Integrating a product of two Meijer $G$ functions

Question

I want to evaluate an integral $I_1$ defined in $Eq.(1)$ as

\begin{align} I_1=\int_{0}^{\infty}\frac{x\exp(-\beta x)K_1(\alpha x)}{1+x}dx\tag{1} \end{align} Where $\alpha\geq0$, $\beta\geq0$, and $K_1(.)$ is modified first order Bessel's function of second kind

In order to evaluate it, I went through transforming $\frac{x}{1+x}$ and $K_1(\alpha x)$ in terms of Meijer functions, which has been verified in MATHEMATICA and MUPAD

\begin{align} \frac{x}{1+x}=G_{1,1}^{1,1}\bigg(x\bigg|_{1}^{1}\bigg)\tag{2} \end{align} and \begin{align} K_1(\alpha x)=\exp (\alpha x)\sqrt{\pi}G_{1,2}^{2,0}\bigg(2\alpha x\bigg|_{1, -1}^{1/2}\bigg)\tag{3} \end{align}

thereby rewriting $Eq.(1)$ as $I_2$ from $Eq.(2)$ and $Eq.(3)$ \begin{align} I_2=\int_{0}^{\infty}\exp(-\phi x)G_{1,1}^{1,1}\bigg(x\bigg|_{1}^{1}\bigg)G_{1,2}^{2,0}\bigg(2\alpha x\bigg|_{1, -1}^{1/2}\bigg)dx\tag{4} \end{align} where $\phi = \beta - \alpha$

Now in order to proceed further I need to have a single Meijer function for the product of above two Meijer's function i.e.

\begin{align} G_{pq}^{mn}\bigg(\theta x\bigg|_{b_1,b_2,...,b_q}^{a_1,a_2,...,a_p}\bigg)=G_{1,1}^{1,1}\bigg(x\bigg|_{1}^{1}\bigg)G_{1,2}^{2,0}\bigg(2\alpha x\bigg|_{1, -1}^{1/2}\bigg)\tag{5} \end{align}

From $Eq.(5)$ and $Eq.(4)$, I can evaluate the above integral as \begin{align} I=\int_{0}^{\infty}\exp(-\phi x)G_{p,q}^{m,n}\bigg(\theta x\bigg|_{b_1,b_2,...,b_q}^{a_1,a_2,...,a_p}\bigg)dx =\frac{1}{\phi}G_{p+1,q}^{m,n+1}\bigg(\frac{\theta}{\phi} \bigg|_{b_1,b_2,...,b_q}^{0,a_1,a_2,...,a_p}\bigg)\tag{6} \end{align}

Solution to integral is from $[Eq.7.813.1]$, Table of Integrals, Series, and Products, I. S. Gradshteyn and I. M. Ryzhik, 8e

Here what I've tried at MATHEMATICA

MeijerG[{{1}, {}}, {{1}, {}}, x] Exp[\[Alpha]x] Sqrt[\[Pi]] MeijerG[{{}, {1/2}}, {{1, -1}, {}}, 2 \[Alpha]x]

which results into

\begin{align} \frac{xK_1(\alpha x)}{1+x} \end{align}

Instead of this I want a single Meijer's G function representation of it.

Answer 1

The general case seems to be insoluble for Mathematica, but in the special case of $\alpha=\beta$, one can obtain a result in terms of Meijer $G$. Note that the integral for $\alpha=\beta$ can be recast as a Mellin convolution:

$$\int_0^\infty x\exp(-\beta x)K_1(\beta x)\frac1{1+\frac1x}\frac{\mathrm dx}{x}$$

and thus:

MellinConvolve[x Exp[-b x] BesselK[1, b x], 1/(1 + x), x, s] /. s -> 1
   Sqrt[π] MeijerG[{{-1}, {1/2}}, {{-1, -1, 1}, {}}, 2 b]

Compare:

With[{b = 5}, 
     NIntegrate[(x E^(-b x) BesselK[1, b x])/(1 + x), {x, 0, ∞}, WorkingPrecision -> 20]]
   0.024024943996405476775

With[{b = 5}, Sqrt[π] MeijerG[{{-1}, {1/2}}, {{-1, -1, 1}, {}}, 2 b]]
   0.024024943996405476768