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.

This is not an answer, so please don't down-vote this.  Nevertheless, here's how this problem can be solved using computer algebra.  In Mathematica:
Assuming[Re[x] > 0, 
  Integrate[Exp[-t]/Sqrt[t + x], {t, 0, \[Infinity]}]] == 
 Assuming[Re[x] > 0, 
  2 Integrate[Exp[-t^2 - 2 t Sqrt[x]], {t, 0, \[Infinity]}]]

(* True *)
