Krdytkyu

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!

Well, we know that the standard modified Bessel function of the first kind is:

$$mathcal{I}_1left(omegacdotleft(t+text{R}right)right)=sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omegacdotleft(t+text{R}right)}{2}right)^{2text{n}}=$$
$$sum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}tag1$$

Now, for the Laplace transform we know that:

$$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}:=int_0^inftyexpleft(-text{s}tright)cdotmathcal{I}_1left(omegacdotleft(t+text{R}right)right)spacetext{d}t=$$
$$int_0^inftyexpleft(-text{s}tright)cdotsum_{text{n}=0}^inftyfrac{left(t+text{R}right)^{2text{n}}}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}spacetext{d}t=$$
$$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotleft{int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}tright}tag2$$

Now, the integral equals (when $Releft(text{s}right)>0spacewedgeReleft(text{R}right)>0$):

$$int_0^inftyexpleft(-text{s}tright)cdotleft(t+text{R}right)^{2text{n}}spacetext{d}t=frac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag3$$

So:

$$mathcal{L}_tleft[mathcal{I}_1left(omegacdotleft(t+text{R}right)right)right]_{left(text{s}right)}=$$
$$sum_{text{n}=0}^inftyfrac{1}{text{n}!cdotGammaleft(1+text{n}right)}cdotleft(frac{omega}{2}right)^{2text{n}}cdotfrac{expleft(text{R}text{s}right)cdotGammaleft(1+2text{n},text{R}text{s}right)}{text{s}^{1+2text{n}}}tag4$$