Strange integrand: $\int \sqrt{x} e^{x}dx$

Question

Is it possible, and if yes, how, to evaluate an integral like $\int \sqrt{x} e^{x}dx$? I have heard of the Gaussian function which integrates to $\sqrt{\pi}$ but what about this? Thank you.

Answer 1

That indefinite integral is not an elementary one: in fact using the substitution $y=\sqrt{x}$ and integrating by parts you obtain: $$\begin{split} \int \sqrt{x}\ e^x\ \text{d}x &\stackrel{x=y^2}{=} \int 2y^2\ e^{y^2}\ \text{d} y \\ &= \int y\ \text{d} \left( e^{y^2}\right)\\ &= y\ e^{y^2} -\int e^{y^2}\ \text{d} y \end{split}$$ where the latter integral is not elementary.

Answer 2

To get $\sqrt{\pi}$ you need to integrate from $0$ to $\infty$, and you need to replace $e^x$ by $e^{-x}$. Specifically, we have $$\int_0^\infty \sqrt{x}e^{-x}dx =\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}.$$

See the following answer for a variety of different proofs: $\int_{-\infty}^{+\infty} e^{-x^2} dx$ with complex analysis