This question follows [this one](https://mathoverflow.net/questions/271936/closed-form-for-int-0t-e-x-fraci-n-alpha-xxdx/271955), where the general problem has apparently no simpler form than the integral one. I focus now on the limit case:
\begin{align}
\int\_0^T e^{-x}\frac{nI\_n(x)}{x}dx=\int\_0^T e^{-x}\frac{I\_{n-1}(x)-I\_{n+1}(x)}{2}dx
\end{align}
where $I\_n(x)$ is the modified Bessel function of the first kind, and $n$ is an integer.


I can get the following result from the series expansion of the modified Bessel function:
\begin{align}
\int\_0^T e^{-x}\frac{I\_{n}(x)}{2}dx&=\frac{(T/2)^{n+1}}{(n+1)!}{}\_2F\_2[\{n+\frac{1}{2},n+1\};\{2n+1,n+2\};-2T] \\
\int\_0^T e^{-x}\frac{nI\_n(x)}{x}dx&=\frac{(T/2)^{n}}{n!}{}\_2F\_2[\{n+\frac{1}{2},n\};\{2n+1,n+1\};-2T]
\end{align}


I have some trouble to transforming this expression. I know from the answer to [this post](https://mathoverflow.net/questions/271936/closed-form-for-int-0t-e-x-fraci-n-alpha-xxdx/271955) that there exist some polynomials of order $n$, $P\_n(T)$ and $Q\_n(T)$, such that:
\begin{align}
\int\_0^T e^{-x}\frac{nI\_n(x)}{x}dx=\Big(1+\frac{e^{-T}}{T^{n-1}}\big(P\_{n-1}(T)I\_0(T)+Q\_{n-1}(T)I\_1(T)\big)\Big)
\end{align}
which looks somehow to the result of an integration by parts. However I don't manage to use the derivation relation for the Bessel function, or the identities (especially number 2) with the hypergeometric function [here](https://en.wikipedia.org/wiki/Generalized_hypergeometric_function#Generalization_of_Kummer.27s_transformations_and_identities_for_2F2) to obtain the expression of the polynomials.


Furthermore, I have some hints (from numerical evaluations of the formula of a related model) that this could be written in terms of error functions, powers and exponentials. I have troubles to use the previous identities.


What would be an effective method to get more informations from this integral?



Integral with Bessel function and hypergeometric function ${}_2F_2$: explicit expression for these polynomials?
From [DLMF](http://dlmf.nist.gov/10.43.E6), one has, for $n=0,1,2...$
$$\int\_{0}^{x}e^{-t}I\_{n}\left(t\right)\mathrm{d}t=xe^{-x}(I\_{0}\left(x\right)+I%
\_{1}\left(x\right))+n(e^{-x}I\_{0}\left(x\right)-1)+2e^{-x}\sum\_{k=1}^{n-1}(n-k%
)I\_{k}\left(x\right)$$
which may be obtained by recurrence, I guess. Then
\begin{align}
F\_n&=\frac{1}{2}\int\_0^T e^{-x}[I\_{n-1}(x)-I\_{n+1}(x)]\,dx\\
&=1-e^{-T}\left[I\_{0}\left( T \right)+I\_{n}\left( T \right)+2\sum\_{k=1}^{n-1}I\_{k}\left(T\right)\right] \label{eq:1}\tag{1}
\end{align} 
This expression can be checked by taking the derivative of the rhs, using the recursion relation for the Bessel function $I'\_n(T)=\frac{1}{2}\left(I\_{n-1}(T)+I\_{n+1}(T)\right)$:
$$F'\_n=\frac{1}{2}e^{-T}[I\_{n-1}(T)-I\_{n+1}(T)]$$
Now, using the recurrence relation for the Bessel functions
\begin{equation}
I\_{k}\left( T \right)=I\_{k-2}\left( T \right)-\frac{2(k-1)}{T}I\_{k-1}\left( T \right)
\end{equation} 
one deduce that
\begin{equation}
I\_k\left( T \right)=T^{1-k}\left[A\_{k-1}\left( T \right)I\_0\left( T \right)+B\_{k-1}\left( T \right)I\_1\left( T \right)\right]
\end{equation} 
where $A\_{k-1}$ and $B\_{k-1}$ are polynomials of degree $k-1$.
Finally, as expected
\begin{equation}
F\_n=1+\frac{e^{-T}}{T^{n-1}}\left[P\_{n-1}\left( T \right)I\_0\left( T \right)+Q\_{n-1}\left( T \right)I\_1\left( T \right)\right]
\end{equation} 
Eq. (\ref{eq:1}) can be evaluated by Maple using ``simplify'' command giving the polynomials obtained in the previous post [Closed form for $\int\_0^T e^{-x}\frac{I\_n(\alpha x)}{x}dx$](https://mathoverflow.net/questions/271936/closed-form-for-int-0t-e-x-fraci-n-alpha-xxdx/271955)


**Edit** (22/08/'17): The polynomials can be expressed using analogous of the [Lommel polynomials](https://en.wikipedia.org/wiki/Lommel_polynomial) for the modified Bessel functions. In Lemma 5 of [this paper](http://www.sciencedirect.com/science/article/pii/S0377042702005964),  

$$
I\_k(T)=\left( \frac{2}{T} \right)^k r\_k^0\left( \frac{T^2}{4} \right)I\_0(T)+\left( \frac{2}{T} \right)^{k-1} s\_k^0\left( \frac{T^2}{4} \right)I\_1(T)
$$
Expressions for the polynomials $r\_k^0(x)$ and $s\_k^0(x)$ are given either in terms of the Lommel polynomials or explicitly.


