2
$\begingroup$

Given the function $$f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)},$$

can we find and inverse Mellin transform for $f(s)$? That is, $$\frac{1}{2 \pi i}\int_{- i\infty}^{ i\infty} 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)} x^{-s-1 } ds$$ for $x>0$.

I was wondering if the integral can be expressed in terms of hypergeometric functions? For example, this is very similar to the Mellin–Barnes integral \begin{align} {}_2F_1(a,b;c;z) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds \end{align}

However, I am not sure how to connect it to my problem!

Thanks.

  • 0
    this is related to Mellin-Barnes integrals – 2017-01-10
  • 0
    @tire I was thinking about it, but we have a factor of $2^{ \frac{it+1}{6}}$ in front. How to deal with that? – 2017-01-10
  • 0
    can't this just be absorbed into the exponential from the ILT? – 2017-01-10
  • 0
    @tired am not very sure what ILT is? – 2017-01-10
  • 0
    @tired when you have time can you put more details on your thoughts about Mellin-Barnes integral? – 2017-01-11

2 Answers 2

2

Using residues, this isn't very hard. Essentially

$$g(x) = \frac{1}{2\pi i}\int x^{-s} f(s)\,ds = \sum_{k} Res_{s=s_k} x^{-s}f(s)$$

where the $s_k$ are the poles of $\Gamma(2s/3)$ (which occur when $s = -3k/2$ for $k \ge 0$) which aren't cancelled by the zeroes of $\Gamma(s/2)$ (which occur when $s = -2k$). Since the poles are simple, the residues aren't too hard to find.

The poles have principal part

$$\frac{3}{2}\frac{(-1)^n}{n!(s+3n/2)}$$

Therefore

$$g(x) = \sum_{n=0}^\infty \frac{3}{2}\frac{(-1)^n}{n!\Gamma(-3n/4)}2^{(1-3n/2)/6}x^{3n/2}$$

That should do it. Some of these terms disappear because $\frac{1}{\Gamma(-3n/4)}$ sometimes vanishes, I'm too lazy to siphon out the terms that do or don't appear :). Note this is an analytic continuation, and the functions $g(\sqrt[3]{x^2})$ and $g(x^2)$ are entire in $x$.

  • 0
    Thanks. I was thinking that there is an expression in terms of hypergeometric functions? – 2017-01-10
  • 0
    That's a good question. I despise hypergeometric functions and how clunky and ugly they are, so I got nothing to add there, lol. – 2017-01-10
  • 0
    I looked around and seems that your approach might be the best. Can you explain to me a few things? 1) Can you tell me about polls of $\Gamma()$ why are they occurring at $s=-3k/2$ ? – 2017-01-13
  • 0
    $\Gamma(s)$ has poles at the non positive integers $s=-n$ for $n \ge 0$, Their principal part is $\frac{(-1)^n}{n!(n+s)}$. Therefore when you take $\Gamma(2s/3)$ its principal parts are $\frac{(-1)^n}{n!(n+2s/3)}$, which when written in a more manageable form equals $\frac{3}{2}\frac{(-1)^n}{n!(s+3n/2)}$. I can't really say it simpler, all you need to do is to read a little bit up on $\Gamma$ function. – 2017-01-13
  • 0
    Thanks. Now I understand this part. Can you give more details on how you compute the Residue? – 2017-01-13
  • 0
    Mathematica seems to be able to calculate this sums in terms of $_2F_2$...Frightening! – 2017-01-13
  • 0
    @Boby It's just Cauchy's integral theorem. I mean, to show it in full is kind of exhausting. If you've studied Complex Analysis its rather straight forward. Just take a box contour which encompasses all the poles, the left side and the top and bottom tend to zero as they grow to infinity, and you're left with this infinite sum equaling the right side which is the contour integral you are trying to solve for. – 2017-01-13
  • 0
    @james.nixon That part I inderstand. How did you find $Res ( x^{-s} 2^{\frac{s}{6}} \frac{\Gamma( \frac{3s}{2})}{\Gamma(\frac{s}{2})} ; -2k)$?? I guess you tried to use the fact that it is a simple pole. Did you use $\lim_{ s \to -2k} (s-2k) \ x^{-s} 2^{\frac{s}{6}} \frac{\Gamma( \frac{3s}{2} )}{\Gamma(\frac{s}{2})} $ ?? – 2017-01-13
  • 0
    Oh, this is much simpler than you're making it. First of all the pole is $-3k/2$, and secondly we use cauchy's integral formula, it's much easier than you're making it. $Res(\frac{f(s)}{s-s_k},s_k) = f(s_k)$ if $f$ is holomorphic in a neighborhood of $s_k$. – 2017-01-13
  • 0
    @james.nixon Thanks. Another, question. What is the value of the series at $x=0$?? Is it $g(x=0)=0??$ Because if it is then $g'(x)<0$. So, the function is negative in the neighbourhood of zero. Right? – 2017-01-13
  • 0
    $g(0) = 0$ and the derivative function is negative in a neighborhood zero – 2017-01-13
  • 0
    @james.nixon Thank you for all your help. Best.. – 2017-01-14
  • 0
    @james.nixon why is the derivative negative around zero? The term $n=1$ is equal to $ \frac{-1}{ \Gamma(-3/4)}\approx 0.2$ and is positive? – 2017-01-16
  • 0
    Oh you're right, forgot the Gamma function is negative. – 2017-01-16
  • 0
    @james.nixon Dear James I was wondering if you can take a look at very similar question that I posted here: http://math.stackexchange.com/questions/2153479/sign-of-integral-of-frac-2-fracit2-3-gamma-fracit-12-3-2 – 2017-02-22
  • 0
    Would you mind re-checking your derivation? I can't seem to reconcile your purported series with the result I got in my answer. – 2017-10-23
2

Let me offer a partial solution, which I might finish later. Repeatedly using the Gauss multiplication formula yields the following identity:

$$2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}=\frac1{\sqrt{\pi } \sqrt[6]{2}}\frac{\Gamma \left(\frac{s}{6}+\frac{11}{12}\right) \Gamma \left(\frac{s}{6}+\frac{2}{3}\right) \Gamma \left(\frac{s}{6}+\frac{5}{12}\right)}{\Gamma\left(\frac{s}{6}+\frac{1}{2}\right) \Gamma \left(\frac{s}{6}+\frac{5}{6}\right)}\left(\sqrt{\frac38}\right)^{-s}$$

and now the inverse Mellin transform can be directly converted into a Meijer $G$-function, yielding

$$\frac{3\sqrt[6]{32}}{\sqrt{\pi}}G_{2,3}^{3,0}\left(\frac{27z^6}{512}\middle| {{\frac12,\frac56}\atop{\frac{5}{12},\frac23,\frac{11}{12}}}\right)$$

This can then be further expanded into a sum of $3$ ${}_2 F_2$ hypergeometric functions, using this formula.