Definite integral of the product of modified Bessel function of the first kind, exponential, and a power term

Question

Is there a solution to the following integral?

$$ \int_0^{\infty} t^{-0.5}e^{-at}I_{l}\left(kt\right)dt,\;\;\;a,k>0 $$

Here, $I_{l}$ is the modified Bessel function of the first kind, and $a,k$ are constants. I have found solutions of similar integrals without the $t^{-0.5}$ term, and with $l=0$ (e.g. here and here). The closest question and explanation I could find was this. However, the subscript $l$ is important for my question because the entire integral sits inside a summation over integer values of $l\in[-\infty,\infty]$.

Any kind of help/suggestions will be greatly appreciated. Thanks!

Answer 1

I assume that $a>k$ and that $l$ is an integer. Then we have, by equation 2.6.52 in F. Oberhettinger, Tables of Mellin Transforms, Springer-Verlag, New York, 1974., \begin{align*} \int_0^{ + \infty } {t^{ - 1/2} e^{ - at} I_l (kt)dt} & = \int_0^{ + \infty } {t^{ - 1/2} e^{ - at} I_{\left| l \right|} (kt)dt} \\ & = \frac{1}{{k^{1/2} }}\int_0^{ + \infty } {t^{ - 1/2 - 1} (te^{ - (a/k)t} I_{\left| l \right|} (t))dt} \\ & = \Gamma \!\left( {\left| l \right| + \frac{1}{2}} \right)\frac{1}{{(a^2 - k^2 )^{1/4} }}P_{ - 1/2}^{ - \left| l \right|}\! \left( {\frac{a}{{\sqrt {a^2 - k^2 } }}} \right). \end{align*} Here $P_\nu^\mu(z)$ is the associated Legendre function of the first kind. Note that the integral does not converge if $a \leq k$.

Answer 2

There are solutions if $a>k$. For simplicity, let $a=\lambda k$

If $$A_n=\int_0^\infty \frac{e^{-\lambda k t}\, I_n(k t)}{\sqrt{t}}\,dt$$ they seem to write $$\frac 12 b_n\sqrt{\pi k (\lambda +1)} \, A_n=P_n(\lambda) \,K\left(\frac{2 }{1+\lambda}\right)-(1+\lambda)\,Q_n(\lambda) \,E\left(\frac{2 }{1+\lambda}\right)$$ where $K(.)$ and $E(.)$ are the the complete elliptic integrals of the first and second kinds.

The $b_n$ correspond to sequence $$\{1,1,3,15,105,315,3465,45045,45045,765765,14549535,\cdots\}$$

(have a look at sequence $A025547$ in $OEIS$).

The first polynomials are given below $$\left( \begin{array}{ccc} n & P_n(\lambda) & Q_n(\lambda) \\ 0 & 1 & 0 \\ 1 & \lambda & 1 \\ 2 & 4 \lambda ^2-1 & 4 \lambda \\ 3 & 32 \lambda ^3-17 \lambda & 32 \lambda ^2-9 \\ 4 & 384 \lambda ^4-304 \lambda ^2+25 & 384 \lambda ^3-208 \lambda \\ 5 & 2048 \lambda ^5-2144 \lambda ^3+411 \lambda & 2048 \lambda ^4-1632 \lambda ^2+147 \\ 6 & 40960 \lambda ^6-53248 \lambda ^4+16428 \lambda ^2-675 & 40960 \lambda ^5-43008 \lambda ^3+8556 \lambda \end{array} \right)$$

Answer 3

$$ \int_0^{\infty}{t^{-\frac{1}{2}}I_l\left( kt \right) e^{-at}dt} \\ =\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}}\int_0^{\infty}{t^{-\frac{1}{2}+2n+l}e^{-at}dt} \\ =\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}\frac{\Gamma \left( 2n+l+\frac{1}{2} \right)}{a^{2n+l+\frac{1}{2}}}} \\ =\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\sum_{n=0}^{\infty}{\frac{\Gamma \left( n+\frac{l}{2}+\frac{1}{4} \right) \Gamma \left( n+\frac{l}{2}+\frac{3}{4} \right)}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k^2}{a^2} \right) ^n} \\ =\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\frac{\Gamma \left( \frac{l}{2}+\frac{1}{4} \right) \Gamma \left( \frac{l}{2}+\frac{3}{4} \right)}{\Gamma \left( l+1 \right)}\,\,_2F_1\left( \frac{l}{2}+\frac{1}{4},\frac{l}{2}+\frac{3}{4};l+1;\frac{k^2}{a^2} \right) \\ Q_{\nu}^{\mu}(z)=\frac{e^{\mu \pi i}}{2^{\nu +1}}\frac{\Gamma (\nu +\mu +1)}{\Gamma \left( \nu +\frac{3}{2} \right)}\frac{\Gamma \left( \frac{1}{2} \right) \left( z^2-1 \right) ^{\frac{\mu}{2}}}{z^{\nu +\mu +1}}F\left( \frac{\nu +\mu}{2}+1,\frac{\nu +\mu +1}{2};\nu +\frac{3}{2};\frac{1}{z^2} \right) \\ Q_{l-\frac{1}{2}}(z)=\frac{1}{2^{l+\frac{1}{2}}}\frac{\Gamma (l+\frac{1}{2})}{\Gamma \left( l+1 \right)}\frac{\Gamma \left( \frac{1}{2} \right)}{z^{l+\frac{1}{2}}}F\left( \frac{l}{2}+\frac{3}{4},\frac{l}{2}+\frac{1}{4};l+1;\frac{1}{z^2} \right) \\ \int_0^{\infty}{t^{-\frac{1}{2}}I_l\left( kt \right) e^{-at}dt} \\ =\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\frac{\Gamma \left( \frac{l}{2}+\frac{1}{4} \right) \Gamma \left( \frac{l}{2}+\frac{3}{4} \right)}{\Gamma \left( l+1 \right)}\,\,_2F_1\left( \frac{l}{2}+\frac{1}{4},\frac{l}{2}+\frac{3}{4};l+1;\frac{k^2}{a^2} \right) \\ =\frac{2}{\sqrt{2k\pi}}Q_{l-\frac{1}{2}}(\frac{a}{k}) $$