Laplace transform of "shifted" modified Bessel function Dear all: i'm trying to derive a closed form the following integral,
$$
X_n(R)=\int_0^\infty \exp(-p\, r)I_n(\omega (r+R))\, dr,
$$
where $I_n$ is the standard modified Bessel function of the first kind and $p>\omega$, $0 \leq \omega<1$. I know that, when there is no shift ($R=0$), the values $X_n(R=0)$ are well-known because they are standard Laplace transforms $L[I_n(\omega\, \cdot)](p)$, see for instance here. 
Yet, for $R \not =0$, i've come up with 2 possibilities,
--  using a "shift theorem" for the Laplace transform, which means that i change variable $r \to r+R$ in order to get an exponential term $\exp(pR)$ outside. However, this has an effect on the interval of integration,
$$
X_n(R)=\exp(pR) \int_R^\infty exp(-pr)I_n(\omega\, r)\, dr,
$$
so that the value of the integral isn't well-known anymore. If $|R| \ll 1$, one may use a polynomial approximation of $I_n$ in order to (approximately) compute the term
$$
Y_n(R)=\int_0^R exp(-pr)I_n(\omega\, r)\, dr \simeq \int_0^R exp(-pr)\frac{x^n}{2^n \,n!} \, dr,
$$
which should be subtracted from the value of the Laplace transform as follows: 
$$
X_n(R)=exp(pR)\Big(L[I_n(\omega\, \cdot)](p)-Y_n(R) \Big).
$$
-- using the "summation formula" for Bessel functions, 
$$
I_n(a+b)=\sum_{k=-\infty}^\infty I_k(a)I_{n-k}(b),
$$
in order to retrieve
$$
X_n(R)=\sum_{k=-\infty}^\infty I_{n-k}(\omega R) \int_0^\infty \exp(-p r) I_{k}(\omega r)\, dr,
$$
which is a doubly.infinite series of Laplace transforms,
$$
X_n(R)=\sum_{k=-\infty}^\infty I_{n-k}(\omega R)\, L[I_{k}(\omega\, \cdot)](p),
$$
and, thanks to the property $I_k(\cdot)=I_{-k}(\cdot)$, this expression factorizes with respect to $L[I_0(\omega\, \cdot)](p)=1/\sqrt{p^2 - \omega^2}$ as follows,
$$
X_n(R)=\frac{1}{\sqrt{p^2 - \omega^2}} \sum_{k=-\infty}^\infty I_{n-k}(\omega R)\, \left(\frac{\omega}{p+\sqrt{p^2 - \omega^2}} \right)^{|k|}.
$$
At this point, I was hoping to use the generating function to get rid of the series, at least in the simplest case $n=0$, but having the absolute value $|k|$ cancels all the negative powers, and makes this approach inconclusive. Especially because someone aleady asked for any good value of "half the sum" of the generating function here.
Did I miss anything ? or did i make errors in my derivations ? Any help on this delicate question will be greatly appreciated ... Happy new year 2019 to everyone!
 A: Well, we know that the standard modified Bessel function of the first kind is:
$$\mathcal{I}_1\left(\omega\cdot\left(t+\text{R}\right)\right)=\sum_{\text{n}=0}^\infty\frac{1}{\text{n}!\cdot\Gamma\left(1+\text{n}\right)}\cdot\left(\frac{\omega\cdot\left(t+\text{R}\right)}{2}\right)^{2\text{n}}=$$
$$\sum_{\text{n}=0}^\infty\frac{\left(t+\text{R}\right)^{2\text{n}}}{\text{n}!\cdot\Gamma\left(1+\text{n}\right)}\cdot\left(\frac{\omega}{2}\right)^{2\text{n}}\tag1$$
Now, for the Laplace transform we know that:
$$\mathcal{L}_t\left[\mathcal{I}_1\left(\omega\cdot\left(t+\text{R}\right)\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\exp\left(-\text{s}t\right)\cdot\mathcal{I}_1\left(\omega\cdot\left(t+\text{R}\right)\right)\space\text{d}t=$$
$$\int_0^\infty\exp\left(-\text{s}t\right)\cdot\sum_{\text{n}=0}^\infty\frac{\left(t+\text{R}\right)^{2\text{n}}}{\text{n}!\cdot\Gamma\left(1+\text{n}\right)}\cdot\left(\frac{\omega}{2}\right)^{2\text{n}}\space\text{d}t=$$
$$\sum_{\text{n}=0}^\infty\frac{1}{\text{n}!\cdot\Gamma\left(1+\text{n}\right)}\cdot\left(\frac{\omega}{2}\right)^{2\text{n}}\cdot\left\{\int_0^\infty\exp\left(-\text{s}t\right)\cdot\left(t+\text{R}\right)^{2\text{n}}\space\text{d}t\right\}\tag2$$
Now, the integral equals (when $\Re\left(\text{s}\right)>0\space\wedge\Re\left(\text{R}\right)>0$):
$$\int_0^\infty\exp\left(-\text{s}t\right)\cdot\left(t+\text{R}\right)^{2\text{n}}\space\text{d}t=\frac{\exp\left(\text{R}\text{s}\right)\cdot\Gamma\left(1+2\text{n},\text{R}\text{s}\right)}{\text{s}^{1+2\text{n}}}\tag3$$
So:
$$\mathcal{L}_t\left[\mathcal{I}_1\left(\omega\cdot\left(t+\text{R}\right)\right)\right]_{\left(\text{s}\right)}=$$
$$\sum_{\text{n}=0}^\infty\frac{1}{\text{n}!\cdot\Gamma\left(1+\text{n}\right)}\cdot\left(\frac{\omega}{2}\right)^{2\text{n}}\cdot\frac{\exp\left(\text{R}\text{s}\right)\cdot\Gamma\left(1+2\text{n},\text{R}\text{s}\right)}{\text{s}^{1+2\text{n}}}\tag4$$
