$\int \frac{e^{-x}}{\sqrt{x}}\,\mathrm dx$

Question

I have been working on this problem but I am not sure in which direction to head with this. Am I supposed to do an integration by parts first?

$$\int \frac{e^{-x}}{\sqrt{x}}\,\mathrm dx$$

This would yield: $$ \begin{align} u&=x^{-1/2}\\ v&=-e^{-x}\\ \mathrm du&=-\frac23x^{-3/2}\\ \mathrm dv&=e^{-x}\,\mathrm dx \end{align} $$

But then what?

Answer 1

Since the problem is tagged improper-integrals, I'll assume the bounds are $0$ and $\infty$.

Hint: Using an appropriate $u$-substitution, we find $$\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,dx=2\int_0^\infty e^{-u^2}\,du=\sqrt{\pi}.$$