Re: int(exp(x)*erf(x),x = a .. b)
From: Urijah Kaplan (uak_at_sas.upenn.edu)
Date: 06/13/04
- Next message: Chris.B: "Re: A little offtopic"
- Previous message: Jack Sarfatti: "/\zpf = (mc^2/e^2)^2(1 + (2/alpha)^2) for electron stability"
- In reply to: John Creighton: "int(exp(x)*erf(x),x = a .. b)"
- Next in thread: Axel Vogt: "Re: int(exp(x)*erf(x),x = a .. b)"
- Reply: Axel Vogt: "Re: int(exp(x)*erf(x),x = a .. b)"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 13 Jun 2004 16:27:19 -0400
John Creighton wrote:
> 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?
If this is any help, on Mathematica 5.01 I get
In[1]:=
Integrate[Erf[x]*E^x, {x, a, b}]
Out[1]=
(-E^(1/4))*Erf[1/2 - a] - E^a*Erf[a] + E^(1/4)*Erf[1/2 - b] +
E^b*Erf[b]
and
In[2]:=
Integrate[Erf[x]*E^x^2, {x, a, b}]
Out[2]=
(1/Sqrt[Pi])*((-a^2)*HypergeometricPFQ[{1, 1}, {3/2, 2}, a^2] +
b^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, b^2])
where HypergeometricPFQ is defined here
<http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F2/>
--Urijah Kaplan
- Next message: Chris.B: "Re: A little offtopic"
- Previous message: Jack Sarfatti: "/\zpf = (mc^2/e^2)^2(1 + (2/alpha)^2) for electron stability"
- In reply to: John Creighton: "int(exp(x)*erf(x),x = a .. b)"
- Next in thread: Axel Vogt: "Re: int(exp(x)*erf(x),x = a .. b)"
- Reply: Axel Vogt: "Re: int(exp(x)*erf(x),x = a .. b)"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
