<s>[INST] Integral equality of 1st intrinsic volume of spheroid 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.
 [/INST] 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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