Does $\int_{-\infty}^{\infty} e^{-\sqrt{|x|}}\,\mathrm{d}x$ converge?

Question

I would like to find out if this integral converges: $$\int_{-\infty}^{\infty} e^{-\sqrt{|x|}}\,\mathrm{d}x$$

Since this is a symmetric function I figured I could focus on only one side of the integral, namely

$\displaystyle\int_{0}^{\infty} e^{-\sqrt{|x|}}\,\mathrm{d}x$ which in this case is equivalent to $\displaystyle\int_{0}^{\infty} e^{-\sqrt{x}}\,\mathrm{d}x$ (since $|x| = x$ when $x > 0$)

Also, $e^{-\sqrt{x}}$ is bounded from 0 to 1 meaning the integral there is a constant, so I will use the integral from 1 to $\infty$.

I know this converges (checked with a calculator) but cannot seem to find an argument for the comparison test to say that since $e^{-\sqrt{x}} < $ "some other function which converges" for $x > 1$, thus $\displaystyle\int_1^{\infty} e^{-\sqrt{x}}\,\mathrm{d}x$ converges.

In other words, I need a function which is always greater than $e^{-\sqrt{x}}$ and whose integral converges. I know that $e^{-x}$ and $e^{-2x}$ both converge, but these are both smaller than $e^{-\sqrt{x}}$ for $x > 1$.

Tips would be appreciated. Thank you.

Answer 1

Hint: for $x > 75$, $\ln(x^{2}) < \sqrt{x}$.

Answer 2

Note that $$\int_0^A e^{-\sqrt x}dx=\int_0^{\sqrt A}2ue^{-u}du$$

Answer 3

Using L'Hospital, we get $$ \begin{align} \lim_{x\to\infty}xe^{-x} &=\lim_{x\to\infty}\frac{x}{e^x}\\ &=\lim_{x\to\infty}\frac1{e^x}\\[4pt] &=0 \end{align} $$ Therefore, $$ \begin{align} \lim_{u\to\infty}\frac14\sqrt{u}e^{-\frac14\sqrt{u}}&=0&&\text{substitute $x=\frac14\sqrt{u}$}\\ \lim_{u\to\infty}\frac1{256}u^2e^{-\sqrt{u}}&=0&&\text{raise to the $4^{\text{th}}$ power}\\ \lim_{u\to\infty}u^2e^{-\sqrt{u}}&=0&&\text{multiply by $256$}\\ \end{align} $$ Thus, for all $x$, $e^{-\sqrt{|x|}}\le1$, and for $x$ sufficiently large, i.e. $|x|\ge16\,\mathrm{W}_{\!-1\!}\left(-\frac14\right)^2\approx74.186688$, $e^{-\sqrt{|x|}}\le\frac1{x^2}$. This implies that $$ \int_{-\infty}^\infty e^{-\sqrt{|x|}}\,\mathrm{d}x $$ converges.