Re: int(exp(x)*erf(x),x = a .. b)
From: A. Dieckmann (dieckman_at_physik.uni-bonn.de)
Date: 06/17/04
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Date: Thu, 17 Jun 2004 14:56:29 +0000 (UTC)
On 14 Jun 2004, John Creighton wrote:
>JohnCreighton_@hotmail.com (John Creighton) wrote in message news:<c5e803c2.0406131138.7dfffea9@posting.google.com>...
>> using MATLAB 6.1.0.450 (R12.1) to access the maple
>> kernel I don't get a solution. for:
>> int(exp(x)*erf(x),x = a .. b)
>> Does anyone know if this is solvable? It looks like you could do parts
>> a few times, and then equate terms. Since:
>>
>> >> int(sym('erf(x)'))
>>
>> ans =
>>
>> x*erf(x)+1/pi^(1/2)*exp(-x^2)
>>
>> What about the integral:
>> int(exp(x^2)*erf(x),x = a .. b)
>>
>> For this I could let u=erf(x)
>> and do substitution. Right?
>>
>> Moreover, If I figure out how to do the integral myself, how do I
>> teach maple to do it?
>
>Okay, Now that the above intgrals were solved how about something more
>general:
>
>int(erf(a*(x-d))*exp(-b*(x-c)^2),x = -inf .. inf)
>
>This is really what I need to solve anyway.
>
>P.S. Thanks for the help. Oh, yeay, I also am interested on how they
>are solved. I figured out how to solve int(sym('erf(x)')) by looking
>at the taylor series. of a exp(-x)^2 and erf(x) and int(erf(x)) and
>just used linear algabra to see what linear combination gave the
>answer.
int(erf(a*(x-d))*exp(-b*(x-c)^2),x = -inf .. inf)=sqrt(pi/b) erf(a*sqrt(b/(a^2 + b))*(c-d)), (b>0)
got it by differentiating the integrand w.r.t. a, then integrated over x=-inf..inf, then substituted a=sqrt(b*z)/sqrt(1-z) and integrated over z and then - most important - checked the result numerically.
Cheers, Andreas
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