Re: troubling simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg91356] Re: troubling simple integral
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sun, 17 Aug 2008 06:41:14 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <g86844$hr7$1@smc.vnet.net>
did wrote:
> Mathematica 6.0.3.0 has trouble computing the quite simple integral
>
> In[2]:= FullSimplify[ Integrate[ (b + k*x) / k^2 * Exp[-k*x] *
> Sin[k*a] * Sin[k*y] , {k, 0,Infinity},
> Assumptions -> x > 0 && Im[a] == 0 && Im[b] == 0 && Im[y] == 0 ]]
>
> Out[2]= -(1/8) I (b + 2 I x Log[1 + (4 a y)/(x^2 + (a - y)^2)])
>
> which is obviously wrong. Repeating the computation, I get different
> results.
>
> What should I do to get a correct answer?
> (This one I can do by hand, but I have more complicated similar ones
> to evaluate.)
Does the following result look better?
In[1]:= Assuming[x > 0 && Im[a] == 0 && Im[b] == 0 && Im[y] == 0,
FullSimplify[
Integrate[(b + k*x)/k^2*Exp[-k*x]*Sin[k*a]*Sin[k*y], {k, 0,
Infinity}]]]
Out[1]= 1/4 (2 b ((-a + y) ArcTan[(a - y)/x] + (a + y) ArcTan[(a + y)/
x] - x ArcTanh[(2 a y)/(a^2 + x^2 + y^2)]) +
x Log[1 + (4 a y)/(x^2 + (a - y)^2)])
Note that *Assuming* passes the assumptions to both Integrate[] *and*
FullSimplify[]. (In your original expression, only Integrate[] could
take into account the assumptions.)
HTH,
-- Jean-Marc