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EDIT: Some additional details and corrections, I would appreciate any information about the highlighted expression.

I try to solve $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$ where $I_n(x)$ is the modified Bessel function of the first kind and $0<\alpha<1$.

My first approach was to turn this integral into an infinite sum to fit a hypergeometric series:

  • Using the infinite series representation of the Bessel function, I got incomplete gamma functions in the sum, which does not sound promising.
  • The multiplication theorem yields an infinite series of integrals where we get rid of the $\alpha$: $$\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx=\alpha^n\sum_{m=0}^{\infty}\frac {\big(\frac {\alpha ^{2}-1}{2}\big)^m}{m!}\int_0^T e^{-x}x^{m-1}I_{n+m}(x)dx$$

According to a table of integrals, the new integrals are: \begin{align} \int_0^T e^{-x}x^{m-1}I_{n+m}(x)dx&=\frac{T^{2m+n}}{2^{m+n}}\frac{\Gamma(2m+n)}{\Gamma(m+n+1)\Gamma(2m+n+1)}\\ &\times{}_2F_2[\{m+n+\frac{1}{2},2m+n\};\{2m+2n+1,2m+n+1\};-2T] \end{align}

Expanding ${}_2F_2$ (let's call $k$ the summation index), we get a double infinite series which might fit the definition of a hypergeometric function of 2 variables. However, I get several Pochhammer symbols with coupled summations indices:

$$\sum_{m,k=0}^{\infty}\frac{(n+\frac{1}{2})_{m+k}(n)_{2m+k}}{(n+1)_{2m+k}(2n+1)_{2m+k}}\frac{X^mY^k}{m!\,k!}$$

which, apparently, does not fit any hypergeometric function definition (at least, this is not an Appell function).

Another approach could be to get inspiration from the limit $T\rightarrow \infty$ which is the Laplace transform of $\frac{I_n(x)}{x}$ (up to a constant) and it has a closed form (according to a table):

$$\int_0^\infty e^{-x}\frac{I_n(\alpha x)}{x}dx=\frac{\big(\frac{\alpha}{1+\sqrt{1-\alpha^2}}\big)^n}{n}$$

However, I don't find any reference on the way to compute this. EDIT: This comes from the recurrence identity $I_{n-1}(x)-I_{n+1}(x)=2n\frac{I_n}{x}$ and the calculation of the Laplace transform of $I_n(x)$ is well documented.

Do you have any information or suggestion about the above formulae ?

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    $\begingroup$ I don't understand the downvotes. On the other hand, I don't understand why the OP expects a closed form, either... $\endgroup$ – Igor Rivin Jun 11 '17 at 13:23
  • $\begingroup$ You are right, it may have no closed form. I asked because this last integral seems to be an unexploited approach. Maybe I should look at the asymptotic behaviour and be satisfied with it. $\endgroup$ – Alexandre Jun 11 '17 at 13:32
  • $\begingroup$ for $\alpha=1$ there is a closed form expression in terms of $I_0(t)$ and $I_1(t)$ $\endgroup$ – Carlo Beenakker Jun 11 '17 at 14:29
  • $\begingroup$ and for small $\alpha$ it's an incomplete gamma function, are these asymptotics of interest? $\endgroup$ – Carlo Beenakker Jun 11 '17 at 14:56
  • $\begingroup$ Great! I found the incomplete gamma for small $\alpha$ from the first infinite series proposed, and I can also get $\alpha=1$ from the second infinite series but it is expressed with the hypergeometric function ${}_2F_2$. Do you have more detail about this closed expression in terms of $I_0$ and $I_1$? $\endgroup$ – Alexandre Jun 11 '17 at 15:53
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as requested by the OP in the comment section:

$$\int_0^T e^{-x}I_n(x)\frac{1}{x}\,dx=\frac{1}{n}+\frac{1}{n T^{n-1}}e^{-T}\left[a_n(T)I_0(T)+b_n(T)I_1(T)\right]$$

the functions $a_n$ and $b_n$ are polynomials of degree $n-1$, I do not have a closed form expression; the first few are:

$$a_1(T)=-1,\;\;a_2(T)=-2T,\;\;a_3(T)=-3 T^2+4 T,$$ $$a_4(T)=-4 T^3+8 T^2-24 T,\;\;a_5(T)=-5 T^4+20 T^3-48 T^2+192 T$$ $$b_1(T)=-1,\;\;b_2(T)=-2T+2,\;\;b_3(T)=-3T^2+4T-8,$$ $$b_4(T)=-4 T^3+12 T^2-16 T+48,\;\;b_5(T)=-5 T^4+20 T^3-88 T^2+96 T-384$$

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  • $\begingroup$ Thanks! I would appreciate any reference or indication on how you got this. The coefficients $b_{n,0}$ seems to be $(-2)^nn!$. $\endgroup$ – Alexandre Jun 11 '17 at 19:33
  • $\begingroup$ Mathematica evaluates the integral for arbitrary real $n$ in terms of a hypergeometric function, which reduces to these explicit expressions for integer $n$. $\endgroup$ – Carlo Beenakker Jun 11 '17 at 20:14
  • $\begingroup$ Your expression seems to come from the formula relating ${}_2F_2$ and ${}_0F_1$ and other similar identities. But the expressions of your polynomials are not clear yet. $\endgroup$ – Alexandre Jun 13 '17 at 10:26

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