I am stuck at an integral $$\int_0^{\frac{1}{3}}\frac{e^{-x^2}}{\sqrt{1-x^2}}dx$$

My attempt is substitute the $x=\sin t$, however there may be no primitive function of $e^{-\sin^2 t}$.

So does this integral has a definitive value? If does, how can we solve it? Thank you!

  • 3
    $\begingroup$ Of course it has a definitive value...it is the area under a continuous curve. $\endgroup$ – fretty Feb 25 '15 at 10:08
  • $\begingroup$ did you try $1-x^2 = t$? $\endgroup$ – Alex Feb 25 '15 at 10:12
  • $\begingroup$ Very likely there is no closed-form expression for the value, which is $\sim 0.3274711440$... $\endgroup$ – Travis Willse Feb 25 '15 at 10:13
  • $\begingroup$ Wolfram-Alfha give no closed form. $\endgroup$ – Emilio Novati Feb 25 '15 at 10:15
  • $\begingroup$ @fretty Yeah...En, my point is the some kind form of the value. $\endgroup$ – gaoxinge Feb 25 '15 at 10:20

It does not seem that a closed form exist. To evaluate the integral, start with a Taylor expansion which gives $$\frac{1}{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3 x^4}{8}+O\left(x^6\right)$$ So you are let with the weighted sum of integrals $$I_n=\int x^{2n}e^{-x^2}\,dx$$ which lead to gamma functions. For the given bounds, the sum seems to converge very quickly to the value given by Travis (only four terms required for six significant digits).

  • $\begingroup$ Still,it wouldn't have close form right? $\endgroup$ – Akshay Bodhare Feb 25 '15 at 10:29
  • $\begingroup$ For me, no closed form ... but I may be wrong !! $\endgroup$ – Claude Leibovici Feb 25 '15 at 10:30
  • $\begingroup$ May I add that you could use $$I_n = (-1)^n\dfrac{d^{n}}{d\alpha^n}\int \mathrm{e}^{-\alpha x^2}dx$$ which depending on limits the closest closed form (if you could call it that) is the cumulative distribution. Though I could be wrong ;). $\endgroup$ – Chinny84 Feb 25 '15 at 10:36
  • $\begingroup$ @Chinny84. Why don't you add that as an answer ? It is very interesting. Cheers :-) $\endgroup$ – Claude Leibovici Feb 25 '15 at 10:40
  • $\begingroup$ Because it only compliments your answer. So I much prefer if the comment is logical to add to your fine answer (I would not of got to the conclusion above without your work). Thanks :) $\endgroup$ – Chinny84 Feb 25 '15 at 10:51

We have: $$ I = \int_{0}^{\arcsin\frac{1}{3}}\exp\left(-\sin^2\theta\right)\,d\theta \tag{1}$$ but since:

$$ \exp(-\sin^2\theta) = \frac{1}{\sqrt{e}}\left(I_0\left(\frac{1}{2}\right)+2\sum_{n\geq 1}I_n\left(\frac{1}{2}\right)\cos(2n\theta)\right)\tag{2} $$ we have: $$ I = e^{-1/2}\left(\arcsin\frac{1}{3}\right) I_0\left(\frac{1}{2}\right)+e^{-1/2}\sum_{n\geq 1}\frac{1}{3n}\, I_n\left(\frac{1}{2}\right)U_{2n-1}\left(\sqrt{\frac{8}{9}}\right)\tag{3}$$ where $I_m$ is a modified Bessel function and $U_k$ is a Chebyshev polynomial of the second kind.


Approach $1$:


$=\int_0^{\sin^{-1}\frac{1}{3}}\dfrac{e^{-\sin^2t}}{\sqrt{1-\sin^2t}}d(\sin t)$




For $n$ is any natural number,

$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.


$=\left[t+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$

$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$


Approach $2$:






$=\dfrac{1}{3}\Phi_1\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{3}{2};\dfrac{1}{9},-\dfrac{1}{9}\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)


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