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I want to calculate the inverse Mellin transform of products of 3 gamma functions. $$F\left ( s \right )=\frac{1}{2i\pi}\int \Gamma(s)\Gamma (2s+a)\Gamma( 2s+b)x^{-s}ds$$ Above contour integral has 3 poles. $$s_{1}=-n$$ $$s_{2}=-\left ( \frac{n+a}{2} \right )$$ $$s_{3}=-\left ( \frac{n+b}{2} \right )$$ Which pole is used in contour integral? $$F\left ( s \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( a-2n \right )\Gamma \left ( -2n+b \right )x^{n}+\sum_{n=0}^{\infty}\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( -\frac{n}{2}-\frac{a}{2} \right)\Gamma \left ( -n-a+b\right )x^{\frac{n}{2}+\frac{a}{2}}+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma \left (-\frac{n}{2}-\frac{b}{2} \right )\Gamma (-n-b+a)x^{\frac{n}{2}+\frac{b}{2}}$$ Is this correct??

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  • $\begingroup$ this is almost correct, you missed factors 1/2 in the second and third sum, see below. $\endgroup$ Commented Apr 22, 2022 at 11:27

2 Answers 2

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To avoid all poles in the Mellin inversion formula you want to integrate along the line $\int_{\gamma-i\infty}^{\gamma+i\infty}ds$ where $\gamma>\max(0,-a/2,-b/2)$; then Mathematica gives the result in terms of the Meijer G-function:

\begin{align} F( x )=&\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s)\Gamma (2s+a)\Gamma( 2s+b)x^{-s}\,ds\\ =&\frac{1}{\pi}2^{a+b-2} G_{0,5}^{5,0}\left(\frac{x}{16}| \begin{array}{c} 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\ \end{array} \right). \end{align} The integral can alternatively be evaluated by contour integration, closing the contour in the left-half complex plane and picking up the poles at $-n$, $-(n+a)/2$, $-(n+b)/2$, $n=0,1,2,\ldots$. For this we assume that $a\neq b$ and $a,b$ are both non-integer --- so that all poles are simple. I then find that

\begin{align} F(x)=&\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( a-2n \right )\Gamma \left ( -2n+b \right )x^{n}\\ &+\frac{1}{2}\sum_{n=0}^{\infty}\frac{\left ( -1 \right )^{n}}{n!}\Gamma \left ( -\frac{n}{2}-\frac{a}{2} \right)\Gamma \left ( -n-a+b\right )x^{\frac{n}{2}+\frac{a}{2}}\\ &+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\Gamma \left (-\frac{n}{2}-\frac{b}{2} \right )\Gamma (-n-b+a)x^{\frac{n}{2}+\frac{b}{2}}. \end{align} I checked numerically that this agrees with the Meijer G-function result -- it differs from the result in the OP by factors 1/2 in the second and third sum.

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  • $\begingroup$ I don't understand your answer. Can you explain for me what did you do? $\endgroup$
    – Pouya
    Commented Apr 21, 2022 at 19:43
  • $\begingroup$ I applied the Mellin inversion formula to the line $\text{Re}\,s=\gamma$ that avoids the three poles you noted in your post; the integral is then evaluated by Mathematica $\endgroup$ Commented Apr 21, 2022 at 19:46
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    $\begingroup$ I do not want to use Mathematica or any computing software. I want it to be calculated manually. $\endgroup$
    – Pouya
    Commented Apr 21, 2022 at 19:51
  • $\begingroup$ I want to know which poles is used? Any 3 poles? $\endgroup$
    – Pouya
    Commented Apr 21, 2022 at 19:57
  • $\begingroup$ @Pouya --- OK, I also calculated the result "manually", as requested; all poles contribute to the contour integral. $\endgroup$ Commented Apr 22, 2022 at 11:26
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This answer is intended to provide additional insight into the initial answer posted by Carlo Beenakker.


The MeijerG function is defined as

$$\text{MeijerG}\left[\{\{a_1..a_n\},\{a_{n+1}..a_p\}\},\{\{b_1..b_m\},\{b_{m+1}..b_q\}\},z,r\right]$$ $$=\frac{r}{2 \pi i} \int \frac{\left(\left(\prod\limits_{i=1}^n \Gamma(1-a_i-r s)\right) \prod\limits_{i=1}^m \Gamma(b_i+r s)\right)}{\left(\prod\limits_{i=n+1}^p \Gamma(a_i+r s)\right) \prod\limits_{i=m+1}^q \Gamma(1-b_i-r s)} z^{-s} \, ds\tag{1}$$

and with the default $r=1$ this becomes

$$\text{MeijerG}\left[\{\{a_1..a_n\},\{a_{n+1}..a_p\}\},\{\{b_1..b_m\},\{b_{m+1}..b_q\}\},z\right]$$ $$=G_{p, q}^{m, n}\left(z\left| \begin{array}{c} a_1 \text{..} a_p \\ b_1 \text{..} b_q \\ \end{array} \right.\right)$$ $$=\frac{1}{2 \pi i} \int \frac{\left(\left(\prod\limits_{i=1}^n \Gamma(1-a_i-s)\right) \prod\limits_{i=1}^m \Gamma(b_i+s)\right)}{\left(\prod\limits_{i=n+1}^p \Gamma(a_i+s)\right) \prod\limits_{i=m+1}^q \Gamma(1-b_i-s)} z^{-s} \, ds\tag{2}$$


