2
$\begingroup$

This question follows this one, 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 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 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?

$\endgroup$
1
$\begingroup$

From DLMF, 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 $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$

Edit (22/08/'17): The polynomials can be expressed using analogous of the Lommel polynomials for the modified Bessel functions. In Lemma 5 of this paper,
$$ I_k(T)=\left( \frac{2}{T} \right)^k r_k^0\left( \frac{T^2}{4} \right)I_0(T)+\left( \frac{2}{T} \right)^{k-1} s_k^0\left( \frac{T^2}{4} \right)I_1(T) $$ Expressions for the polynomials $r_k^0(x)$ and $s_k^0(x)$ are given either in terms of the Lommel polynomials or explicitly.

$\endgroup$
0
$\begingroup$

In addition to this excellent answer, I would like to answer this secondary question:

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 the answer for this. The study of this formula was an asymptotic limit of the expression of this other related question: \begin{align} \int_0^T e^{-x}\frac{nI_n(x)}{x}dx=\lim_{\alpha \rightarrow1}\int_0^T e^{-x}\frac{nI_n(\alpha x)}{x}dx \end{align} and $\alpha \rightarrow1$ in my specific model also meant (not explicitly here) that $x$ and $n$ are large. Then I needed to take the following asymptotic formula using Hankel's expansion of the Bessel function: \begin{align} e^{-x}I_n(x)&=\frac{1}{\sqrt{2\pi x}}\sum_{k=0}^\infty (-1)^k \frac{\prod_{l=1}^k \big((2n)^2-(2l-1)^2\big)}{k!\,(8x)^k} \sim \frac{e^{-\frac{n^2}{2x}}}{\sqrt{2\pi}} \end{align} The gaussian function appears and the error function too once integrated. This is a continuous limit for the discrete integers $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.