Why does the integral $\int_{x = 0}^{x = \infty } {\frac{{\sqrt{ax}{K_1}\left( {\sqrt {ax} } \right)}}{{x + 1}}} dx$ result in the Meijer G function?

Asked Oct 02 '20 at 12:20

Active Oct 02 '20 at 12:20

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By using Mathematica, I am able to obtain the result of $\int_{x = 0}^{x = \infty } {\frac{{\sqrt {ax} {K_1}\left( {\sqrt {ax} } \right)}}{{x + 1}}} dx$ where $a>0$ as

But I am not sure why does the Meijer G function pops up.

Could you kindly explain the reason for this ?

Thank you !

asked Oct 02 '20 at 12:20

Tuong Nguyen Minh

Essentially it's the convolution theorem for the Mellin transform. The Mellin transforms of the factors are $$\mathcal Mx \mapsto (x + 1)^{-1} = \Gamma(s) \Gamma(1 - s), \ \mathcal Mx \mapsto K_1(\sqrt {a x}) = (a/4)^{-s} \Gamma(1/2 + s) \Gamma(-1/2 + s)/2.$$ – Maxim Oct 03 '20 at 02:02
Could you elaborate more, I still do not understand the relation between the Meilin transform and the appearance of the Meijer G function ? – Tuong Nguyen Minh Oct 03 '20 at 08:29

Why does the integral $\int_{x = 0}^{x = \infty } {\frac{{\sqrt{ax}{K_1}\left( {\sqrt {ax} } \right)}}{{x + 1}}} dx$ result in the Meijer G function?

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