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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.

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    $\begingroup$ The integrand is not real in the interval specified. $\endgroup$ – Torsten Hĕrculĕ Cärlemän Dec 21 '13 at 9:31
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    $\begingroup$ In what context did this integral arise? $\endgroup$ – E.O. Dec 21 '13 at 9:51
  • $\begingroup$ The integrand equals 0 when x=1 so it cannot be done $\endgroup$ – user85798 Dec 21 '13 at 10:20
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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...
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\ds{x = 1}$.

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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.

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