express as contour integral $ f(x) = \int_0^\infty dt \; e^{-t/g} \; \frac{1}{\sqrt{1 - 2 t x}} $ Let $0 < x < 1$, I have to compute this Laplace transform:
$$ f(x) = \int_0^\infty dt \; e^{-t/g} \; \frac{1}{\sqrt{1 - 2 t  x}} $$
I am not 100% this interal is defined.  If $t > \frac{1}{x}$ this integral is imaginary.  If we ignore that issue, let $s = tx$ and $ds = x \, dt$.
$$ \frac{1}{x} \int_0^\infty ds \; e^{-s/gx} \; \frac{1}{\sqrt{1 - 2 s}} $$
This is not going to work since $1-2s < 0$ for $s > \frac{1}{2}$ and the square root becomes negative.  In fact as $s$ goes from $0$ to infinity, $\frac{1}{\sqrt{1-2s}}$ goes from $1 \to \infty$ along the real axis and $\infty \to 0$ along the imaginary axis.
The circle is finished by another function: $\frac{1}{\sqrt{1+2s}}$ which moves from $1 \to 0$ along the real axis.

How about if we tried a complex integral:
$$\int_0^\infty dt \; e^{-t/g} \left[ \frac{1}{\sqrt{1 - (2t+i\epsilon)x}} - \frac{-i e^{\frac{1}{2g}} }{\sqrt{1 - (-2t + i\epsilon)x}} \right]$$
Can this be expressed as a complex integral with a simple pole?  Then I can take the residue.  The $\epsilon > 0$ is just to avoid branch cut issues.
The answer should simply be $e^{-\frac{x}{2g}}$
 A: Ok, as it stands the integral is not well defined. To give it a meaning  we somehow need to regularize it. One way of doing that is to define the value of the integral as ($\delta\rightarrow 0_+$)
$$
I\equiv\frac{1}{2}\left(\int_0^{\infty+i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt+\int_{\infty-i\delta}^0e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt\right)
$$
This somehow resembles the definition of the principal value commonly used in complex analysis $P\frac{1}{x}=\frac{1}{2}\left(\frac{1}{x+i\delta}+\frac{1}{x-i\delta}\right)$ extended to the case of a non isolated singularity (a branch cut). Furthermore we may interpret this as employing the standard time ordering procedure of Quantum Field Theory (The above is somehow equivalent to add a retared and advanced ''propagator'' which gives a ''time orderd'' propagator)
Now, because the square root is well defined for $0<t<x/2$ this part of the integrals will cancel out for sure, and we are left with
$$
I=\frac{1}{2}\left(\int_{x/2+i\delta}^{\infty+i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt-\int_{x/2-i\delta}^{\infty-i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt\right)
$$
Taking the principal branch of the logarithm we may obtain
$$
I=i\int_{x/2}^{\infty}e^{-t/g}\frac{1}{\sqrt{2xt-1}}
$$
This integral is quiet standard to solve taking $2xt-1=q^2$ we obtain 
$$
I=\frac{i}{x}e^{-1/(2gx)}\int_{0}^{\infty}e^{-q^2/(2gx)}dq
$$
which is a standard Gaussian integral yielding
$$
I=\frac{i \sqrt{\pi g}}{\sqrt{2x}}e^{-1/(2gx)}
$$
