Closed form of, or series for $\int_{\epsilon-i\infty}^{\epsilon+i\infty}\frac{e^{az+b^2z^2}}{\sin\pi z}\,dz$ I've been trying to find a closed form expression/series expansion for the following integral without success:
$$F(a,b)=\int_{\epsilon-i\infty}^{\epsilon+i\infty} e^{az+b^2z^2}\Gamma(z)\Gamma(1-z)\,dz=\pi\int_{\epsilon-i\infty}^{\epsilon+i\infty}\frac{e^{az+b^2z^2}}{\sin\pi z}\,dz$$
for some $\epsilon\in(0,1)$. Any input is greatly appreciated!
 A: Integrals of this type are systematically studied by L. J. Mordell. One of them is $$I(z,\tau)=\int_{-\infty}^\infty\frac{\exp(i\pi\tau w^2-2zw)}{\exp(2\pi w)-1}\,dw,\qquad(z,\tau\in\mathbb{C},\ \Im\tau>0)$$ where the path of integration encircles $w=0$ from below (we may assume that it is $\int_{-\infty-i\epsilon}^{\infty-i\epsilon}$ along the straight line), so that our $F(a,b)$ is basically $2\pi I\left(-\frac{\pi+ia}{2},\frac{ib^2}{\pi}\right)$. The relevant result is \begin{align*}
I(z,\tau)&=\color{blue}{\frac{\eta(z/\tau,-1/\tau)+i\tau\eta(z,\tau)}{\tau\theta(z,\tau)}},
\\i\theta(z,\tau)&:=\sum_{n\in\mathbb{Z}}(-1)^n q^{(n+1/2)^2}e^{(2n+1)iz},
\\i\eta(z,\tau)&:=\sum_{n\in\mathbb{Z}}(-1)^n\frac{q^{(n+1/2)^2}}{1+q^{2n+1}}e^{(2n+1)iz},
\end{align*} where $q=e^{i\pi\tau}$ (and $q^\alpha$ means $e^{i\pi\tau\alpha}$); the "theta" is well-known as $\theta_1$ or $\vartheta_{11}$ (I'm adhering to the DLMF notation which seems more common), while the function I've denoted by $\eta$ is in the class of so-called mock theta functions. The approach is to obtain functional equations connecting each of $I(z\pm\pi,\tau)$ and $I(z\pm\pi\tau,\tau)$ with $I(z,\tau)$, see that these are satisfied by the proposed result, and finally prove that the solution is unique (similarly to how it's done for elliptic functions, using contour integration of logarithmic derivatives).

 Of course, a naive attempt to use partial fractions for $\pi/\sin\pi z$, resulting in $$F(a,b)=i\pi e^{-c^2}\left(E(-c)+\sum_{n=1}^{\infty}(-1)^n\big(E(nb-c)-E(nb+c)\big)\right),\qquad c=\frac{a}{2b},$$ where $E(z)=e^{z^2}\operatorname{erfc}z=(2/\sqrt\pi)\int_0^\infty e^{-x^2-2zx}\,dx$${}=E_{1/2}(-z)$, is possible, but this series, in addition to consisting of "somewhat more special" functions, converges fairly slowly.

