3
$\begingroup$

I'm looking to compute an exact integral formula for the reciprocal of the double factorial function, $(2n-1)!!$, or just as easily for the reciprocal gamma function, $\Gamma\left(n+\frac{1}{2}\right)$. I found the post located here and that formula works well for me when, for example, I take $c := 1$.

However, there is another known formula that I'm looking to replicate, or at least find a suitable analog to. Namely, that for integers $n \geq 0$ we have that (this formula is found in the appendices of the Concrete Mathematics book, for example): \begin{align*} \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-n\imath t} e^{e^{\imath t}} dt & = \frac{1}{n!}. \end{align*} I have had a look around google and found page 3 of this article and the Hankel loop contour described here, though I am still struggling to find the analog to this formula for the double factorial function case. I believe that the integral formula above is derived from the contour integral for the reciprocal gamma function, but when I perform a change of variable in the loop contour formula and plugin $z \mapsto n + \frac{1}{2}$, Mathematica computes the following result when $n = 3$ (the expected result is acceptably $\frac{8}{105}$, or ideally $\frac{1}{105}$): \begin{align*} \frac{1}{4\sqrt{\pi}} \int_{-\pi}^{\pi} e^{-(n_3+\frac{1}{2})\imath t} e^{e^{\imath t}} dt & = -\frac{1}{21 e \sqrt{\pi}} + \frac{8}{105} \operatorname{erf}(1). \end{align*} The result is obviously close to the intended formula, so I'm thinking that perhaps it's an issue with the bounds on the integral. I would like to keep the bounds of integration finite as in the factorial function formula if possible. Does anyone have any thoughts, advice, or solutions for this problem?

$\endgroup$
  • $\begingroup$ Just to be clear about another integral representation that does work for this, the following integral obtained from the post here correctly generates the reciprocals of the double factorial function:$\frac{1}{2^{n+1} \sqrt{\pi}} \int_{-\infty}^{\infty} (1+\imath t)^{-(n+\frac{1}{2})} e^{1+\imath t} dt$. $\endgroup$ – mds May 10 '17 at 17:38
  • $\begingroup$ Is this a question that is better asked on Math Overflow? $\endgroup$ – mds May 11 '17 at 21:25
  • $\begingroup$ For completeness, there is also this reference on computing the reciprocal of the gamma function via complex contour integration. $\endgroup$ – mds May 11 '17 at 23:41
1
$\begingroup$

Update: The solution to my original question, which was asking if given an OGF $F(z)$ for some sequence $\{f_n\}_{n \geq 0}$, whether there is an integral transform that generalizes the known OGF-to-EGF transform given by $$\widehat{F}(z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} F\left(z e^{-\imath t}\right) e^{e^{\imath t}} dt,$$ can be answered with Fourier series and integral representations for the Hadamard product of two generating functions. I'm actually writing up a short note one this now, but the integral transform in the previous question is given by $$\sum_{n \geq 0} \frac{f_n z^n}{(2n+1)!!} = \frac{1}{2\sqrt{2\pi}} \int_{-\pi}^{\pi} F\left(z e^{-\imath t}\right) e^{\frac{1}{2}\left(e^{\imath t} -\imath t\right)} \operatorname{erf}\left(\frac{e^{\imath t/2}}{\sqrt{2}}\right) dt.$$

$\endgroup$
  • $\begingroup$ The draft of the article posted on arXiv (arxiv.org/abs/1809.03933, math.CO) has now actually been accepted for publication in a special issue of the journal Axioms. The modified draft of the article contains some more useful related integral transforms for anyone who finds this answer interesting or of use in some application. $\endgroup$ – mds May 18 at 17:31

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.