The Gaussian integral is
$$ \begin{align} \sqrt{\pi} &= \int_{-\infty}^{\infty} e^{-x^2}\,\mathrm d x \\ &= \int_{0}^{\infty} 2\, e^{-x^2}\,\mathrm d x \end{align} $$
since the integrand is even.
We also have $\Gamma(1/2) = \sqrt \pi$, so with the definition of the Gamma function it follows that
$$ \int_{0}^{\infty} \left( 2\, e^{-x^2} - \frac{e^{-x}}{\sqrt x} \right) \,\mathrm d x \,\,\, = \,\,\, 0 $$
although the integrand is nonzero almost everywhere.
My question: Can this be easily derived by looking at the last integral alone? Is there a transformation that turns the integral into something that is "obviously" zero?
Hint : try this variable change $u=x^2$ on the gaussian integrale