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.):


where $a$ and $b$ are real numbers and $b>0$.


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.


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