Integral of $\frac{e^{-x^2}}{\sqrt{1-x^2}}$ I am stuck at an integral $$\int_0^{\frac{1}{3}}\frac{e^{-x^2}}{\sqrt{1-x^2}}dx$$
My attempt is substitute the $x=\sin t$, however there may be no primitive function of $e^{-\sin^2 t}$.
So does this integral has a definitive value? If does, how can we solve it? Thank you!
 A: It does not seem that a closed form exist. To evaluate the integral, start with a Taylor expansion which gives $$\frac{1}{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3 x^4}{8}+O\left(x^6\right)$$ So you are let with the weighted sum of integrals $$I_n=\int x^{2n}e^{-x^2}\,dx$$ which lead to gamma functions. For the given bounds, the sum seems to converge very quickly to the value given by Travis (only four terms required for six significant digits).
A: We have:
$$ I = \int_{0}^{\arcsin\frac{1}{3}}\exp\left(-\sin^2\theta\right)\,d\theta \tag{1}$$
but since:
$$ \exp(-\sin^2\theta) = \frac{1}{\sqrt{e}}\left(I_0\left(\frac{1}{2}\right)+2\sum_{n\geq 1}I_n\left(\frac{1}{2}\right)\cos(2n\theta)\right)\tag{2} $$
we have:
$$ I = e^{-1/2}\left(\arcsin\frac{1}{3}\right) I_0\left(\frac{1}{2}\right)+e^{-1/2}\sum_{n\geq 1}\frac{1}{3n}\, I_n\left(\frac{1}{2}\right)U_{2n-1}\left(\sqrt{\frac{8}{9}}\right)\tag{3}$$
where $I_m$ is a modified Bessel function and $U_k$ is a Chebyshev polynomial of the second kind.
A: Approach $1$:
$\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}dx$
$=\int_0^{\sin^{-1}\frac{1}{3}}\dfrac{e^{-\sin^2t}}{\sqrt{1-\sin^2t}}d(\sin t)$
$=\int_0^{\sin^{-1}\frac{1}{3}}e^{-\sin^2t}~dt$
$=\int_0^{\sin^{-1}\frac{1}{3}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}dt$
$=\int_0^{\sin^{-1}\frac{1}{3}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}\right)dt$
For $n$ is any natural number,
$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int_0^{\sin^{-1}\frac{1}{3}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}\right)dt$
$=\left[t+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{-1}\dfrac{1}{3}}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{2\sqrt2(-1)^n(2n)!((k-1)!)^2}{4^{n-k+1}9^k(n!)^3(2k-1)!}$
Approach $2$:
$\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}dx$
$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{1-x}}d(\sqrt{x})$
$=\dfrac{1}{2}\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{x}\sqrt{1-x}}dx$
$=\dfrac{1}{2}\int_0^1\dfrac{e^{-\frac{x}{9}}}{\sqrt{\dfrac{x}{9}}\sqrt{1-\dfrac{x}{9}}}d\left(\dfrac{x}{9}\right)$
$=\dfrac{1}{6}\int_0^1\dfrac{e^{-\frac{x}{9}}}{\sqrt{x}\sqrt{1-\dfrac{x}{9}}}dx$
$=\dfrac{1}{3}\Phi_1\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{3}{2};\dfrac{1}{9},-\dfrac{1}{9}\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)
