1
$\begingroup$

I am trying to get the solution for the integral given below

$$ I = \int_0^\infty t^{-1} \exp(At)\exp(-Bt^2) \large{G}_{0,2}^{2,0}\left( Ct \left| \begin{array}{cc} - \\ \alpha, \beta \end{array} \right. \right) \ dt $$

where A, B, and C are constants, and $\alpha$ and $\beta$ are positive parameters. It's known that the $\exp$ function can be written in the form of Miejer G-function by the using the relation given as

$$ \exp(t)= \large{G}_{0,1}^{1,0}\left( -t \left| \begin{array}{cc} - \\ 0 \end{array} \right. \right)$$

Any help will be appreciated. Thanks

$\endgroup$
3
$\begingroup$

Not an answer, but an expression for the integral which may be easier to handle: from the Wolfram function site or from (9.34.3) in Gradshteyn and Rydzhik, \begin{equation} {G}_{0,2}^{2,0}\left( t \left| \begin{array}{cc} - \\ \alpha, \beta \end{array} \right. \right) =2t^{\tfrac{\alpha+\beta}{2}}K_{\alpha-\beta}\left( 2\sqrt{t} \right) \end{equation} where $K_\nu(.)$ is the modified Bessel function. The integral becomes (with $a=A/C$, $b=B/C^2$) \begin{equation} I=2\int_0^\infty \exp(-bt^2+at)K_{\alpha-\beta}\left( 2\sqrt{t} \right) t^{\tfrac{\alpha+\beta}{2}-1}\,dt \end{equation}

$\endgroup$
1
$\begingroup$

This is not an answer, but a hint. Your Integral may be expressed as a Meijer G-Function of two Variables, which is discussed by Agarwal in 1964 for the first time. A more general form is the H-Fox-Function of two Variables which can be found in the standard publication of Mathai: https://www.researchgate.net/publication/266566090_The_H-function_Theory_and_Applications. An example of how to calculate your integral can be found here: https://www.researchgate.net/publication/269504875_Capacity_of_k_-_m_Shadowed_Fading_Channels. You may find also the source code for implementing the function in Mathematica in the publication above, I already implemented it, too, but it works only for one variable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.