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

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

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