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This question is related to the following formulas for the reciprocal gamma function $\frac{1}{\Gamma(s)}$ where formula (2) represents the analytic continuation of the sum over $k$ in formula (1). The parameter $f$ in formula (1) is assumed to be a positive integer.


(1) $\quad\frac{1}{\Gamma(s)}=\underset{N,f\to\infty\land M(N)=0}{\text{lim}}\left(e^2\,2^{-s}\sum\limits_{n=1}^N\mu(n)\,n^{s-1}\sum\limits_{k=1}^{f\,n}\frac{(n+i \pi k)^s+(n-i \pi k)^s}{\left(n^2+\pi^2 k^2\right)^s}\right),\quad\Re(s)>0$

(2) $\quad\frac{1}{\Gamma[s]}=\underset{N\to\infty\land M(N)=0}{\text{lim}}\left(e^2\,(2 \pi)^{-s}\sum\limits_{n=1}^N\mu(n)\,n^{s-1}\left(i^s \zeta\left(s,1+\frac{i n}{\pi}\right)+(-i)^s \zeta\left(s,1-\frac{i n}{\pi}\right)\right)\right),\quad s\in\mathbb{C}$


A conditional convergence requirement of the two formulas above is $M(N)=0$ where $M(N)=\sum_{n\le N}\mu(n)$ is the Mertens function. Note the condition $M(N)=0$ can be met for arbitrarily large magnitudes of $N$ since the Mertens function has an infinite number of integer zeros. See https://oeis.org/A028442 for the zeros of the Mertens function.


Formulas (1) and (2) above for $\frac{1}{\Gamma(s)}$ are illustrated in Figures (1) to (6) following the questions below.


Question (1): Is it true formulas (1) and (2) for $\frac{1}{\Gamma(s)}$ converge for $\Re(s)>0$ and $s\in\mathbb{C}$ respectively?


In response to the answer posted by reuns, I added information on the derivation of formula (1) for $\frac{1}{\Gamma(s)}$ to the end of my question below. I don't believe the derivation in the answer posted by reuns is equivalent to my derivation as I don't believe an expression in terms of $\zeta(s,nr/2\pi)$ is equivalent to an expression in terms of $\left(i^s \zeta\left(s,1+\frac{i n}{\pi}\right)+(-i)^s \zeta\left(s,1-\frac{i n}{\pi}\right)\right)$.


The Mellin and Laplace transforms of $\frac{1}{\Gamma(s)}$ defined in formula (1) above are defined in formulas (3) and (4) below. I find it interesting that the $\Gamma(z)$ term appears in the Mellin transform of $\frac{1}{\Gamma(s)}$ illustrated in formula (3) below. I'm not sure the Laplace transform defined in formula (4) below is valid as Mathematica indicates the transform integral of the underlying term of the series in formula (1) for $\frac{1}{\Gamma(s)}$ is only valid for $n\le 1$. Formulas (3) and (4) are illustrated in Figures (7) and (8) following the question below.


(3) $\quad\mathcal{M}_s\left[\frac{1}{\Gamma (s)}\right](z)=\int\limits_0^\infty \frac{1}{\Gamma(s)} s^{z-1}\,ds=\\$ $\qquad\underset{N,f\to\infty\land M(N)=0}{\text{lim}}\left(e^2\,\Gamma(z)\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f n}\left(\left(\log(2)-\log\left(\frac{n}{n+i \pi k}\right)\right)^{-z}+\left(\log(2)-\log\left(\frac{n}{n-i \pi k}\right)\right)^{-z}\right)\right)$

(4) $\quad\mathcal{L}_s\left[\frac{1}{\Gamma (s)}\right](z)=\int\limits_0^\infty \frac{1}{\Gamma(s)} e^{-z s}\,ds=\\$ $\qquad\underset{N,f\to\infty\land M(N)=0}{\text{lim}}\left(e^2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f n} \left(\frac{1}{z+\log\left(2-\frac{2 i \pi k}{n}\right)}+\frac{1}{z+\log\left(2+\frac{2 i \pi k}{n}\right)}\right)\right)$


Question (2): Are there a closed form expressions for the Mellin and Laplace transforms of $\frac{1}{\Gamma(s)}$? If not, are there other formulas for the Mellin and Laplace transforms of $\frac{1}{\Gamma(s)}$ that can be used to compare against formulas (3) and (4) above?


