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.
 A: Let's decompose your integral in three terms :
\begin{align}
I&:=\int_0^3 \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx=I_1+I_2+I_3\\
&=\int_0^1 \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx+\int_1^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx\\
&=\int_0^{\pi/2} \frac{e^{-\sin(t)^2}}{\sqrt{1-\sin(t)^2}} \,d(\sin(t))+\int_0^\infty \frac{e^{-\cosh(u)^2}}{i\sqrt{\cosh(u)^2-1}} \,d\left(\cosh(u)\right)-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx\\
&=\int_0^{\pi/2} e^{\cos(2t)/2-1/2}\,dt-i\int_0^\infty e^{-\cosh(2u)/2-1/2} \,du-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx\\
&=\frac 1{2\,\sqrt{e}}\left(\pi\,\operatorname{I}_0\left(\frac 12\right)-i\operatorname{K}_0\left(\frac 12\right)\right)-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx\\
\end{align}
Using the integral for $\operatorname{I}$, the integral for $\operatorname{K}$.
Wolfram Alpha proposes to simplify this as $\;\displaystyle I=-\frac {i}{2\,\sqrt{e}}\,\operatorname{K}_0\left(-\frac 12\right)-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx\;$ (probably using this relation for $m=1,n=0$) but there is a sign error in its result for the real part.
In summary (for $x>1$ the terms are imaginary since $1-x^2<0$)  : 


*

*the first term is real $\quad\displaystyle I_1=\frac {\pi}{2\,\sqrt{e}}\,\operatorname{I}_0\left(\frac 12\right)\approx 1.013219033$

*the second is imaginary $\displaystyle I_2=-\frac {i}{2\,\sqrt{e}}\,\operatorname{K}_0\left(-\frac 12\right)\approx -i\cdot 0.2803442545$

*the remaining term is imaginary too and rather small : $$I_3=-\int_3^\infty \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx=\frac {i}{\sqrt{e}}\int_{\operatorname{arccosh}(3)}^\infty e^{-\cosh(2u)/2} \,du\approx i\cdot 0.000006566431462$$ I dont think it may be written in 'closed-form' except possibly as an 'incomplete modified Bessel function' or something like that (i.e. nearly equivalent to the integral definition...).
An approximation is obtained with $\displaystyle I_3\approx \frac {i}{2\,\sqrt{e}}\operatorname{Ei}\left(-\frac{e^{2\,\operatorname{arccosh}(3)}}4\right)$ since $\displaystyle \int_a^\infty e^{-\frac 14 e^{2u}} \,du=\frac 12\operatorname{Ei}\left(-\frac{e^{2a}}4\right)\approx\frac{e^{-e^{2a}/4}}{2\,e^{2a}/4}\;$ (with $\operatorname{Ei}$ the exponential integral).
(btw $\,e^{2\,a}=e^{2\,\operatorname{arccosh}(3)}=17+12\sqrt{2}$)
and we may get many better ones but not the asked closed form...

A: If you meant $\displaystyle\int_0^{\color{red}1}\frac{e^{-x^2}}{\sqrt{1-x^2}}dx$, then the answer is $\dfrac\pi{2\sqrt e}\cdot \text{Bessel I}\left(0,\dfrac12\right)$. See Bessel function.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\int_{0}^{3}{\expo{-x^{2}} \over \root{1 - x^{2}}}\,\dd x
=
\int_{0}^{1}{\expo{-x^{2}} \over \root{1 - x^{2}}}\,\dd x
-
\ic\int_{1}^{3}{\expo{-x^{2}} \over \root{x^{2} - 1}}\,\dd x
\\[3mm]&=
\int_{1}^{0}{\expo{-\pars{1 - z^{2}}^{2}} \over \root{1 - \pars{1 - z^{2}}^{2}}}
\pars{-2z\,\dd z}
-
\ic\int_{0}^{\root{2}}{\expo{-\pars{1 + z^{2}}^{2}}
\over \root{\pars{1 + z^{2}}^{2} - 1}}\pars{2z\,\dd z}
\\[3mm]&=
\underbrace{2\int_{0}^{1}{\expo{-\pars{z^{2} - 1}^{2}} \over \root{2 - z^{2}}}\,\dd z
}_{\ds{\approx 1.01322}}\
-\
\underbrace{2\ic\int_{0}^{\root{2}}{\expo{-\pars{z^{2} + 1}^{2}}
\over \root{z^{2} + 2}}\,\dd z}
_{\ds{\approx 0.280338\,\ic}}
\end{align}
The numerical value $\ds{1.01322 - 0.280338\,\ic}$ was found with Mathematica. The original integral was not calculated by Mathematica due to the integrable singularity at $\ds{x = 1}$.
A: You can express in terms of Incomplete Bessel Functions:
$\int_0^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx+\int_1^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx-i\int_1^3\dfrac{e^{-x^2}}{\sqrt{x^2-1}}~dx$
$=\int_0^\frac{\pi}{2}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)-i\int_0^{\cosh^{-1}3}\dfrac{e^{-\cosh^2x}}{\sqrt{\cosh^2x-1}}~d(\cosh x)$
$=\int_0^\frac{\pi}{2}e^\frac{\cos2x-1}{2}~dx-i\int_0^{\ln(3+2\sqrt 2)}e^{-\frac{\cosh2x+1}{2}}~dx$
$=e^{-\frac{1}{2}}\int_0^\pi e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)-ie^{-\frac{1}{2}}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~d\left(\dfrac{x}{2}\right)$
$=\dfrac{e^{-\frac{1}{2}}}{2}\int_0^\pi e^\frac{\cos x}{2}~dx-\dfrac{ie^{-\frac{1}{2}}}{2}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~dx$
$=\dfrac{\pi e^{-\frac{1}{2}}}{2}I_0\left(\dfrac{1}{2}\right)-\dfrac{ie^{-\frac{1}{2}}}{2}J\left(\dfrac{1}{2},0,2\ln(3+2\sqrt 2)\right)$
