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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.

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    $\begingroup$ As written, this double integral is divergent. $\endgroup$ – Olivier Oloa Dec 19 '16 at 9:39
  • $\begingroup$ @Olivier: Yes. I want to calculate the asymptotic form as $x\to 0$. $\endgroup$ – GotchaP Dec 19 '16 at 9:43
  • $\begingroup$ Are you interested in an asymptotics of $\int_0^M du\int_0^{\infty}dv\, \frac{1}{(u+v)^2}\exp\left(-\frac{x}{u+v}\right)$ as $x \to 0^+$? $\endgroup$ – Olivier Oloa Dec 19 '16 at 9:48
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    $\begingroup$ @Olivier: Yes I am. $\endgroup$ – GotchaP Dec 19 '16 at 10:10
  • $\begingroup$ Ok, please see my answer below. $\endgroup$ – Olivier Oloa Dec 19 '16 at 10:56
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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 ($a=0=-n$).

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  • $\begingroup$ $$ \frac{1}{x}\left( M-e^{-\frac{x}{M}}M \right)=1-\frac{1}{2}\frac{x}{M}+\frac{1}{6}\frac{x^2}{M^2}+\cdots$$ And from 8.4.15 of this, $$ \Gamma(0,\frac{x}{M})=-\gamma -\log \frac{x}{M} +\left( \sum_{k=1}^{\infty}\frac{1}{k!}\frac{1}{k} \right)\frac{x}{M} $$ Right? $\endgroup$ – GotchaP Dec 20 '16 at 12:27
  • $\begingroup$ When I printed the web page, $(-z)^k$ wasn't correctly printed. Then the correct expression is $\Gamma(0,\frac{x}{M})=-\gamma -\log \frac{x}{M} +\left( \sum_{k=1}^{\infty}\frac{1}{k!}\frac{1}{k} \frac{(-x)^k}{M^k}\right)$. $\endgroup$ – GotchaP Dec 20 '16 at 12:44
  • $\begingroup$ @GotchaP Yes, this time it's Ok ;) $\endgroup$ – Olivier Oloa Dec 20 '16 at 12:46
  • $\begingroup$ The correct result is $$-\log x-\gamma-1+\log M+\frac{1}{2}\frac{x}{M}-\frac{x^2}{12 M^2}+O(x^3)$$ $\endgroup$ – GotchaP Dec 21 '16 at 0:58
  • $\begingroup$ @GotchaP I don't understand your $\color{red}{-}1$. The first term in the expansion of $\frac{1}{x}\left( M-e^{-\frac{x}{M}}M \right)$ as $x \to 0$ is $1$, not $\color{red}{-}1$. $\endgroup$ – Olivier Oloa Dec 21 '16 at 1:03

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