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The picture above is from uninstallation tool of fake antivirus in Korea. The "official" uninstallation tool will not proceed anymore unless user input the correct answer. (Nobody succeeded this) Due to its preposterousness, the image has been used as "meme" for malwares.

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

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

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    $\begingroup$ I get erf vibes so my guess is that there's no closed form for the result. $\endgroup$ – Oria Gruber Oct 15 '17 at 13:22
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    $\begingroup$ wtf kind of program gives you integrals to use? $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 13:23
  • $\begingroup$ @DanielC WolframAlpha/Matlab gives $0.3274711...$, but none of the answers like 0.327, 0.3274, 0.32747 worked. The developer answered "they are approximated results, not the accurate answers," from angry e-mails from users. Actually, the company went bankrupt in 2011, so we would never know what the "right answer" the program wanted was. $\endgroup$ – fiverules Oct 15 '17 at 13:29
  • $\begingroup$ Maybe if the bounds were from $0$ to $1$ I would have hope, but the choice of bounds here implies either there is a very special trick (one that WA does not know) or that the anti-derivative can be found (which I doubt). $\endgroup$ – Simply Beautiful Art Oct 15 '17 at 13:34
  • $\begingroup$ @SimplyBeautifulArt. If the upper bound was $1$, it would reduce to something proportional to Bessel $\frac {\pi}{2 \sqrt e} I_0(1/2)$ $\endgroup$ – Claude Leibovici Oct 15 '17 at 13:56
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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

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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

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