Figure (1) below illustrates formula (1) for $\frac{1}{\Gamma(s)}$ for real $s$, and figures (2) and (3) below illustrate the real and imaginary parts of formula (1) for $\frac{1}{\Gamma(s)}$ evaluated along the line $s=1+i t$. All three plots are evaluated at $f=4$ and the orange and green curves represent the evaluation limits $N=39$ and $N=101$ respectively overlaid on the reference function in blue.


Illustration of formula (1) for 1/Gamma(s)

Figure (1): Illustration of formula (1) for $\frac{1}{\Gamma(s)}$


Illustration of formula (1) for Re(1/Gamma(1+i t))

Figure (2): Illustration of formula (1) for $\Re\left(\frac{1}{\Gamma(1+i t)}\right)$


Illustration of formula (1) for Im(1/Gamma(1+i t))

Figure (3): Illustration of formula (1) for $\Im\left(\frac{1}{\Gamma(1+i t)}\right)$


Figure (4) below illustrates formula (2) for $\frac{1}{\Gamma(s)}$ for real $s$, and figures (5) and (6) below illustrate the real and imaginary parts of formula (2) for $\frac{1}{\Gamma(s)}$ evaluated along the line $s=i t$. In all three plots the orange and green curves represent the evaluation limits $N=39$ and $N=101$ respectively overlaid on the reference function in blue.


Illustration of formula (2) for 1/Gamma(1+i t)

Figure (4): Illustration of formula (2) for $\frac{1}{\Gamma(s)}$


Illustration of formula (2) for Re(1/Gamma(i t))

Figure (5): Illustration of formula (2) for $\Re\left(\frac{1}{\Gamma(i t)}\right)$


Illustration of formula (2) for Im(1/Gamma(i t))

Figure (6): Illustration of formula (2) for $\Im\left(\frac{1}{\Gamma(i t)}\right)$


Figures (7) and (8) below illustrates formulas (3) and (4) which are the Mellin and Laplace transforms of formula (1) for $\frac{1}{\Gamma(s)}$. Both plots are evaluated at $f=4$ and the orange and green curves represent the evaluation limits $N=39$ and $N=101$ respectively.


Illustration of formula (3) which is the Mellin transform of formula (1)

Figure (7): Illustration of formula (3) which is the Mellin transform of formula (1) for $\frac{1}{\Gamma(s)}$


Illustration of formula (4) which is the Laplace transform of formula (1)

Figure (8): Illustration of formula (4) which is the Laplace transform of formula (1) for $\frac{1}{\Gamma(s)}$


In response to the answer posted below by reuns below, actually I derived formula (1) from the relationship

$$y^{-s}=e^2\int\limits_1^\infty x^{-3}\,\delta(\log(x)-1)\frac{\left(\frac{y}{\log(x)}\right)^{-s}}{\log(x)}\,dx\tag{5}$$

using the nested Fourier series representation of $\delta(x-1)$. This leads to

$$y^{-s}=\underset{N,f\to\infty\land M(N)=0}{\text{lim}}\left(e^2\,2^{-s}\,y^{-s}\,\Gamma(s)\sum\limits_{n=1}^N\mu(n)\,n^{s-1}\sum\limits_{k=1}^{f\,n}\frac{(n+i \pi k)^s+(n-i \pi k)^s}{\left(\pi^2 k^2+n^2\right)^s}\right)\tag{6}$$

which is valid for $\Re(s)>0$. Dividing both sides by $y^{-s}\,\Gamma(s)$ leads to

$$\frac{1}{\Gamma(s)}=\underset{N,f\to\infty\land M(N)=0}{\text{lim}}\left(e^2\,2^{-s}\sum\limits_{n=1}^N\mu(n)\,n^{s-1}\sum\limits_{k=1}^{f\,n}\frac{(n+i \pi k)^s+(n-i \pi k)^s}{\left(\pi^2 k^2+n^2\right)^s}\right)\tag{7}$$

which is also valid for $\Re(s)>0$ and identical to formula (1) above.

