I was trying to figure the volume of a convex body and I ran into the following integral:
$$\displaystyle\int_0^{\frac{\pi}{2}}\left( \int_{-1}^1\int_{-1}^1 |x\cos(\theta)+y\sin(\theta)|\,dxdy \right)^{-2}d\theta.$$
So you have a geometric idea we are integrating on the first quadrant of the 2-dimensional sphere and on centered square of side 2. As it's the volume of a convex body I know it must be finite but I'm incapable to solve it. I will appreciate any help. Thank you in advance!
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\pi/2}\bracks{\int_{-1}^{1}\int_{-1}^{1} \verts{\vphantom{\large A}x\cos\pars{\theta} + y\sin\pars{\theta}} \,\dd x\,\dd y}^{-2}\,\dd\theta \\[5mm] = &\ \int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta \over \ds{\bracks{\int_{-1}^{1}\int_{-1}^{1} \verts{x + y\tan\pars{\theta}}\,\dd x\,\dd y}^{2}}} \,\,\,\stackrel{t\ =\ \tan\pars{\theta}}{=} \int_{0}^{\infty}{\dd t \over \ds{\bracks{\int_{-1}^{1}\int_{-1}^{1}\verts{x + ty}\,\dd x\,\dd y}^{2}}} \\[5mm] = &\ 2\int_{0}^{1}{\dd t \over \ds{\bracks{\int_{-1}^{1}\int_{-1}^{1}\verts{x + ty}\,\dd x\,\dd y}^{2}}} \,,\quad \pars{\substack{\ds{by\ splitting\ over\ \pars{0,1}\ and\ \pars{1,\infty}} \\[2mm] \ds{Set\ t\ \mapsto\ {1 \over t}\ in\ the\ integral\ over\ \pars{1,\infty}}}} \label{1}\tag{1} \end{align}
Hereafter, I'll evaluate the "$\ds{xy}$-integral":
\begin{align} \left.\int_{-1}^{1}\int_{-1}^{1}\verts{x + ty}\,\dd x\,\dd y\, \right\vert_{\ 0\ <\ t\ <\ 1} & = \overbrace{2\int_{0}^{1}\int_{0}^{1}\pars{x + ty}\,\dd x\,\dd y} ^{\ds{=\ 1 + t}}\ +\ 2\int_{0}^{1}\int_{0}^{1}\verts{x - ty}\,\dd x\,\dd y \end{align} The second integral is evaluated as follows: \begin{align} &\left.2\int_{0}^{1}\int_{0}^{1}\verts{x - ty}\,\dd x\,\dd y\, \right\vert_{\ 0\ <\ t\ <\ 1} = 2\int_{0}^{1}\int_{-ty}^{1 - ty}\verts{x}\,\dd x\,\dd y \\[5mm] = &\ 2\int_{0}^{1}\bracks{% \int_{-ty}^{0}\pars{-x}\,\dd x\,\dd y + \int_{0}^{1 - ty}x\,\dd x\,\dd y} = 2\int_{0}^{1}\pars{t^{2}y^{2} - ty + {1 \over 2}}\,\dd y = {2 \over 3}\,t^{2} - t + 1 \\[5mm] &\ \mbox{such that}\quad \bbx{\left.\int_{-1}^{1}\int_{-1}^{1}\verts{x + ty}\,\dd x\,\dd y\, \right\vert_{\ 0\ <\ t\ <\ 1} = {2 \over 3}\pars{t^{2} + 3}} \end{align}
\eqref{1} becomes:
\begin{align} &\int_{0}^{\pi/2}\bracks{\int_{-1}^{1}\int_{-1}^{1} \verts{\vphantom{\large A}x\cos\pars{\theta} + y\sin\pars{\theta}} \,\dd x\,\dd y}^{-2}\,\dd\theta = {9 \over 2}\int_{0}^{1}{\dd t \over \pars{t^{2} + 3}^{2}} \\[5mm] = &\ \bbx{{\root{3} \over 24}\,\pi + {3 \over 16}} \approx 0.4142 \end{align}
Tricky question. If we exploit $$ |K|=\int_{-\infty}^{+\infty}\frac{1-\cos(K s)}{\pi s^2}\,ds \tag{1} $$ we get that $$ \iint_{[-1,1]^2}\left|ax+by\right|\,dx\,dy = \frac{4}{\pi}\int_{-\infty}^{+\infty}\frac{a b s^2-\sin(a s)\sin(b s)}{a b s^4}\,ds \tag{2}$$ so by computing the RHS of $(2)$ through Fourier or Laplace transforms: $$ \iint_{[-1,1]^2}\left|ax+by\right|\,dx\,dy = \frac{2}{3\max(a,b)}\left[a^2+b^2+2\max(a,b)^2\right]\tag{3} $$ for any $a,b\geq 0$. By setting $a=\cos\theta$ and $b=\sin\theta$, for any $\theta\in\left(0,\frac{\pi}{2}\right)$ we have $$ \iint_{[-1,1]^2}\left|x\cos\theta+y\sin\theta\right|\,dx\,dy = \frac{2}{3\max(\sin\theta,\cos\theta)}\left[1+2\max(\cos\theta,\sin\theta)^2\right]\tag{4}$$ and the original integral equals
$$ I=\int_{0}^{\pi/4}\left[\frac{3\cos\theta}{2(1+2\cos^2\theta)}\right]^2\,d\theta+\int_{\pi/4}^{\pi/2}\left[\frac{3\sin\theta}{2(1+2\sin^2\theta)}\right]^2\,d\theta \tag{5}$$ and by symmetry and the substitution $\theta=\arctan u $ it follows that: $$\boxed{ I = \color{red}{\frac{1}{48}\left(9+2\pi\sqrt{3}\right)}}\tag{6} $$