In my answer to this question, I come across the following case of the Meijer G-function:

$$F(b)=G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right), b>0$$

and based on my experiments, $F(b)$ have the following closed form:

$$F(b) \stackrel?=\frac{\Gamma(b)}{b}\left(-\gamma-\psi(b)+\frac{2^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-1\right)+b{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-1\right)\right)$$

Is there any chance of proving this?

Edit: Using the definition of the Meijer G-Function, $F(b)$ have the integral representation: $$F(b)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}dx.$$

Edit 2: I've found a further generalization: $$F(b,z)=G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}e^{(b/2+ix)\log z}dx\\ \stackrel?=\frac{\Gamma(b)z^b}{b}\left(-\log z-\gamma-\psi(b)+\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\right), b\not\in\{0,-1,-2,\dots\},z\neq0,-1.$$ Edit 3: Further simplified the ${_2F_1}$ part.

  • $\begingroup$ Is there any chance of proving this? - Yes, of course. This guy can help. I've heard he's pretty good. :-) $\endgroup$
    – Lucian
    Mar 17 '14 at 16:48
  • $\begingroup$ @Lucian does this Ramanujan fellow have an account on MSE? $\endgroup$ Mar 20 '14 at 12:47
  • $\begingroup$ @OlivierBégassat I think he'd be inactive here, even if he had :) $\endgroup$
    – Ruslan
    Mar 20 '14 at 13:18
  • 2
    $\begingroup$ @Ruslan Or he is posting under a pseudonym, known to us as Cleo. $\endgroup$
    – Chen Wang
    Mar 20 '14 at 13:25

Case $F(1,1)$ is this claim: $$ 2 \int_{-\infty}^\infty \frac{dx}{(4 x^2+1)\cosh(\pi x)} = 2\log 2 . $$ Maybe start with a proof of that.

added March 24

We will attempt this $F(1,1)$ case using residues. I changed variables slightly to make the problem $$ \int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)\cosh(\pi x/2)} = 2\log 2 . $$ The function $$ F(z) := \frac{1}{(z^2+1)\cosh(\pi z/2)} $$ is analytic in the complex plane except for poles at $(2n+1)i$, $n \in \mathbb Z$. Let $M$ be a large integer, and consider the contour integral of $F$ around the square with vertices $$ -4 M ,\quad 4 M ,\quad 4M + 8 M i,\quad -4 M\pi + 8 M i . $$ The integral along the bottom side is $$ \int_{-4M}^{4M} \frac{dx}{(x^2+1)\cosh(\pi x/2)} $$ so the limit as $M \to \infty$ is the integral to be computed. We can do estimates for the other three sides to show their integrals go to zero as $M \to \infty$. (Postponed...)

The poles for $F(z)$ inside the square are at $z=(2n+1) i$, for $n=0,1,\dots, 4M-1$. Now $z=i$ is a double pole, the residue there is $$ \mathrm{Res}_{z=i} F(z) = \frac{-i}{2\pi} . $$ For $n>0$, the residue is $$ \mathrm{Res}_{z=(2n+1) i} F(z) = \frac{(-1)^n i}{2 n (n+1)\pi} $$ Now as $M \to \infty$, the square encloses more and more of these poles, so in the limit the sum is the sum of all the residues in the upper half-plane: $$ \frac{-i}{2\pi}+\sum_{n=1}^\infty \frac{(-1)^n i}{2n(n+1)\pi} =\frac{-i\log 2}{\pi} $$ so finally the integral around the contour (which converges to the integral we want) also converges to $2\pi i$ times this sum of residues, $$ \int_{-\infty}^{+\infty} F(z)\,dz = 2 \log 2 . $$


Using the method in the other answer, assuming it is enough to sum the residues in the upper half-plane as in that case, I get $$ \frac{1}{2\pi}\,\int _{-\infty }^{\infty }\!{\frac {\Gamma \left(b/2+ix \right) \Gamma \left( b/2-ix \right) {{\rm e}^{ \left( b/2+ix \right) \ln \left( z \right) }}}{{b}^{2}/4+{x}^{2}}}{dx} ={\frac {\Gamma \left( b \right) \ln \left( z \right) }{b}}-{ \frac {\Psi \left( b \right) \Gamma \left( b \right) }{b}}-{\frac { \gamma\,\Gamma \left( b \right) }{b}}+{\frac {\Gamma \left( b \right) }{{b}^{2}}}+ \frac{\Gamma \left( b \right) b}{z(b+1)} \;{\mbox{$_4$F$_3$}\left(1,1,b+1,b+1;\,2,2,b+2;\,-\frac{1}{z}\right)} $$

Numerically, this seems to agree with the formula given in the question.

