Let $a \ge 0$, $\theta \ge 0$, $\zeta \ge 0$ and $x_0 \in {\mathbb R}$. Consider the following integral:
\begin{eqnarray} {\mathcal I}_{a,\theta,\zeta}(x_0):= \int\limits_0^\infty (z+a)^\theta z^\zeta \exp\left( - \frac{x_0^2}{z} - z \right) dz \tag{1} \end{eqnarray}
This integral, or its partial derivatives, appears in the context of evaluating path integrals for the fractional Brownian motion (see here). By using the Slater theorem we have evaluated that integral as follows:
\begin{eqnarray} &&{\mathcal I}_{a,\theta,\zeta}(x_0)= \\ &&\sum\limits_{n=0}^\infty \frac{1}{n!} \left(-x_0^2\right){}^n \left( \right. \\ &&\left. a^{\zeta +\theta -n+1} \cdot \Gamma (-n+\zeta +1) U(-n+\zeta +1,-n+\zeta +\theta+2,a) + \right. \\ &&\left. \left(x_0^2\right){}^{\zeta +1} a^{\theta -n} \cdot \Gamma (-n-\zeta -1) U(-n,-n+\theta +1,a) \right. \\ && \left. \right) \tag{2} \end{eqnarray}
Now, as always, the code below verifies the result numerically:
{a, \[Theta], x0, \[Zeta]} =
RandomReal[{1, 2}, 4, WorkingPrecision -> 50]; \[Gamma] = 1/7;
(*Slater theorem*)
NIntegrate[ (z + a)^\[Theta] z^\[Zeta] Exp[-( (x0^2)/ z) - z], {z,
0, Infinity}]
NIntegrate[
a^((I s + \[Gamma]) + 1 + \[Zeta] + \[Theta])
HypergeometricU[(I s + \[Gamma]) + 1 + \[Zeta],
2 + (I s + \[Gamma]) + \[Zeta] + \[Theta],
a] x0^-2^(I s + \[Gamma]) Gamma[(I s + \[Gamma]) +
1 + \[Zeta]] Gamma[(I s + \[Gamma])], {s, -Infinity,
Infinity}]/(2 Pi )
M = 20;
Take[Accumulate@
Table[(-x0^2)^n/
n! (a^((-n) + 1 + [Zeta] + [Theta])
HypergeometricU[(-n) + 1 + [Zeta],
2 + (-n) + [Zeta] + [Theta], a] Gamma[(-n) + 1 + [Zeta]] +
a^((-n - 1 - [Zeta]) + 1 + [Zeta] + [Theta])
HypergeometricU[(-n - 1 - [Zeta]) + 1 + [Zeta],
2 + (-n - 1 - [Zeta]) + [Zeta] + [Theta],
a] x0^2^(1 + [Zeta]) Gamma[(-n - 1 - [Zeta])]), {n, 0,
M}], -5] // MatrixForm
My question is can we derive this result in a different way, for example by using fractional calculus.
