{1-x^{2}}}\right]\\ &=\int_{0}^{1}\mathrm{d}t\,\left[\frac{\pi}{2at^{2}}-\frac{\pi\left(1-a^{2}t^{2}\right)}{2at^{2}\sqrt{1-a^{2}t^{2}}}\right]\\ &=\frac{\pi}{2a}\int_{0}^{1}\mathrm{d}t\,\frac{1-\sqrt{1-a^{2}t^{2}}}{t^{2}}\\ &=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}t\,\frac{a}{1+\sqrt{1-a^{2}t^{2}}}\\ &=\frac{\pi}{2}\left[\arcsin{\left(at\right)}-\frac{at}{1+\sqrt{1-a^{2}t^{2}}}\right]_{0}^{1}\\ &=\frac{\pi}{2}\left[\arcsin{\left(a\right)}-\frac{a}{1+\sqrt{1-a^{2}}}\right].\blacksquare\\ \end{align}$$

Note that setting $a=1$ yields the conjectured value given by the OP:

$$\mathcal{I}{\left(1\right)}=\frac{\pi}{2}\left[\arcsin{\left(1\right)}-1\right]=\frac{\pi^{2}}{4}-\frac{\pi}{2}.$$

Solution 2:

To complement the existing answers, here is a geometrical way.

Consider the following integral on the surface of a unit sphere (positive octant):

$$\mathcal{I}=\int\limits_{x^2+y^2+z^2=1,~x,y,z\ge0}d\Omega~\sin^{-1}(y).$$

where $\Omega$ is the solid angle (or area). By using the spherical coordinate system: $x=\sin\theta\cos\phi,~y=\sin\theta\sin\phi,~z=\cos\theta,$ where $0\le\theta,\phi\le\pi/2,$ we see this is the original integral that we need to evaluate.

Note $\mathcal{I}$ is invariant under rotation, so we also have

$$\mathcal{I}=\int\limits_{x^2+y^2+z^2=1,~x,y,z\ge0}d\Omega~\sin^{-1}(z).$$

Again using the spherical coordinates, we get

$$\mathcal{I}=\int_0^{\pi/2}d\phi\int_0^{\pi/2}d\theta~\sin\theta~\sin^{-1}(\cos\theta)=\frac{\pi}{2}\int_0^{\pi/2}d\theta~\sin\theta\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}\left(\frac{\pi}{2}-1\right)$$

as desired, where we have used integration by parts for the last step.

For the general case that David H. considered, we have when $r\le1$:

$$\mathcal{I}(r)=\int\limits_{x^2+y^2+z^2=r^2,~x,y,z\ge0}d\Omega~\sin^{-1}(y).$$

Similarly we get

$$\mathcal{I}(r)=\int_0^{\pi/2}d\phi\int_0^{\pi/2}d\theta~\sin\theta~\sin^{-1}(r\cos\theta)=\frac{\pi}{2r}\int_0^rdt\sin^{-1}(t)~=\frac{\pi}{2r}\left(r\sin^{-1}(r)+\sqrt{1-r^2}-1\right)=\frac{\pi}{2}\left(\sin^{-1}(r)+\frac{\sqrt{1-r^2}-1}{r}\right).$$

Again we have used integration by parts for the last integral.

Solution 3:

As mickep has observed \begin{equation*} I_{1} = \int_0^{\pi/2}\int_0^{\pi/2}\sin(x)\arcsin(\sin(x)\sin(y))\,dxdy = \dfrac{1}{2} \int_{0}^{\pi/2}\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\dfrac{\cos^{2}x}{\sin x}\, dx. \end{equation*} We intend to finish the solution without using series expansion. Instead we will use complex integration and Cauchy integral theorem.

After integration by parts we have \begin{gather*} I_{1} = \dfrac{1}{2}\int_{0}^{\pi/2}\left(\dfrac{1}{\sin x}-\sin x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\, dx = \dfrac{1}{2}\left[\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\right]_{0}^{\pi/2}\\ - \int_{0}^{\pi/2}\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\dfrac{1}{\cos x}\, dx = - \underbrace{\int_{0}^{\pi/2}\dfrac{\ln\left(\tan\dfrac{x}{2}\right)}{\cos x}\, dx}_{I_{2}} - \dfrac{\pi}{2}. \end{gather*} Finally we will prove that $I_2 = -\dfrac{\pi^{2}}{4}.$ Since $\displaystyle \tan^{2}\dfrac{x}{2} = \dfrac{1-\cos x}{1+\cos x}$ we get \begin{gather*} I_{2}= \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\cos x}\ln\left(\dfrac{1-\cos x}{1+\cos x}\right)\, dx = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\sin x}\ln\left(\dfrac{1-\sin x}{1+\sin x}\right)\, dx = \left[t = \tan\dfrac{x}{2}\right]\\ = \int_{0}^{1}\dfrac{1}{t}\ln\left(\dfrac{1-t}{1+t}\right)\, dt = \left[\ln(t)\ln\left(\dfrac{1-t}{1+t}\right)\right]_{0}^{1} - 2\int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt. \end{gather*} But \begin{equation*} \int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt = \int_{1}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt . \end{equation*} Thus $ \displaystyle I_{2} = -\int_{0}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt.$ If $\log $ is the principal branch of the logarithm then the function $ f(z) = \dfrac{\log(z)}{z^{2}-1}$ has a removable singularity at $z=1$. If we integrate $f(z)$ ''around quadrant 1'' and use Cauchy integral theorem we get \begin{equation*} \int_{0}^{\infty}\dfrac{\ln(x)}{x^{2}-1}\, dx - \int_{0}^{\infty}\dfrac{\ln(|iy|)+i\pi/2}{(iy)^{2}-1}i\, dy = 0. \end{equation*} Consequently \begin{equation*} -I_{2} + i\int_{0}^{\infty}\dfrac{\ln(y)}{y^{2}+1}\, dy -\dfrac{\pi}{2}\underbrace{\int_{0}^{\infty}\dfrac{1}{y^{2}+1}\, dy}_{ = \pi/2} = 0.\tag{1} \end{equation*} From the real part in (1) we finally get $I_2 = -\dfrac{\pi^2}{4}$.