Since $e^{-x^{2}/2}$ is an even function. $\int_{0}^{\infty}e^{-x^{2}/2} dx=\sqrt{\pi/2}.$
Similarly, as $|x|^{1/2}e^{|x|}$ an even function. $\int_{-\infty}^{\infty}|x|^{1/2}e^{|x|} dx=2\int_{0}^{\infty}\sqrt{x}e^{x} dx$
$\mathbf{My \ attempt}$
Let $x=t^{2}/2$ $\Rightarrow dx=t\ dt$
Therefore, $2\int_{0}^{\infty}\sqrt{x}e^{x} dx=\sqrt{2}\int_{0}^{\infty}t^{2}e^{t^{2}/2}dt$.
Which is where I get stuck. I believe there's a misprint in the question. The power of the $|x|$ should be $-\frac{1}{2}$ and not $\frac{1}{2}$.
Because the I'll get $2\int_{0}^{\infty}\sqrt{x}e^{x} dx=\sqrt{2}\int_{0}^{\infty}e^{t^{2}/2}dt=\sqrt{2} \cdot \sqrt{\pi/2}=\sqrt{\pi}$.
But I may be wrong. Please point me in the right direction. Thanks in advance.