# An integral formula for the reciprocal gamma function

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?

• 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:$\frac{1}{2^{n+1} \sqrt{\pi}} \int_{-\infty}^{\infty} (1+\imath t)^{-(n+\frac{1}{2})} e^{1+\imath t} dt$. – mds May 10 '17 at 17:38
• Is this a question that is better asked on Math Overflow? – mds May 11 '17 at 21:25
• For completeness, there is also this reference on computing the reciprocal of the gamma function via complex contour integration. – mds May 11 '17 at 23:41

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 $$\sum_{n \geq 0} \frac{f_n z^n}{(2n+1)!!} = \frac{1}{2\sqrt{2\pi}} \int_{-\pi}^{\pi} F\left(z e^{-\imath t}\right) e^{\frac{1}{2}\left(e^{\imath t} -\imath t\right)} \operatorname{erf}\left(\frac{e^{\imath t/2}}{\sqrt{2}}\right) dt.$$