Let's say we have
$\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}dx$.
I'd like to find the antiderivative and then analyze the integral going from 0 to N as N goes to infinity. However, I don't know how to find this particular primitive, and the online calculators do something that we haven't gone over in class (they use the Gauss error function), which leads me to suspect that there's another way of doing this.
Does anyone have an idea? Any help is appreciated, thank you.
Finding an antiderivative is going to be difficult. Seeing if it converges will not, $$ \int_0^\infty \frac{e^{-x}}{\sqrt{x}}\mathrm dx= \int_0^1 \frac{e^{-x}}{\sqrt{x}}\mathrm dx+\int_1^\infty \frac{e^{-x}}{\sqrt{x}}\mathrm dx $$ Now, for $0\leq x\leq 1$, we have $$ \frac{e^{-x}}{\sqrt{x}}\leq \frac{1}{\sqrt{x}} $$ and for $1\leq x$, we have $$ \frac{e^{-x}}{\sqrt{x}}\leq e^{-x} $$
Alternatively, enforce the substitution $u=\sqrt{x}$, then your integral is $$ \int_0^\infty \frac{e^{-x}}{\sqrt{x}}\mathrm dx= 2\int_0^\infty e^{-u^2}\mathrm du=\sqrt{\pi} $$