The integral I'm working on is $$I(x)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}ds\frac{1}{s}e^{x(s-s^{1/2})}\ .$$
I'm told the path of integration is a vertical line to the right of the origin.
First I need to find the branch point, which I'm not sure about. My guess is $s=0$, though that is also a first order pole.
Then I need to find a branch cut which doesn't intersect the path of integration. If the branch point is at $s=0$, I think the cut would go up the imaginary axis.
Then I need to evaluate $I(x)$ assuming $x<0$. I'm not sure how to tackle this at all, especially since I don't know where the branch points/cuts are.
Any help?
There is a "closed" form solution to this integral in terms of the error function. For $x>0$ we find $$ I(x)=1-\text{Erf}\left(\frac{\sqrt{x}}{2}\right), $$ whereas $I(x)=0$ for $x<0$.
Proof:
First of all the $s=0$ point is a branch point and a first order pole at the same time.
This integral is the inverse Laplace transform (ILT) of the function $\tilde I(s)=e^{-x\sqrt{s}}/s$, which is analytic everywhere except (i) at $s=0$ and (ii) the chosen branch cut of the square root. Taking the principal branch of the square root the cut is on the negative real line (connecting the two branch points $s=0$ and $s=\infty$).
1) For $x<0$ you can close the Bromwich contour on the right half plane with an infinitely large semicircle and since the function is analytic inside this closed contour the integral is zero due to Cauchy's theorem, $I(x)=0$. Note that this is not surprising since in the theory of (one-sided) Laplace transforms we consider functions that are by definition zero on the negative reals.
2) For $x>0$ we can deform the Bromwich contour into the keyhole contour (the infinite arcs will give zero contribution as their radius $R$ goes to infinity). The integral on the keyhole contour reads \begin{align} I(x)&=\frac{1}{2\pi i}\int_{-\infty}^{-\delta} \frac{du}{u-i\epsilon}e^{xu}e^{-x\sqrt{u-i\epsilon}}- \frac{1}{2\pi i}\int_{-\infty}^{-\delta} \frac{du}{u+i\epsilon}e^{xu}e^{-x\sqrt{u+i\epsilon}}\\ &+\frac{1}{2\pi}\int_{-\pi}^{\pi}d\phi\exp(x[\delta e^{i\phi}-\sqrt{\delta}e^{i\phi/2}]). \end{align} The first integral runs under the negative real axis ($\epsilon\to0+$), the second runs slightly above the real line and the last term accounts for the integral around the origin on a small circle of radius $\delta\to0+$. It is important to note that in the keyhole contour $\epsilon$ and $\delta$ are not independent: $0<\epsilon\le\delta$. Therefore we have to first take the limit $\epsilon\to0+$ and then $\delta\to0+$. The third term yields 1 in this limit while the first two terms can be rewritten as \begin{align} I(x)&=1+\frac{1}{\pi}\int_\delta^\infty dt e^{-xt}\cos(x\sqrt{t})\frac{\epsilon}{t^2+\epsilon^2}- \frac{1}{\pi}\int_\delta^\infty dte^{-xt}\sin(x\sqrt{t})\frac{t}{t^2+\epsilon^2}, \end{align} where we used $\sqrt{u\pm i\epsilon}=\pm i\sqrt{-u}$ for $u<0$ and then applied the $u=-t$ substitution. If we now take $\epsilon\to0+$ in the second term the integrand will contain a delta function, $\pi\delta(t)$, but since $t=0$ is not in the domain of integration this integral is just zero (note, this was a point where the order of limits was crucial!). In the last term we can safely take both limits (order does not matter) because the resulting $1/\sqrt{t}$ singularity is integrable and we get \begin{align} I(x)&=1-\frac{1}{\pi}\int_0^\infty\frac{dt}{t}e^{-xt}\sin(x\sqrt{t})\\ &=1-\text{Erf}\left(\frac{\sqrt{x}}{2}\right). \end{align} The last integral can be looked up from tables or done with Mathematica.