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)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right) \end{alignat}

How can we solve?

This question is related with this post (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)$), or the solutions in three-dimensional space for this post (https://physics.stackexchange.com/questions/61498/a-four-dimensional-integral-in-peskin-schroeder#)

Thank you so much


$$\int \frac{1}{(u+v)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right)\,dv=-\frac{\sqrt{\pi } }{\sqrt{x}}\,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{u+v}}\right)$$

$$\int_0^{\infty} \frac{1}{(u+v)^{\frac{3}{2}}}\exp\left(-\frac{x}{u+v}\right)\,dv=\frac{\sqrt{\pi }}{\sqrt{x}}\,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)$$

$$\int \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=(u+2 x)\, \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)+\frac{2 \sqrt{u} \sqrt{x} }{\sqrt{\pi }} e^{-\frac{x}{u}}$$ $$\int_0^M \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=(M+2 x) \,\text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)+\frac{2 \sqrt{M} \sqrt{x} }{\sqrt{\pi }}e^{-\frac{x}{M}}-2 x$$ which does not converge if $M\to \infty$. $$I_2(x)=\frac{\sqrt{\pi }}{\sqrt{x}}\int_0^M \text{erf}\left(\frac{\sqrt{x}}{\sqrt{u}}\right)\,du=2 \sqrt{M} e^{-\frac{x}{M}}-2 \sqrt{\pi } \sqrt{x}+\frac{\sqrt{\pi } (M+2 x) }{\sqrt{x}}\text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)\tag 1$$ Using series, we should end with $$I_2(x)=4 \sqrt{M}-2 \sqrt{\pi } \sqrt{x}+\frac{4 x}{3 \sqrt{M}}-\frac{2 x^2}{15 M\sqrt{M}}+O\left(x^3\right)$$


Concerning the derivatives of $I_2(x)$, using $(1)$, we should have $$\frac{d}{dx} I_2(x)=\frac{\sqrt{M} e^{-\frac{x}{M}}}{x}+\frac{\sqrt{\pi } \left((M-2 x) \text{erfc}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)-M\right)}{2 x^{3/2}}$$

$$\frac{d^2}{dx^2} I_2(x)=-\frac{3 \sqrt{M} e^{-\frac{x}{M}}}{2 x^2}+\frac{\sqrt{\pi } \left((3 M-2 x) \text{erf}\left(\frac{\sqrt{x}}{\sqrt{M}}\right)+2 x\right)}{4 x^{5/2}}$$

Expanding as series around $x=0$, this would lead to $$\frac{d}{dx} I_2(x)=-\frac{\sqrt{\pi }}{\sqrt{x}}+\frac{4}{3 \sqrt{M}}-\frac{4 x}{15 M^{3/2}}+O\left(x^{2}\right)$$ $$\frac{d^2}{dx^2} I_2(x)=\frac{\sqrt{\pi }}{2 x^{3/2}}-\frac{4}{15 M^{3/2}}+\frac{4 x}{35 M^{5/2}}+O\left(x^{2}\right)$$

  • $\begingroup$ Thanks you very much. I would like to ask, if the above results on the $x$ derivative, this is reasonable? $\endgroup$ – The Deer Hunter May 14 '17 at 13:48
  • $\begingroup$ @Samurai. I am not sure I understand. Do you want the asymptotics of $\frac{d}{dx}I_2(x)$ ? Tell me if this is the case. I should look at it tomorrow morning (here it is almost 5:00pm French time). Cheers. $\endgroup$ – Claude Leibovici May 14 '17 at 14:25
  • $\begingroup$ nice answer @ClaudeLeibovici (+1) $\endgroup$ – tired May 14 '17 at 16:36
  • $\begingroup$ strange i get a leading term $M \sqrt{\pi/x}$.. i will come back later (first some sports..) $\endgroup$ – tired May 14 '17 at 16:51
  • 1
    $\begingroup$ @tired. You are right : first, some sports ! Second : may be some mistakes on my side (I shall redo the work tomorrow morning - it is 8:00pm here). Third : I shall wait for your work. Cheers :-) $\endgroup$ – Claude Leibovici May 14 '17 at 17:51

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