$\sin ix$ integral identity If you plug in $\sin ix$ to wolframalpha, you get this really weird integral identity back:
$$\sin(ix) = \frac{x}{4\sqrt{\pi}}\int_{-i\infty+\gamma}^{i\infty+\gamma}\frac{e^{s+\frac{x^2}{4s}}}{s^{3/2}}ds, \ \gamma > 0$$ 
There is a whole package of weirdness contained here, from super weird  and ambiguous notation (it feels like something Euler would write) on the limits of integration to that the answer is independent of $\gamma$ but only holds for its positive values (something to do with residues?). Even disregarding all that, it's still looks like a very unexpected equation. Does anyone know where this originates from?
 A: Take the definition of inverse Laplace transform
$$
f(t)=\mathcal{L}^{-1}\{F(s)\}(t)=\frac{1}{2\pi\mathrm{i}}\lim_{T\to\infty}\int_{\gamma-\mathrm{i}T}^{\gamma+\mathrm{i}T}e^{st} F(s)ds\ ,
$$
meaning that you are integrating over a vertical line $\mathrm{Re}(s)=\gamma$ in the complex plane such that $\gamma$ is greater than the real part of all the singularities of $F$.
Comparing with
$$
\frac{x}{4\sqrt{\pi}}\int_{-i\infty+\gamma}^{i\infty+\gamma}\frac{e^{s+\frac{x^2}{4s}}}{s^{3/2}}ds, \qquad \gamma > 0\ ,
$$
you may identify this as the inverse Laplace transform of a function
$$
F(s;x)=\frac{2\pi\mathrm{i}x}{4\sqrt{\pi}}\frac{e^{\frac{x^2}{4s}}}{s^{3/2}}\ ,
$$
evaluated at $t=1$.
Using the integral
$$
\int_0^\infty dt\ e^{-s t}\frac{2 \sinh \left(\sqrt{t} x\right)}{\sqrt{\pi } x}=\frac{e^{\frac{x^2}{4 s}}}{s^{3/2}}\qquad s>0\ ,
$$
we identify the sought inverse Laplace transform as
$$
f(t;x)=\frac{2\pi\mathrm{i}x}{4\sqrt{\pi}}\frac{2 \sinh \left(\sqrt{t} x\right)}{\sqrt{\pi } x}\ ,
$$
which, evaluated at $t=1$ yields
$$
f(1;x)=\mathrm{i}\sinh(x)=\sin(\mathrm{i}x)\ .
$$
