Is there any closed form for this integral?

Question

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$$\int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-x^2}} dx$$

Anyhow, is there any closed form for the result of this definite integral? How can one compute this without calculators like Wolframalpha?

Answer 1

I have generated a few series for the integral based on the following: $$ I = \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-x^2}} dx $$ $$ I = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_{0}^{1/3} \frac{x^{2n}}{\sqrt{1-x^2}} dx $$ letting $x^2=u$ $$ I = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_{0}^{1/9} \frac{u^n}{\sqrt{1-u}} \frac{du}{2\sqrt{u}} $$ $$ I = \sum_{n=0}^\infty \frac{(-1)^n}{2n!}\int_{0}^{1/9} u^{n-\frac{1}{2}}(1-u)^{\frac{-1}{2}}\;du $$ The incomplete beta function, a generalisation of the beta function, is defined as $$ B(x;\,a,b) = \int_0^x t^{a-1}\,(1-t)^{b-1}\,dt. $$ $$ I = \frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{n!}B\left(\frac{1}{9};n+\frac{1}{2},\frac{1}{2}\right) $$ I can't get any further this route but it seems to hold numerically. It seems we can also write which seems to hold numerically $$ I= \frac{1}{2\sqrt{\pi}}\sum_{n=0}^\infty \frac{\Gamma(n+\frac{1}{2})(\Gamma(n+\frac{1}{2})-\Gamma(n+\frac{1}{2},\frac{1}{9}))}{n!} $$ which contains the incomplete gamma function. This comes from the Mellin transform $$ \int_0^\infty \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{1-a x^2}} dx \; da = \int_{0}^{1/3} \frac{\Gamma(s)\Gamma(\frac{1}{2}-s)}{\sqrt{\pi}}\frac{\exp(-x^2)}{(-x^2)^s} dx = \frac{(-1)^{-s} \Gamma \left(\frac{1}{2}-s\right) \Gamma (s) \left(\Gamma \left(\frac{1}{2}-s\right)-\Gamma \left(\frac{1}{2}-s,\frac{1}{9}\right)\right)}{2 \sqrt{\pi }} $$ by doing a double Mellin transform I managed to generate the following sum which also seems to hold numerically $$ I = \sum_{s=0}^\infty \frac{(-1)^s}{3^{1+2s}(1+2s)s!}\;_2F_1\left(\frac{1}{2},\frac{1}{2}+s;\frac{3}{2}+s;\frac{1}{9}\right) $$ this regenerates the sum containing the incomplete gamma function if the sum from inside the hypergeometric function is swapped with the outside sum.

We can also introduce a parameter and take the inverse Laplace transform $$ I(a) = \int_{0}^{1/3} \frac{e^{-x^2}}{\sqrt{a-x^2}} dx $$ $$ \mathcal{L}^{-1}_{a \to s}[I(a)] = \int_0^\frac{1}{3} \frac{e^{-x^2+sx^2}}{\sqrt{\pi s}} \; dx = \frac{\mathrm{erfi}\left(\frac{\sqrt{s-1}}{3}\right)}{2\sqrt{s(s-1)}} $$ so we can rewrite the original integral as $$ I(a=1)=\frac{1}{2}\int_0^\infty \frac{\mathrm{erfi}\left(\frac{\sqrt{s-1}}{3}\right)}{\sqrt{s(s-1)}}e^{-s} \; ds $$ which again seems to hold out numerically

Answer 2

Only a half answer: $$ \int_0^{\frac{1}{3}} \frac{\exp \left(-x^2\right)}{\sqrt{1-x^2}} \, dx=\int_0^{\frac{1}{9}} \frac{\exp (-x)}{2 \sqrt{(1-x) x}} \, dx=\int_0^{\infty } \frac{\exp \left(-\frac{1}{x+9}\right)}{2 (x+9) \sqrt{x+8}} \, dx= $$

Using the Laplace Transform we can write:

$$ \int_0^{\infty } \left(\mathcal{L}_x^{-1}\left[\exp \left(-\frac{1}{x+9}\right)\right](s)\right) \left(\mathcal{L}_x\left[\frac{1}{(x+9) \sqrt{x+8}}\right](s)\right) \, ds= $$

$$ \int_0^{\infty } \left(e^{-9 s} \left(-\frac{J_1\left(2 \sqrt{s}\right)}{\sqrt{s}}+\delta (s)\right)\right) \left(2 e^{9 s} \pi \left(\frac{\text{erfc}\left(\sqrt{s}\right)}{2}-2 T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)\right)\right) \, ds= $$ $$ \int_0^{\infty } \left(-\frac{\pi J_1\left(2 \sqrt{s}\right) \text{erfc}\left(\sqrt{s}\right)}{\sqrt{s}}+\pi \delta (s) \text{erfc}\left(\sqrt{s}\right)+\frac{4 \pi J_1\left(2 \sqrt{s}\right) T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)}{\sqrt{s}}-4 \pi \delta (s) T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)\right) \, ds= $$

$$ -\tan ^{-1}\left(2 \sqrt{2}\right)+\frac{\pi I_0\left(\frac{1}{2}\right)}{2 \sqrt{e}}+\int_0^{\infty } \frac{2 \pi J_1\left(2 \sqrt{s}\right) T\left(\sqrt{2} \sqrt{s},2 \sqrt{2}\right)}{\sqrt{s}} \, ds= $$ $$ -\tan ^{-1}\left(2 \sqrt{2}\right)+\frac{\pi I_0\left(\frac{1}{2}\right)}{2 \sqrt{e}}+4 \pi \int_0^{\infty } J_1(2 x) T\left(\sqrt{2} x,2 \sqrt{2}\right) \, dx $$

where T(x,a) is the Owen's T-function and J(1,x) is Bessel function of the first kind

EDITED:

Maybe not exist closed form solution and answer is on window box:

noclose

Answer 3

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

$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{1-x}}~d(\sqrt{x})$

$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{2\sqrt{x}\sqrt{1-x}}~dx$

$=\int_0^1\dfrac{e^{-\frac{x}{9}}}{2\sqrt{\dfrac{x}{9}}\sqrt{1-\dfrac{x}{9}}}~d\left(\dfrac{x}{9}\right)$

$=\int_0^1\dfrac{e^{-\frac{x}{9}}}{6\sqrt{x}\sqrt{1-\dfrac{x}{9}}}~dx$

$=\dfrac{1}{3}\Phi_{1}\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{3}{2};\dfrac{1}{9},-\dfrac{1}{9}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)