determine whether $\int_0^\infty e^{-ax^2}dx$ converges or diverges

Question

How would you determine whether $\int_0^\infty e^{-ax^2}dx$ converges or diverges for some real constant $a$ such that $a>0$

I've tried comparing the function to $e^{-x^2}$ but $ e^{-ax^2}$ is only smaller then $e^{-x^2}$ when $a<1$

Answer 1

If $a>0$ then $e^{-ax^2}$ is positive on $\mathbb{R}^+$ but it is bounded by $1$ on $[0,1]$ and by $e^{-ax}$ on $[1,+\infty)$, hence $$\int_{0}^{+\infty} e^{-ax^2}\,dx \leq 1+\int_{1}^{+\infty}e^{-ax}\,dx = 1+\frac{1}{a e^a}.$$ Alternative way: the LHS is just a constant times $\frac{1}{\sqrt{a}}$ by a straightforward substitution.

Answer 2

Use the limit comparison test. Indeed $$ \exp(-ax^2)/x^{-2}\to0 $$ as $n\to\infty$ whence there exists an $N$ such that $x\geq N$ implies that $\exp(-ax^2)<x^{-2}$. Thus $$ \int_{1}^\infty \exp(-ax^2)<\infty $$ by comparison with $x^{-2}$ and hence your integral converges too (as obviously the integral on $[0,1]$ is finite.