Asymptotics of a double integral: $ \int_0^{\infty}du\int_0^{\infty}dv\, \frac{1}{(u+v)^2}\exp\left(-\frac{x}{u+v}\right)$

Question

I want to calculate the asymptotic form as $x \to 0$ of the following integral. \begin{alignat}{2} I_2(x) &=&& \int_0^{\infty}du\int_0^{\infty}dv\, \frac{1}{(u+v)^2}\exp\left(-\frac{x}{u+v}\right) \\ &=&& \frac{\partial^2}{\partial x^2} \int_0^{\infty}du\int_0^{\infty}dv\, \exp\left(-\frac{x}{u+v}\right) \end{alignat} How can we solve?
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Thanks.

Answer 1

Assume $M\ge1, x>0$. One may obtain, as $x \to 0^+$,

$$ \int_0^M\!\!\int_0^\infty \frac{e^{\Large -\frac{x}{u+v}}}{(u+v)^2} \:du\:dv=-\log x-\gamma+1+\log M+\frac{x}{2M}-\frac{x^2}{12 M^2}+O(x^3). $$ where $\gamma$ is the Euler-Mascheroni constant.

Hint. The above integrand is positive, one may then write $$ \begin{align} \int_0^M\!\int_0^\infty \frac{e^{\Large -\frac{x}{u+v}}}{(u+v)^2} \:du\:dv&=\int_0^M\left(\int_0^\infty \frac{e^{\Large -\frac{x}{u+v}}}{(u+v)^2} \:du\right)dv \\&=\int_0^M\left[\frac{e^{-\frac{x}{u+v}}}{x}\right]_0^\infty dv \\&=\frac1x\int_0^M\left(1-e^{-\large\frac{x}{v}}\right) dv \\&=\frac1x\int_{1/M}^\infty\left(1-e^{-\large xt}\right)\frac{dt}{t^2} \\&=\frac1x \left(M-e^{-\frac{x}{M}} M+x\: \Gamma\left(0,x/M\right)\right) \\&=-\log x-\gamma+1+\log M+\frac{x}{2M}-\frac{x^2}{12 M^2}+O(x^3) \end{align} $$ where we have used the incomplete gamma function $$ \Gamma(a,z):=\int_z^\infty t^{a-1}e^{-t}dt $$ and its asymptotics expansion ($a=0=-n$).