# 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.

• The integrand is not real in the interval specified. – Torsten Hĕrculĕ Cärlemän Dec 21 '13 at 9:31
• In what context did this integral arise? – E.O. Dec 21 '13 at 9:51
• The integrand equals 0 when x=1 so it cannot be done – user85798 Dec 21 '13 at 10:20

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...


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.

You can express in terms of Incomplete Bessel Functions:

$$\int_0^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$$

$$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx+\int_1^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$$

$$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx-i\int_1^3\dfrac{e^{-x^2}}{\sqrt{x^2-1}}~dx$$

$$=\int_0^\frac{\pi}{2}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)-i\int_0^{\cosh^{-1}3}\dfrac{e^{-\cosh^2x}}{\sqrt{\cosh^2x-1}}~d(\cosh x)$$

$$=\int_0^\frac{\pi}{2}e^\frac{\cos2x-1}{2}~dx-i\int_0^{\ln(3+2\sqrt 2)}e^{-\frac{\cosh2x+1}{2}}~dx$$

$$=e^{-\frac{1}{2}}\int_0^\pi e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)-ie^{-\frac{1}{2}}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~d\left(\dfrac{x}{2}\right)$$

$$=\dfrac{e^{-\frac{1}{2}}}{2}\int_0^\pi e^\frac{\cos x}{2}~dx-\dfrac{ie^{-\frac{1}{2}}}{2}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~dx$$

$$=\dfrac{\pi e^{-\frac{1}{2}}}{2}I_0\left(\dfrac{1}{2}\right)-\dfrac{ie^{-\frac{1}{2}}}{2}J\left(\dfrac{1}{2},0,2\ln(3+2\sqrt 2)\right)$$