Hi,

You might also look into the following reference:

\> Wolfram Functions: Bessel.

\>\> Product Integrals

\>\>\> Power Function

\>\>\> Exponential Function

A lot of integrals are already written down, usually in terms of a linear argument. I agree with Blinder, you may want to start by changing variables: $\sqrt{x}=y$.

Although, it's also useful to look at the series expansion:

Bessel Series Expansion

As it is an even function, the integrand should expand in whole powers of $x$. Then it's possible to integrate term by term,

Integrate[ Normal[Series[(Exp[-(x + s^2)/(2*a^2)])*(BesselI[0, 
      Sqrt[x]*s/a^2]), {x, 0, 10}]], {x, 0, g}]
N[% /. {a -> 1, s -> 2, g -> 3}]

Depending on the limit $g$, this is an okay method numerically. It also lends some insight in terms of symbols. The functional part has a reliable lexicographic ordering, so we just need to figure out sequence or sequences of denominators. It's useful to compare with OEIS, for example: A002866, and A187735 which appear correct to a factor for cases with $s^0$ and $s^2$ .

If you get stuck with transforms, it can help to calculate a few coefficients.

Brad

It is best to use NIntegrate in yours case.

f[a_?NumericQ, s_?NumericQ, g_?NumericQ] := 
NIntegrate[(Exp[-(x + s^2)/(2 a^2)]) (BesselI[0, Sqrt[x]*s/a^2]), {x,0, g}]

f[1., 2, 3]
(*0.580509*)

Analytical solution maybe not exist.

Integral represented by an infinite sum.

func = (Exp[-(x + s^2)/(2 a^2)]) (BesselI[0, Sqrt[x]*s/a^2])
sol = InverseZTransform[func /. x -> 1/x, x, n]
sol2 = Integrate[sol*x^n, {x, 0, g}, Assumptions -> n >= 0]
(*(2^-n a^(-2 n) E^(-(s^2/(2 a^2))) g^(1 + n) HypergeometricU[-n, 1, s^2/(2 a^2)])/((1 + n) Gamma[1 + n]^2)*)

Sum[sol2, {n, 0, 1000}] /. a -> 1 /. s -> 2 /. g -> 3 // N
(*0.580509*)