Can anyone help with this integral?

[ricardop]

Hello
I need to evaluate this integral:

$$\int_0^1e^{ax^2}erf(bx)dx$$

where "erf" is the error function:$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^x\exp(-t^2)dt$.
I can't find this integral in Gradshteyn, Abramowitz, Prudnikov, etc.
Neither Mathematica nor Maple do this integral (definite or indefinite).
The best I can do is by series...
Can anyone help with this integral?

Thanks

 

[CompuChip]

I can't find this integral in Gradshteyn, Abramowitz, Prudnikov, etc.
Neither Mathematica nor Maple do this integral (definite or indefinite).

So it is very likely that there is no closed form solution. So what kind of an answer would you like then?

 

[Pere Callahan]

If a happens to be equal b^2, then Mathematica gives the answer

$$\frac{b}{\sqrt{\pi}}\operatorname{HypergeometricPFQ}[\{1,1\},\{\frac{3}{2},2\},b^2]$$

 

[ricardop]

I'm looking for a closed form... with a not necessarily equal b.

 

[ricardop]

If a happens to be equal b^2, then Mathematica gives the answer

$$\frac{b}{\sqrt{\pi}}\operatorname{HypergeometricPFQ}[\{1,1\},\{\frac{3}{2},2\},b^2]$$

Yes, I knew it.
This is in Prudnikov tables.
But a not necessarily equal b.
Thanks

 

[ricardop]

So it is very likely that there is no closed form solution. So what kind of an answer would you like then?

I'm looking for a closed form.
Thanks

 

[CompuChip]

So if you checked several references an none of them gives one, maybe it doesn't exist :)
You can define it, though.
RicardoP's function
$$\mathcal P(a, b) := <br /> \int_0^1e^{ax^2} \operatorname{erf}(bx)dx$$
Then the closed form solution is
$$\mathcal P(a, b)$$

 

[maze]

You may be able to do a change of variables to make the limits from 0 to infinity, then try some sort of complex contour integral. I haven't tried this yet but its worth a shot.

 

[ricardop]

You may be able to do a change of variables to make the limits from 0 to infinity, then try some sort of complex contour integral. I haven't tried this yet but its worth a shot.

I can change x=1/y, and the interval goes from 1 to \infty, but not from 0 to \infty, or -\infty to \infty.

I tryed several changes of variable...
I changed the order of integration (this is a double integral).
I tryed complex contour integral too, but without sucess.

Thanks!

 

[ricardop]

So if you checked several references an none of them gives one, maybe it doesn't exist :)

This integral is very alike the integrals in the article:
"Integration of a Class of Transcendental Liouvillian Functions with Error-Functions, Part II."
Paul H. Knowles, J. Symb. Comp. 16 (1993) 227.
http://dx.doi.org/10.1006/jsco.1993.1042"
I'm working on it...

 

[ricardop]

The "Cumulative Bivariate Normal Probability Function" is defined by:
$$L(h,k,\rho)=\int_h^\infty dx\int_k^\infty dy\;g(x,y,\rho)$$
where
$$g(x,y,\rho)=\frac{1}{\sqrt{1-\rho^2}}Z(x)Z\left(\frac{y-\rho x}{\sqrt{1-\rho^2}}\right)$$
is the "Bivariate Normal Probability Function" and
$$Z(x)=\frac{\exp(\frac{-x^2}{2})}{\sqrt{2\pi}}$$
is the "Normal Probability Function", in Abramowitz & Stegun's notation.

With these definitions we have the answer
$$\int_0^1 dx\exp(-ax^2)\mathrm{erf}(bx)=\frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{erf}(\sqrt{a})-\frac{1}{\sqrt{a\pi}}\arctan\left(\frac{\sqrt{a}}{b}\right)+2\sqrt{\frac{\pi}{a}}L\left(\sqrt{2a},0,\frac{-b}{\sqrt{a+b^2}}\right)$$

I want to thank Axel for the hint
http://groups.google.com.au/group/sci.math.symbolic/browse_thread/thread/ddbbb1f3ab27b3c4"

 

[Pere Callahan]

Is this any more useful than using RicardoP's function as suggested by CompuChip?
Granted, you can now make use of some known properties of this "Cumulative Bivariate Normal Probability Function"...

 

[ricardop]

Is this any more useful than using RicardoP's function as suggested by CompuChip?
Granted, you can now make use of some known properties of this "Cumulative Bivariate Normal Probability Function"...

obvious...