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Topic: integral challenge (was: BesselI integral identity
Replies: 3   Last Post: Mar 2, 2007 3:09 AM

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Paul Abbott

Posts: 1,437
Registered: 12/7/04
Re: integral challenge (was: BesselI integral identity
Posted: Mar 2, 2007 3:09 AM
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In article <270220070817132851%edgar@math.ohio-state.edu.invalid>,
"G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote:

> > > > > Integrate[Exp[a x] BesselI[0, b Sqrt[1 - x^2]], {x, -1, 1}] ==
> > > > > 2 Sinh[Sqrt[a^2 + b^2]] / Sqrt[a^2 + b^2]

>
> Now that we have a reference, can someone show us how to
> derive it using one of the CA programs?


I cannot obtain it directly using Mathematica. The best I have is this:

[1] BesselI[0, z] is a hypergeometric function:

Hypergeometric0F1[1, z^2/4] // FullSimplify

BesselI[0, z]

[2] Interchanging the order of summation and integration, the integrals
can be computed

Assuming[ k >= 0,
Integrate[Exp[a x] (b^2 (1 - x^2) / 4)^k, {x, -1, 1}] ] //
FunctionExpand // Simplify

2^(1/2 - k) a^(-1/2 - k) b^(2 k) Sqrt[Pi] BesselI[1/2 + k, a] *
Gamma[1 + k]

[3] The (hypergeometric) sum now reads

Sum[(b^2/(2 a))^k / k! (Sqrt[2 Pi/a] BesselI[k+1/2, a]),
{k, 0, Infinity}]

And this sum can be derived from the generating function (Abramowitz and
Stegun 10.2.31) by differentiation,

Sinh[Sqrt[z^2 + 2 t z]]/Sqrt[z^2 + 2 t z] ==
Sum[t^n /n! (Sqrt[2 Pi/z] BesselI[n + 1/2, z]), {n, 0, Infinity}]

Cheers,
Paul

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