Evaluation of an integral involving hyperbolic sine and exponential I am wondering if the following integral can be reduced to either a closed form involving elementary functions, or well-known special functions (such as $\operatorname{erf}$, Bessel functions, etc.):
$$\int_{-1}^{1}e^{-ax}\sinh(b\sqrt{1-x^2})dx$$
where $a$ and $b$ are real numbers and $b>0$.
 A: Not an answer, just thoughts. Maybe these facts can somehow help.
$$\int_{-1}^{1}e^{-ax}\sinh(b\sqrt{1-x^2})dx=\mathrm{I}(a,b)$$
$$\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial a^2}=\int_{-1}^{1}x^2e^{-ax}\sinh(b\sqrt{1-x^2})dx$$
$$\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial b^2}=\int_{-1}^{1}(1-x^2)e^{-ax}\sinh(b\sqrt{1-x^2})dx$$
So:
$$\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial b^2}=\mathrm{I}(a,b)-\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial a^2}$$
Or
$$\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial a^2}+\frac{\partial^2 \ \mathrm{I}(a,b)}{\partial b^2}-\mathrm{I}(a,b)=0$$
$$ ( \nabla^2 -1) \mathrm{I}(a,b)  =  0,$$
where $\nabla^2=\frac{\partial^2 }{\partial a^2}+\frac{\partial^2 }{\partial b^2}$
I guess this is the Helmholtz equation (with eigenvalue $k=i$).
We can add some boundary conditions:
$$\mathrm{I}(a,0)=\frac{\partial^2 \ \mathrm{I}(a,0)}{\partial a^2}=\ldots=\frac{\partial^n \ \mathrm{I}(a,0)}{\partial a^n}=0$$
So one can say that the integral that is needed to be computed is the solution of the following partial differential equation with the given boundary conditions. 
