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.
 A: Enforcing the substitution $t=(s+\sqrt x)^2-x$ in the first integral, then we have
$$\begin{align}
\int_0^\infty\frac{e^{-t}}{\sqrt{t+x}}\,dt&=\int_0^\infty \frac{2(s+\sqrt x)e^{-(s+\sqrt x)^2+x}}{s+\sqrt x}\,ds\tag 1\\\\
&=2\int_0^\infty e^{-s^2-2s\sqrt x}\,ds\\\\
&=2\int_0^\infty e^{-t^2-2t\sqrt x}\,dt\\\\
\end{align}$$
as was to be shown!

The OP had asked a good question in a comment after the initial post regarding the implication of the alternative substitution $s+\sqrt x=-\sqrt{t+x}$ instead of $s+\sqrt x =\sqrt{t+x}$ as used in arriving at $(1)$.  
So, let's see what happens in this alternative substitution.  Instead of $(1)$, we would have obtained
$$\begin{align}
\int_0^\infty\frac{e^{-t}}{\sqrt{t+x}}\,dt&=\int_{-2\sqrt x}^{-\infty} \frac{2(s+\sqrt x)e^{-(s+\sqrt x)^2+x}}{-(s+\sqrt x)}\,ds\tag 2\\\\
&=2\int_{-\infty}^{-2\sqrt x}e^{-(s+\sqrt x)^2+x}\tag 3
\end{align}$$
Then, substituting $s=-2\sqrt x-t$ into $(3)$ yields
$$\begin{align}
\int_0^\infty\frac{e^{-t}}{\sqrt{t+x}}\,dt&=2\int_0^\infty e^{-(t+\sqrt x)^2+x}\,dt\\\\
&=2\int_0^\infty e^{-t^2-2t\sqrt x}\,dt
\end{align}$$
as expected!
A: This isn't so much of an answer, as showing something kind of nice.
Let's let both these functions be represented as $f(x)$.
Now, the first is $f(x) = \int_0^\infty \frac{e^{-t}}{\sqrt{t-x}}dt$.
Then, we see that $f(0) = t^{-1/2}*e^{-t}dt = \sqrt{\pi}$, as this is the gamma function.
Now, the second function is $g(x) = 2 \int_0^\infty e^{-t^2-2t\sqrt{x}}dt$, and at $g(0)$, this becomes $2 \int_0^\infty e^{-t^2}dt$ which also happens to be $\sqrt{\pi}$, as it is one of the steps in solving the gamma function.
Therefore, there is at least one point where they are equal :)
A: 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 *)
A: 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:
$$
=  2e^x\int_{\sqrt{x}}^{\infty}e^{-u^2}du
$$
A: I know that an answer has been accepted already, but I'd like to present an alternate solution nonetheless.
We have the equation as follows:
$$\int_0^\infty{e^{-t}\over \sqrt{t+x}}dt = 2\int_0^\infty e^{-t^2-2t\sqrt x}dt$$
First, we multiply both sides by $e^{-x}$.
$$\int_0^\infty e^{-t-x} * (t+x)^{-1/2} dt = 2\int_0^\infty e^{-t^2-2t\sqrt x - x}dt$$
Which simplifies to 
$$\int_0^\infty e^{-(t+x)} * (t+x)^{-1/2} dt = 2\int_0^\infty e^{-(t+\sqrt x)^2}dt$$
On the LHS, we do a substitution $u = t+x, du = dt$ and the lower bound becomes $x$. On the RHS, we do a substitution $s = t + \sqrt x, ds = dt$ and the lower bound becomes $\sqrt x$.
$$\int_x^\infty e^{-(u)} * (u)^{-1/2} du = 2\int_{\sqrt x}^\infty e^{-(s^2)}ds$$
Note that these are equal to 
$$ \Gamma (1/2) - \int_0^x e^{-(u)} * (u)^{-1/2} du = \Gamma (1/2) -2\int_{0}^{\sqrt x}e^{-(s^2)}ds$$
Therefore it suffices to take find if the second and fourth term are equal; Simplifying and changing the variable again on the LHS with $p^2 = u, du = 2pdp$, the bounds change to $0$ and $\sqrt x$ we get
$$\int_{0}^{\sqrt x} e^{-(p^2)} * (p^2)^{-1/2} * 2p dp = 2\int_{0}^\sqrt x e^{-(s^2)}ds$$
Which simplifies to 
$$2\int_{0}^{\sqrt x} e^{-(p^2)} dp = 2\int_{0}^\sqrt x e^{-(s^2)}ds$$
Therefore, the two integrals are equal!
