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Computations suggest that

$$\int_{0}^{\infty}\int_{0}^{\infty} \sqrt{x+y^2} \cdot e^{-\frac{1}{2}(\frac{x}{s}+s^2y^2)}dxdy=\frac{2}{s}+\frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}.$$

The question is how to prove this equality.

Background: this is the mean width of an ellipsoid with semiaxes $s,s,1/s^2$ and this almost completes the proof of the uniqueness hypothesis of an ellipsoid with given intrinsic volumes.

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  • 2
    $\begingroup$ although the formula is far from being beautiful, the question looks normal for me $\endgroup$ – Fedor Petrov Oct 22 '18 at 10:44
  • $\begingroup$ @Matt, probably you make a mistake, because: wolframalpha.com/input/… $\endgroup$ – Alexander Tarasov Oct 24 '18 at 6:11
  • $\begingroup$ You are correct, I misread $\sqrt{x+y^2}$ because I am so used to seeing $\sqrt{x^2+y^2}$. $\endgroup$ – Matt F. Oct 24 '18 at 13:37
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First of all make a replacement $x=t^2$ and go to polar coordinates \begin{equation*} \int_{0}^{\infty}\int_{0}^{2\pi} \cos(\phi)r^3 \cdot e^{-\frac{\frac{r^2\cos^2(\phi)}{s}+r^2\sin^2(\phi)s^2}{2}}drd\phi. \end{equation*} After than notice that we can take integral by radius. Let $R=\frac{1}{2}(\cos^2(\phi)\frac{1}{s}+\sin^2(\phi)s^2)$ then \begin{equation*} \left. \int_{0}^{\infty} r^3 \cdot e^{-r^2R}dr= -\frac{1}{2R}\int_{0}^{\infty} r^2\cdot de^{-r^2R}=-\frac{1}{2R}r^2e^{-r^2R}\right|_{0}^{\infty}+\frac{1}{R}\int_{0}^{\infty} re^{-r^2R}dr= \end{equation*} \begin{equation*} =\frac{1}{R}\int_{0}^{\infty}re^{-r^2R}dr=\left.-\frac{e^{-r^2R}}{2R^2}\right|_0^{\infty}=\frac{1}{2R^2} \end{equation*}

Back to the main integral \begin{equation*} \int_0^{2\pi} \frac{\cos(\phi)}{R^2}d\phi=\int_0^{2\pi} \frac{\cos(\phi)}{(\frac{1}{2}\cos^2(\phi)\frac{1}{s}+\frac{1}{2}\sin^2(\phi)s^2)^2}d\phi \propto \end{equation*} \begin{equation*} \propto s^2\int_0^{2\pi} \frac{\cos(\phi)}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\phi= \end{equation*} \begin{equation*} =s^2\int_0^{2\pi} \frac{1}{(\cos^2(\phi)+\sin^2(\phi)s^3)^2}d\sin(\phi)=s^2\int_{-1}^1 \frac{1}{(1-t^2+s^3t^2)^2}dt \end{equation*}

It is easy to chek that

\begin{equation*} \int\frac{1}{(1+x^2a)^2}dx=\frac{x}{2(ax^2+1)}+\frac{\arctan(\sqrt{a})}{2\sqrt{a}}, \end{equation*}

So we finaly obtain for $a=s^3-1$ when $F(s)$ - is the main width of spheroid with semiaxes $\sqrt{s},\sqrt{s},\frac{1}{s}$.

\begin{equation} F(s)=Const\cdot s^2\left(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right)=Const\cdot \left(\frac{1}{s}+\frac{s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}\right) \end{equation}

And fact that $F(s)$ decrease when $s\in(0,1)$ and increase when $s\in(1,\infty)$ means that there are only two spheroids with volume$=1$ and mean width$=W>F(1)$.

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we pose $x=t^2$

$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{t^2+y^2} e^{\frac{t^2}{s}+s^2 y^2} 2tdxdt $

by change of variable ,we get: $\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy =2\int^{\frac{\pi}{2}}_0\int^{\infty}_0 r^3 \cos (\theta) e^{\frac{-1}{2}(\frac{1}{s} \cos ^2\theta+s^2 sin^2 \theta )r^2} drd\theta $

we pose $A= \frac{1}{2}(\frac{1}{s} \cos^2 (\theta)+s^2 \sin^2(\theta) )$

$\int^{+\infty}_0 r^3 e^{-Ar^2} dr=\frac{-r^2}{2A} e^{-Ar^2}\Bigg |^{+\infty}_0+\int^{+\infty}_0 \frac{r}{A}e^{-Ar^2}dr = \frac{-1}{2A^2} e^{-Ar^2} \Bigg |^{+\infty}_0=\frac{1}{2A^2}$

$$\int^{\frac{\pi}{2}}_0\frac{\cos \theta}{2 A}d\theta =2 s^2 \int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta$$

$\int^{\frac{\pi}{2}}_0 \frac{\cos \theta}{\cos^2 \theta+s^3 \sin^2 \theta}d\theta =\int^1_0 \frac{1}{(1+(s^3-1)t^2)^2}dt$

we pose $a=s^3-1$

$\int^1_0 \frac{1}{(1+at^2)^2}dt =\Bigg [ \frac{t}{2(1+at^2)} + \frac{\arctan(\sqrt{a}t)}{2\sqrt{a}}\Bigg ]^1_0 =\frac{1}{2}(\frac{1}{s^3}+\frac{\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}})$

So

$$\int^{+\infty}_0 \int^{+\infty}_0 \sqrt{x+y^2} e^{\frac{x}{s}+s^2 y^2}dxdy=\frac{2}{s}+ \frac{2s^2\arctan(\sqrt{s^3-1})}{\sqrt{s^3-1}}$$

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