I try to calculate by fixing $v$ first: $$ I=\iint\limits_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{e^{-\frac{u}{v^2}}\mathrm{d}v\mathrm{d}u}=\int_0^1{\left( \int_0^1{\left( e^{-\frac{1}{v^2}} \right) ^u\mathrm{d}u} \right) \mathrm{d}v}=\int_0^1{\left( -v^2\left( e^{-\frac{1}{v^2}} \right) ^u \right)\Bigg|_{u=0}^{u=1}\mathrm{d}v}=\int_0^1{-v^2\left( e^{-\frac{1}{v^2}}-1 \right) \mathrm{d}v}. $$
Now the problem is $\int_0^1 v^2 e^{-(1/v^2)}\mathrm{d}v$. I've tried to make a substitution $s=1/v$, but it simply doen't work. It seems to me that the integral is not calculable... Does anyone have an idea about this? Thank you in advance!
\begin{align} I&=\int_1^{\infty} t^{-2} \int_0^1 e^{-ut^2}\,dudt\tag{$t=v^{-1}$}\\ &=\int_1^{\infty} t^{-4} \int_0^{t^2} e^{-s}\,dsdt\tag{$s=ut^{2}$}\\ &=\int_1^{\infty} t^{-4}\left(1-e^{-t^2}\right)\,dt\\ &=\frac13 - \frac{1}{3e}+\frac23\int_1^{\infty} t^{-2}e^{-t^2}\,dt\tag{Integration by parts}\\ &=\frac13 - \frac{1}{3e}+\frac2{3}\left(\frac1e -\sqrt\pi\operatorname{erfc}(1)\right)\tag{Integration by parts}\\ &=\frac13 + \frac{1}{3e}-\frac{2\sqrt\pi}{3}\operatorname{erfc}(1), \end{align} where $\operatorname{erfc}$ is the complementary error function.