# Getting rid of square root via integration

How do we prove, for positive $D$, this result?

$$e^{-2\sqrt D} \sqrt{\pi} = \int_0^\infty s^{-1/2} e^{-(s+D/s)} ds$$

Let $x=y^{2}$ \begin{align} \int\limits_{0}^{\infty} \frac{1}{\sqrt{x}} \mathrm{e}^{-(x+D/x)} dx &= 2\int\limits_{0}^{\infty} \mathrm{e}^{-(y^{2}+D/y^{2})} dy \\ &= \sqrt{\pi} \mathrm{e}^{-2\sqrt{D}} \end{align}
To evaluate the $y$ integral, we used the Cauchy-Schlomilch Transformation, Theorem 2.1 and the actual $y$ integral is evaluated in Example 3.1, equation 3.3.