Evaluate this double integral

Question

Evaluate $\displaystyle\int_{0}^{\infty}\int_{0}^{u}ue^{-u^2/y}\,dy\,du$

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How do I solve this double integral. I also tried reversing the order but all in vain. Even after reversing the order I am unable to solve this

Answer 1

The integral is over $(y,u)$ such that $0 \le y \le u < \infty$. So reversing the order gives $$\int_{0}^{\infty}\int_{y}^{\infty}ue^{-u^2/y}\,du\,dy.$$

For the inner integral, substitute $v = \dfrac{u^2}{y}$, $dv = \dfrac{2u}{y}\,du$ to get $$\int_{0}^{\infty}\int_{y}^{\infty}\dfrac{y}{2}e^{-v}\,dv\,dy.$$

Now, the inner integral should be easy to evaluate. Once this is done, you should be left with something like $\int_{0}^{\infty} \text{const} \cdot ye^{-y}\,dy$, which can be evaluated via integration by parts.