Showing the following two integrals are equal I would like to show that
$$\int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+x}} dt = 2\int_{0}^{\infty}e^{-t^{2}-2t\sqrt{x}}dt,\quad  x>0.$$
I haven't been able to have very much success with this integral. So far I have made a couple observations:
1) I think there is an obvious step of completing the square on the RHS.
2) The presence of the 2 and a Gaussian like integral suggests to me that there is some way to integrate the RHS by extending the limits of integration to all of $\mathbb{R}$ and then using a trick similar to how a Gaussian is integrated using polar coordinates.
Aside from these observations, I have made no progress. In particular, I am find the $\sqrt{t+x}$ term on the LHS hard to be especially difficult to handle.

transform the integral on the left:
$$
\int_{0}^{\infty} \frac{e^{-t}}{\sqrt{t+x}} dt =e^x\int_{0}^{\infty} \frac{e^{-(t+x)}}{\sqrt{t+x}} dt 
$$
now with $u^2=t+x$, so $dt=2udu$
$$
=2e^x\int_{\sqrt{x}}^{\infty} e^{-u^2} du 
$$
now the integral on the right:
$$
2\int_{0}^{\infty}e^{-t^{2}-2t\sqrt{x}}dt =  2e^x\int_{0}^{\infty}e^{-(t-\sqrt{x})^2}dt
$$
now with $u=t-\sqrt{x}$ this becomes:
$$
=  2e^x\int_{\sqrt{x}}^{\infty}e^{-u^2}du
$$