  • $\begingroup$ Using the poles in the lower half-plane gives my conjectured form. Maybe the equivalence is related to some obscure ${_4F_3}$ transformation formula. $\endgroup$
    – Chen Wang
    Mar 25 '14 at 1:48

Inspired by GEdgar's answer for the $F(1,1)$ case, I give an answer (to my own question!) for the general case.

We start from the integral representation $$ F(b,z)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}z^{\frac{b}{2}+ix}dx. $$

We consider the function $G(w)=\frac{\Gamma(\tfrac{b}{2}+iw)\Gamma(\tfrac{b}{2}-iw)}{\tfrac{b^2}{4}+w^2}z^{\frac{b}{2}+iw}dw$. $G(w)$ have double poles at $w=\pm\frac{b}{2}i$, and single poles at $w=\pm(\frac{b}{2}+n)i$ for all positive integer $n$.

Now let $N$ be a large integer, and consider the integral path around the rectancle with vertices $-N,N,N-(N+\frac{b+1}{2})i,-N-(N+\frac{b+1}{2})i$. It is possible to prove that the integrals on the three sides go to zero, and therefore we have $$F(b,z)=-i(\text{sum of residues of $G(w)$ in the lower half-plane}).$$

It remains to calculate the residues at each pole.

In fact, we have $$ \begin{align*} -i\operatorname{Res}_{w=-\frac{b}{2}i}G(w)&=\frac{z^b\Gamma(b)}{b}\left(\frac1b-\log z-\gamma-\psi(b)\right)\\ -i\operatorname{Res}_{w=-(\frac{b}{2}+n)i}G(w)&=\frac{(-1)^{n+1}z^{b+n}\Gamma(b+n)}{n~n!(b+n)}. \end{align*} $$

Comparing with the conjectured closed form, we only need to prove that $$ \frac1b-\sum_{n=1}^{\infty}\frac{(-z)^nb\Gamma(b+n)}{n~n!\Gamma(b)(b+n)}\stackrel?=\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right).$$

Now note that $$ \frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\ =b^{-1}{_2F_1}\left(\begin{array}cb,b\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\ =\underbrace{\sum_{m=0}^{\infty}\frac{\Gamma(b+m)^2\Gamma(b+1)}{b\Gamma(b)^2\Gamma(b+m+1)m!}(-z)^m}_{n=m}-\underbrace{\sum_{m=0}^{\infty}\frac{b\Gamma(m+1)^2\Gamma(b+m+1)\Gamma(2)^2}{\Gamma(1)^2\Gamma(b+1)\Gamma(m+2)^2m!}(-z)^{m+1}}_{n=m+1}\\ =\sum_{n=0}^{\infty}\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\sum_{n=1}^{\infty}\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\\ =\frac1b+\sum_{n=1}^{\infty}\left(\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\right)\\ =\frac1b+\sum_{n=1}^{\infty}\frac{\Gamma(b+n)(-z)^n}{n!\Gamma(b)}\left(\frac{1}{b+n}-\frac{1}{b}\right)\\ =\frac1b-\sum_{n=1}^{\infty}\frac{b\Gamma(b+n)(-z)^n}{n(b+n)n!\Gamma(b)}. $$

Edit: Ninja'd by 2 minute.

  • $\begingroup$ To show that the integrals on the other three sides go to zero, you need to do some estimates of gamma functions in the complex plane. In my original case, I could use addition formulas for trig functions to estimate the integral on the sides of the square. $\endgroup$
    – GEdgar
    Mar 25 '14 at 17:29

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