Integral $\int_{0}^{3} \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx$ I recently got stuck on evaluating the following integral,
$$ \int_{0}^{3} \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx. $$
Is it possible to evaluate this integral in a closed form? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.

You can express in terms of Incomplete Bessel Functions:
$\int_0^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx+\int_1^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx-i\int_1^3\dfrac{e^{-x^2}}{\sqrt{x^2-1}}~dx$
$=\int_0^\frac{\pi}{2}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)-i\int_0^{\cosh^{-1}3}\dfrac{e^{-\cosh^2x}}{\sqrt{\cosh^2x-1}}~d(\cosh x)$
$=\int_0^\frac{\pi}{2}e^\frac{\cos2x-1}{2}~dx-i\int_0^{\ln(3+2\sqrt 2)}e^{-\frac{\cosh2x+1}{2}}~dx$
$=e^{-\frac{1}{2}}\int_0^\pi e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)-ie^{-\frac{1}{2}}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~d\left(\dfrac{x}{2}\right)$
$=\dfrac{e^{-\frac{1}{2}}}{2}\int_0^\pi e^\frac{\cos x}{2}~dx-\dfrac{ie^{-\frac{1}{2}}}{2}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~dx$
$=\dfrac{\pi e^{-\frac{1}{2}}}{2}I_0\left(\dfrac{1}{2}\right)-\dfrac{ie^{-\frac{1}{2}}}{2}J\left(\dfrac{1}{2},0,2\ln(3+2\sqrt 2)\right)$