From the definition above it can be seen that

$$f(x)=\frac{1}{\pi}\ 2^{a+b-2}\ G_{0,5}^{5,0}\left(\frac{x}{16}| \begin{array}{c} 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\ \end{array} \right)$$ $$=\frac{1}{\pi} 2^{a+b-2} \frac{1}{2 \pi i}\int \Gamma(s+0)\ \Gamma\left(\frac{a}{2}+s\right)\ \Gamma\left(\frac{a+1}{2}+s\right)\ \Gamma\left(\frac{b}{2}+s\right)\ \Gamma\left(\frac{b+1}{2}+s\right) \left(\frac{x}{16}\right)^{-s} \, ds$$ $$=\frac{1}{2 \pi i} \int \Gamma(s)\ \Gamma(a+2 s)\ \Gamma(b+2 s)\ x^{-s} \, ds\tag{3}$$


Mathematica also gives the following result for the inverse Mellin transform

$$f(x)=\mathcal{M}_s^{-1}[\Gamma(s)\ \Gamma(a+2 s)\ \Gamma(b+2 s)](x)\tag{4}$$ $$=\frac{1}{2 \pi i} \int \Gamma(s)\ \Gamma(2s+a)\ \Gamma(2s+b)\ x^{-s}\,ds$$ $$=\pi ^3 2^{-a-b-4} \left(4^{a+b+2} \csc (\pi a) \csc (\pi b) \, _0\tilde{F}_4\left(;\frac{1}{2}-\frac{a}{2},1-\frac{a}{2},\frac{1}{2}-\frac{b}{2},1-\frac{b}{2};-\frac{x}{16}\right)-2 \csc (\pi (b-a)) \left(4^{b+1} x^{a/2} \csc \left(\frac{\pi a}{2}\right) \, _0\tilde{F}_4\left(;\frac{a+2}{2},\frac{1}{2},\frac{1}{2} (a-b+1),\frac{1}{2} (a-b+2);-\frac{x}{16}\right)-4^{a+1} x^{b/2} \csc \left(\frac{\pi b}{2}\right) \, _0\tilde{F}_4\left(;\frac{b+2}{2},\frac{1}{2} (-a+b+1),\frac{1}{2} (-a+b+2),\frac{1}{2};-\frac{x}{16}\right)+4^b x^{\frac{a+1}{2}} \sec \left(\frac{\pi a}{2}\right) \, _0\tilde{F}_4\left(;\frac{a+3}{2},\frac{3}{2},\frac{1}{2} (a-b+2),\frac{1}{2} (a-b+3);-\frac{x}{16}\right)-4^a x^{\frac{b+1}{2}} \sec \left(\frac{\pi b}{2}\right) \, _0\tilde{F}_4\left(;\frac{b+3}{2},\frac{1}{2} (-a+b+2),\frac{1}{2} (-a+b+3),\frac{3}{2};-\frac{x}{16}\right)\right)\right)$$

which is valid for $\max\left(0,-\frac{a}{2},-\frac{b}{2}\right)<\Re(s)$.


Mathematica also expands the MeijerG result as follows.

$$f(x)=\frac{1}{\pi }\ 2^{a+b-2}\ G_{0,5}^{5,0}\left(\frac{x}{16}| \begin{array}{c} 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\ \end{array} \right)=\tag{5}$$ $$\frac{\pi ^2 x^{a/2} \csc \left(\frac{\pi a}{2}\right) \csc \left(\frac{1}{2} \pi (a-b)\right) \sec \left(\frac{1}{2} \pi (a-b)\right) \, _0F_4\left(;\frac{1}{2},\frac{a}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{1}{2},\frac{a}{2}-\frac{b}{2}+1;-\frac{x}{16}\right)}{4 \Gamma \left(\frac{a}{2}+1\right) \Gamma (a-b+1)}-\frac{\pi ^2 x^{b/2} \csc \left(\frac{\pi b}{2}\right) \csc \left(\frac{\pi a}{2}-\frac{\pi b}{2}\right) \sec \left(\frac{1}{2} \pi (a-b)\right) \, _0F_4\left(;\frac{1}{2},\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{1}{2},-\frac{a}{2}+\frac{b}{2}+1;-\frac{x}{16}\right)}{4 \Gamma \left(\frac{b}{2}+1\right) \Gamma (-a+b+1)}+\frac{\pi ^2 x^{\frac{a+1}{2}} \sec \left(\frac{\pi a}{2}\right) \csc \left(\frac{1}{2} \pi (a-b)\right) \sec \left(\frac{1}{2} \pi (a-b)\right) \, _0F_4\left(;\frac{3}{2},\frac{a}{2}+\frac{3}{2},\frac{a}{2}-\frac{b}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{4 \Gamma \left(\frac{a}{2}+\frac{3}{2}\right) \Gamma (a-b+2)}-\frac{\pi ^2 x^{\frac{b+1}{2}} \sec \left(\frac{\pi b}{2}\right) \csc \left(\frac{\pi a}{2}-\frac{\pi b}{2}\right) \sec \left(\frac{1}{2} \pi (a-b)\right) \, _0F_4\left(;\frac{3}{2},\frac{b}{2}+\frac{3}{2},-\frac{a}{2}+\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{4 \Gamma \left(\frac{b}{2}+\frac{3}{2}\right) \Gamma (-a+b+2)}+\frac{\pi ^2 \csc \left(\frac{\pi a}{2}\right) \csc \left(\frac{1}{2} \pi (a+1)\right) \csc \left(\frac{\pi b}{2}\right) \csc \left(\frac{1}{2} \pi (b+1)\right) \, _0F_4\left(;\frac{1}{2}-\frac{a}{2},1-\frac{a}{2},\frac{1}{2}-\frac{b}{2},1-\frac{b}{2};-\frac{x}{16}\right)}{4 \Gamma (1-a) \Gamma (1-b)}$$

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