0
$\begingroup$

I'm struggling with the following integral:

$$ I = \int_{a}^{b} {\frac{\mathrm{Erf}\left(\,{x/c}\,\right)}{\,\sqrt {\,{1 - {x^2}}\,}\,}\,\mathrm{d}x} $$

Honestly, I do not know any approaches to solve it, except trying with Mathematica and searching for a possible solution in tables of integrals involving the Erf function, but all failed.

Can somebody give me a hint?

Thank you very much.

Best regards.

$\endgroup$
0
$\begingroup$

As Robert Israel answered, it does not seem that the antiderivative could be computed even using special functions.

However, considering $$I=\int_0^a\frac{\text{erf}\left(\frac{x}{c}\right)}{\sqrt{1-x^2}}\,dx$$ hoping that $a$ is not too large, you could expand the integrand as a truncated Taylor series and integrate termwise. Otherwise, numerical integration would be required.

You would get something like $$\frac{\text{erf}\left(\frac{x}{c}\right)}{\sqrt{1-x^2}}=\frac{2 x}{\sqrt{\pi } c}+\frac{\left(3 c^2-2\right) x^3}{3 \sqrt{\pi } c^3}+\frac{\left(45 c^4-20 c^2+12\right) x^5}{60 \sqrt{\pi } c^5}+\frac{\left(525 c^6-210 c^4+84 c^2-40\right) x^7}{840 \sqrt{\pi } c^7}+O\left(x^9\right)$$

Let us try using $a=\frac 12$ for various values of $c$ $$\left( \begin{array}{ccc} c & \text{exact} & \text{approximation} \\ 1 & 0.1450370 & 0.14500970 \\ 2 & 0.0747909 & 0.07477398 \\ 3 & 0.0501538 & 0.05014195 \\ 4 & 0.0376931 & 0.03768400 \\ 5 & 0.0301833 & 0.03017601 \\ 6 & 0.0251659 & 0.02515974 \\ 7 & 0.0215775 & 0.02157225 \\ 8 & 0.0188842 & 0.01887956 \\ 9 & 0.0167883 & 0.01678417 \end{array} \right)$$

Another solution would be to use, as Robert Israel answered, integration by parts $$J=\int\frac{\text{erf}\left(\frac{x}{c}\right)}{\sqrt{1-x^2}}\,dx=\sin ^{-1}(x) \text{erf}\left(\frac{x}{c}\right)-\frac{2}{\sqrt{\pi } c}\int { e^{-\frac{x^2}{c^2}} \sin ^{-1}(x)}\,dx$$ $$K=\int { e^{-\frac{x^2}{c^2}} \sin ^{-1}(x)}\,dx=c\int e^{-t^2} \sin ^{-1}(c t)\,dt$$ Using the Taylor expansion $$\sin ^{-1}(c t)=\sum^{\infty}_{n=0} \frac{(2n)!\,c^{2n+1}}{4^n (n!)^2 (2n+1)} t^{2n+1}\qquad \text{for}\qquad |ct|\leq 1$$ and use $$\int t^{2n+1}e^{-t^2}\,dt=-\frac{1}{2} \Gamma \left(n+1,t^2\right)$$

$\endgroup$
  • $\begingroup$ if you change your integration limits to (-a,a) the result becomes super simple :) $\endgroup$ – tired Aug 9 '17 at 6:05
  • $\begingroup$ @tired. Very nice observation. $\endgroup$ – Claude Leibovici Aug 9 '17 at 6:48
  • $\begingroup$ Thank for your help. I'm actually deriving the PDF of a random variable. The lower integral limit is -1, and the upper limit can be any. Therefore, your approach may not valid. $\endgroup$ – BinhDDT Aug 9 '17 at 11:36
0
$\begingroup$

Since $a$ and $b$ are unspecified, you're looking for an antiderivative. Integration by parts gives

$$ \arcsin(x) \text{erf}(x/c) - \frac{2}{c \sqrt{\pi}} \int \arcsin(x)\; e^{-x^2/c^2}\; dx $$

But this does not seem to have an elementary antiderivative.

$\endgroup$
  • $\begingroup$ Thank you very much for your help. $\endgroup$ – BinhDDT Aug 9 '17 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.