Re: Integral dt exp(-a t^2) / (b^2 - t^2)^(2/3)
- From: "Alec Mihailovs" <alec@xxxxxxxxxxxxx>
- Date: Thu, 14 Apr 2005 16:33:59 GMT
"G. A. Edgar" <edgar@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:140420050912218179%edgar@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> > does anybody have an idea how to solve analytically the following
>> > integral:
>> >
>> > Integral{0,+inf} dt exp(-a t^2) / (b^2 - t^2)^(2/3)
>> >
>> > where a and b are real numbers > 0.
>> >
>
> I think the OP intends the real 2/3 power, not a complex branch.
> So the answer should be real.
In this case, the integral from 0 to b is real, and the integral from b to
infinity given by hypergeometric formulas equals to the 'real' integral from
b to infinity multiplied by (-1)^(-2/3)=-1/2-sqrt(3)/2*I. Thus, if Ans is
the answer given by the hypergeometric formulas, the 'real' answer is
Re(Ans)-Im(Ans)*sqrt(3).
Alec Mihailovs
http://math.tntech.edu/alec/
.
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