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

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?

Please use $\LaTeX$ mode by putting equations within $ ... $. — bluenote10, Nov 26 '13 at 16:38
It's an incomplete Gamma function. It doesn't have a closed form. http://en.wikipedia.org/wiki/Incomplete_gamma_function — Stephen Montgomery-Smith, Nov 26 '13 at 16:39
@StephenMontgomery-Smith: I'd say that whether or not it has a closed form depends on the bounds... See my answer. — Clayton, Nov 26 '13 at 16:46

score 3 · Answer 1 · answered Nov 26 '13 at 16:45

3

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

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

answered Nov 26 '13 at 16:45

Clayton

1 Answers1