# Definite integral of exponential, power and Bessel function : a hypergeometric function?

I am studying the following integral: $\int_0^T \frac{e^{-x}}{x}I_n(\alpha x)dx$

I have discovered some things about it but I'm not yet satisfied with it.

The first thing I tried is using the infinite serie representation of the modified Bessel function of the first kind. This leads to lower incomplete gamma functions in the serie, which simplifies when $T\rightarrow \infty$. Writing this serie with Pochhammer symbol, we identify the following Gauss hypergeometric function. A calculator is also giving me a closed form in this limit:

$\int_0^\infty\frac{e^{-x}}{x}I_n(\alpha x)dx =\frac{\big(\frac{\alpha}{2}\big)^n}{n}\,{}_2F_1[\frac{n}{2},\frac{n+1}{2};n+1;\alpha^2]\\ =\frac{\big(\frac{\alpha}{1+\sqrt{1-\alpha^2}}\big)^n}{n}$

First, I am quite curious about how we can get this last result, from the previous one or directly from the integral.

Then, for a finite $T$, this first approach doesn't seem promising because the incomplete gamma function is a product of an exponential and Kummer's hypergeometric function. So I tried to use the multiplication theorem to get an infinite sum of simpler integrals (without the $\alpha$), which are products of a power of $T$ and the hypergeometric function ${}_2F_2$. This sounds better if we can identify the (double) serie with a bivaluated hypergeometric function. However, I get several Pochhammer symbols with coupled summations indices in this way: $(n)_{2i+j}$

It doesn't seem to be possible to uncouple the indices or get the form $(n)_{i+j}$ to fit the hypergeometric functions definition.

Do you have any remark or suggestion about the infinite limit or the finite $T$ ?

Thank you !

The infinite limit is equivalent to the "well-known" fact that the Laplace transform of $I_n(x)/x$ is $\frac{1}{n} \left(s + \sqrt{s^2-1}\right)^{-n}$.
• Thank you for your answer, but do you have any idea how to compute it ? It doesn't seem to be trivial. If we write the Bessel function as an integral and then swap the integrals, we are left with $\int_0^\infty\frac{e^{Ax}}{x}dx$ which is not defined. – Alexandre Jun 10 '17 at 10:16