<s>[INST] Solving an integral with power of $\exp$ Is it possible to get a "closed" form for the following integral 
$$\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-\frac{2}{3}\vert x\vert^{\beta}}e^{-\frac{1}{3}\vert y-x\vert^{\beta}}\frac{1}{\vert y\vert^{\gamma}}dxdy,$$
where $1<\beta\le 2\;,0<\gamma<1?$
The case where $\beta=2$ does not seems difficult because of the identity $(x-y)^2=x^2-2xy-y^2$ but as I am interested for the general case and $\beta=2$ will not lead to any idea so I don't think it is "interesting".


*

*We can't expand the two $\exp$ in power series and switch integrals because the result will not converges.

*I am interested in any nice form i.e. using well known functions as for exemple hypergeometric functions.

*In fact we can replace $\frac{2}{3}$ and $\frac{1}{3}$ by $a-b,b$: so is it possible to get a bound in term of $a,b$ ? It will be sufficient for me as well.
 [/INST] We have that $\frac{2}{3} = a-b$ and $\frac{1}{3} = b$ so we will try to deal with it as generally as possible (since those specific numbers are unlikely to have any symmetry associated with them).
We can convert the integral into polar coordinates and get an integral of the form
$$\int_{\alpha_1}^{\alpha_2} \int_0^\infty e^{-r^\beta f_\beta(\theta)} \frac{r^{1-\gamma}}{|\pm \sin \theta|^\gamma} drd\theta$$
where 
$$f_\beta(\theta) = \begin{cases}
(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta & -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{4} \\
(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta & \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2} \\
\end{cases} $$
There is a symmetry in the regions. There are six distinct angular regions to integrate on but they come in symmetric pairs, so we will only have three angular integrals to do.
Next we will use the substitution $\chi = r^\beta f_\beta(\theta)$ to get
$$\int_{\alpha_1}^{\alpha_2} \int_0^\infty e^{-\chi} \chi^{\frac{2-\gamma}{\beta}-1} \frac{1}{\beta(f_\beta(\theta))^{\frac{2-\gamma}{\beta}}|\pm \sin \theta|^\gamma} d\chi d\theta$$
Using this form, we can say the original expression evaluates to
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) (I_1+I_2+I_3)$$
where
$$\begin{align}
& I_1 = \int_0^{\frac{\pi}{4}} \frac{[(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \\
& I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{[(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \\
& I_3 = \int_0^{\frac{\pi}{2}} \frac{[(a-b)\sin^\beta\theta + b(\sin\theta + \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\cos^\gamma \theta} d\theta \\
\end{align}$$
If anyone has any idea what these last three integrals could possibly evaluate to, let me know.

$\mathbf{\text{EDIT}}$: In lieu of a formal solution to these integrals, we can bound them. Take the min of each of the "numerators"(since they are raised to negative powers): 
$$\begin{align}
& I_1 = \int_0^{\frac{\pi}{4}} \frac{[(a-b)\cos^\beta\theta + b(\cos\theta - \sin\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta  \leq \frac{2^{1-\frac{\gamma}{2}}}{(a-b)^{\frac{2-\gamma}{\beta}}} \int_0^{\frac{\pi}{4}} \frac{1}{\sin^\gamma \theta} d\theta \\
& I_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{[(a-b)\cos^\beta\theta + b(\sin\theta - \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\sin^\gamma \theta} d\theta \leq \frac{2^{\frac{\gamma}{2}}}{(\min(a-b,b))^{\frac{2-\gamma}{\beta}}}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d\theta \\
& I_3 = \int_0^{\frac{\pi}{2}} \frac{[(a-b)\sin^\beta\theta + b(\sin\theta + \cos\theta)^\beta]^{\frac{\gamma-2}{\beta}}}{\cos^\gamma \theta} d\theta \leq b^{\frac{\gamma-2}{\beta}}\int_0^{\frac{\pi}{2}} \frac{1}{\cos^\gamma \theta} d\theta \\
\end{align}$$
For the first two equations, both of the terms in the sum are monotonic, so either of the endpoints has to be the min. For the third integral, this isn't the case, but we can take the min of the two functions separately and them together since everything is positive.
Bounding the remaining integrals with singularities:
$$\int_0^{\frac{\pi}{4}} \frac{1}{\sin^\gamma \theta} d\theta = \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{-\gamma}}{\sqrt{1-x^2}}dx \leq \sqrt{2} \int_0^{\frac{1}{\sqrt{2}}} x^{-\gamma} dx = \frac{2^{\frac{\gamma}{2}}}{1-\gamma} $$
$$\int_0^{\frac{\pi}{2}} \frac{1}{\cos^\gamma \theta} d\theta = \int_0^1 \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx$$
$$ = \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx + \int_{\frac{1}{\sqrt{2}}}^1 \frac{x^{-\gamma}}{\sqrt{1-x^2}} dx \leq \frac{2^{\frac{\gamma}{2}}}{1-\gamma} + 2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi$$
All in all, we have bounded our original integral by
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) \left( \frac{2(a-b)^{\frac{\gamma - 2}{\beta}}+2^{\frac{\gamma}{2}}b^{\frac{\gamma - 2}{\beta}}}{1-\gamma} + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{[\min(a-b,b)]^{\frac{2-\gamma}{\beta}}} + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{b^{\frac{2 - \gamma}{\beta}}} \right)$$

$\mathbf{\text{EDIT}}$: Courtesy of Mathematica, those simplified integrals had closed forms, which we can use to get an even tighter bound:
$$\frac{2}{\beta}\Gamma\left( \frac{2-\gamma}{\beta} \right) \left( \frac{\sqrt{2}(a-b)^{\frac{\gamma - 2}{\beta}}}{1-\gamma}{}_2 F_1\left(\frac{1}{2},\frac{1-\gamma}{2},\frac{3-\gamma}{2},\frac{1}{2} \right) + \frac{2^{\frac{\gamma}{2}-2}\hspace{3 pt}\pi}{[\min(a-b,b)]^{\frac{2-\gamma}{\beta}}} + \frac{\sqrt{\pi}}{2b^{\frac{2 - \gamma}{\beta}}}\frac{\Gamma\left( \frac{1-\gamma}{2} \right)}{\Gamma\left( 1-\frac{\gamma}{2} \right)} \right) $$
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