Hi,

It is a quite urgent question.
i need an analytic computation of the following integral:

int(exp(-x^2)*erf(a*x+b), x=0..c)

I have the value of the following integral if it can help you:

int(x*exp(-x^2)*erf(a*x+b), x=0..c)

and this website Link Eqs. (27), (32):
http://mathworld.wolfram.com/Erf.html

The case of a=1 and b=0 is also known.
The analytic formulation of this integral as a fucntion of a,b, and c is needed.

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Request for Question Clarification by welte-ga on 21 Dec 2005 18:45 PST

I don't believe that a closed form solution to this problem exists...
You may need to do a numerical approximation.

      -welte-ga

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Clarification of Question by torrent-ga on 21 Dec 2005 19:58 PST

Hi,

Numerical solution is not enough, i need analytic solution and i'm
quite sure it can be done to a certain level.
Please make more efforts, the listining price could be doubled if i
get "Exactly" what i want.

Thanks.

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Clarification of Question by torrent-ga on 23 Dec 2005 09:11 PST

Hi,

Can someone clarify for me, what Mr berkeleychocolate-ga suggests as a
solution below, in the comments section.

Thank you.

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Clarification of Question by torrent-ga on 30 Dec 2005 09:55 PST

Hi,

Is there someone still working on my problem?

Thanks.

This looks tough. "The Integrator" on the Wolfram site is very
powerful and cannot do this one. On the other hand, your link points
to an analytic expression for a similar indefinite integral that
Mathematica cannot do. Good luck!

It is unreasonable to expect a solution in terms of elementary
functions since the integral of ?x^2 is involved. But there is a
solution in terms of  F(d) =  erf(d) = 2/sqrt(pi) * the integral from
0 to d of ?t^2 dt. Note F(d) is intimately related to the cdf phi of
the standard normal distribution for which tables of values are
ubiquitous. In fact F(d) = 2 phi(d* sqrt(2))-1.

The following is a messy calculation, relying on changing the order in
a double integration, integration by parts and changes of variables.
First let u=ax+b to express the given integral in terms of three
integrals with constant limits ( b and ac+b) of the functions (I)
exp(-(u/a)^2) * F(u), (II) exp(2bu/a)*F(u), and (III) a constant times
F(u).

To find the integral from 0 to d of F(x) dx substitute in the
definition of F(x), change the order of integration, simplify a bit to
get d*F(d)- (1-exp(-d^2))/sqrt(pi). Applying this formula twice one
gets the integral from c to d of the same thing for arbitrary c and d.

For (II) with the limits of 0 to d, do the same thing. Write as a
double integral and change the order. One gets a term involving F(d)
and an integral from 0 to d of exp(2kt-t^2), where k =b/a. Complete
the square and change variables (s=t-k) to reduce this integral to
F(d-k) +F(d). Apply this twice to get an integral with arbitrary
limits.

To find (I) consider the integral from 0 to d of F(kt)*exp(-(mt)^2)
dt. Integrate by parts with u=F(kt) to get a formula for this integral
in terms of F(mt)*exp(-(kt)^2) dt. That is, k and m got reversed.
Repeat this formula with k and m reversed and solve the equation for
the integral of F(kt)*exp(-(mt)^2) dt. I got
(2*sqrt(pi)-pi)/(m*(4-pi))*F(kd)*F(md). Then let m=1.

Finally substitute it all back in to get the final formula in terms of F.

Hi Berkeleychocolate-ga,

Thank you for your help, but i'm a little confused with your explanations.
Indeed, how can you "express the given integral in terms of three
integrals with constant limits ( b and ac+b) of the functions (I)
exp(-(u/a)^2) * F(u), (II) exp(2bu/a)*F(u), and (III) a constant times
F(u)."?
I think it becomes one integral of a product:
"exp(-(u/a)^2)*exp(2bu/a)*F(u)*constant"

How can you separate integration of this product? and where is the
Double integration?

Can you please explain more clearly.
Thank you again for your efforts.
I'm waiting for your response, Thanks.

Sorry about the big blunder with addition and multiplication. You are
right, of course. When I checked my calculations there was another
error in the formula for (I). The corrected formula is, using
integration by parts, (all integrals are from 0 to a constant m) that 
c*the integral of F(ax+b) *exp(-(cx+d)^2) dx + a * the integral of
F(cx+d)*exp(-(ax+b)^2) dx = sqrt(pi)/2 [F(am+b)F(cm+d) ? F(b)F(d)].

From this formula with a=c and b=d one can get a formula involving
only F for the integral of F(ax+b)*exp(-(ax+b)). But it gives no
information using it twice to get a formula in the integral of F(ax+b)
*exp(-(cx+d)^2). That?s all I can offer now.

P.S. The double integral arises when one is integrating a function
involving erf and replaces erf with its definition as an integral.

are you still interested in a solution? I might have an answer as a
series expansion.

Hi berkeleychocolate-ga,

yes i'm still looking for a solution in a simplified form.
I've already use an expansion of erf(x), to compute this integral, but
i didn't get a simple closed form solution. Instead of that i get
three summations over, k j and n, i think.

Could you please writre your solution, perhaps it is better than mine.

Thanks.

since  d{[SQRT(pi)/2]*erf(ax+b)*erf(cx+d)}/dx =
a*EXP(-(ax+b)^2)*erf(cx+d) + c*EXP(-(cx+d)^2)*erf(ax+b) , we will get
berkeleychocolate's result, and turn the original integral into
another. But I don't have a clue to the case a=1 or b=0.
Torrent, would you please showing the solution of the special case a=1 and b=0?
I guess that if the answer could be shown as an analytic form, it will
look like the product of erf.

Hi,

I guess no one has a clue to my integral.
For the a=1 and b=0, it's really simple.
f(x) = exp(-x^2/2)
and integral of f(x) is: F(x) = erf(x/sqrt(2))  (there is some coefficient there)
Thus, integral of f(x)*F(x) is F(x)^2, which will be something like
(erf(x/sqrt(2)))^2.
Or try Maple or Matlab, it gives you the solution :)

Thanks.

Oops, stupid I...
I think "a=1" and "b=0" are two different cases.
Since d{[SQRT(pi)/2]*erf(ax+b)*erf(cx+d)}/dx =
a*EXP(-(ax+b)^2)*erf(cx+d) + c*EXP(-(cx+d)^2)*erf(ax+b)
Let a=c=1, b=d=0 we can get the answer of "a=1 and b=0".
But I think there is no closed form in general case, 
even the expression of the ERF.
Only the most special case could be solved.