The $M(N)=0$ restriction is because the nested Fourier series representation of $\delta(x-1)$ only converges at $x=0$ when $M(N)=0$. For more information on derivation of formulas via Mellin convolutions using the nested Fourier series representation of $\delta(x-1)$, see the answer I posted to my own question at https://math.stackexchange.com/q/2380164.

The analytic continuation

$$\sum _{k=1}^{\infty } \frac{(n+i \pi k)^s+(n-i \pi k)^s}{\left(\pi^2 k^2+n^2\right)^s}=\pi^{-s} \left(i^s \zeta\left(s,1+\frac{i n}{\pi}\right)+(-i)^s \zeta\left(s,1-\frac{i n}{\pi}\right)\right)\tag{8}$$

leads to formula (2) above for $\frac{1}{\Gamma(s)}$. I don't believe the derivation in the answer posted by reuns is equivalent to my derivation as I don't believe an expression in terms of $\zeta(s,nr/2\pi)$ is equivalent to an expression in terms of $\left(i^s \zeta\left(s,1+\frac{i n}{\pi}\right)+(-i)^s \zeta\left(s,1-\frac{i n}{\pi}\right)\right)$.

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  • $\begingroup$ (1) and (2) appear without citation. Do you intend to represent that they are your original work? $\endgroup$ Dec 18, 2019 at 0:28
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    $\begingroup$ @EricTowers Yes, I derived formula (1) a couple of years ago, and recently derived formula (2) from formula (1). To my knowledge these two formulas were not previously known to anyone but myself. $\endgroup$ Dec 18, 2019 at 0:44

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What you have done is

$$\frac1{\Gamma(s)}=\frac{\sin(\pi s)}{\pi}\Gamma(1-s)=\frac{-1}{2i\pi}\int_{C_r} (-z)^{-s}e^{-z}dz$$ $$=\frac{-1}{2i\pi}\int_{C_r} (-z)^{-s} (f(z)+\lim_{N\to \infty}\sum_{n\ge N,M(N)=0} \mu(n) \frac1{e^{zn}-1})dz$$ $$=F_r(s)+\frac{-1}{2i\pi} \lim_{N\to \infty}\sum_{n\ge N,M(N)=0} \mu(n)n^{s- 1}\int_{C_{rn}} (-z)^{-s} \frac1{e^{z}-1}dz$$ $$= F_r(s)-\lim_{N\to \infty}\sum_{n\ge N,M(N)=0} \mu(n)n^{s- 1}\sum_{|2i\pi m|> rn}Res((-z)^{-s} \frac1{e^{z}-1},2i\pi m)$$ The last residue theorem steps applies only for $\Re(s) > 1$, giving an expresion in term of $\zeta(s,nr/2\pi)$, this expression stays true for all $s$.

$C_r$ is the contour $+\infty\to e^{i\pi}r\to e^{-2i\pi }\infty$ enclosing $[0,\infty)$ clockwise with a circle of radius $r$ around $0$, for $n$ large enough $C_{rn}$ encloses the first few poles of $\frac1{e^{z}-1}$.

The $M(N)=0$ restriction is because $e^{-z}=\sum_{n\ge 1}\mu(n)\frac1{e^{nz}-1}$ is valid only for $\Re(z) > 0$, for $\Re(z) < 0$ it is $-e^{z}=\sum_{n\ge 1}\mu(n)(\frac1{e^{nz}-1}+1)$.

Whence $$F_r(s) =\frac{1}{2i\pi} \int_{|z|=r,\Re(z) < 0} (-z)^{-s} (-e^z-e^{-z})dz\ne 0$$ which means that your formula is incorrect.

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  • $\begingroup$ I don't believe your derivation is equivalent to my derivation as I don't believe an expression in terms of $\zeta(s,nr/2\pi)$ is equivalent to an expression in terms of $\left((-i)^s \zeta \left(s,1-\frac{i n}{\pi }\right)+i^s \zeta \left(s,\frac{i n}{\pi }+1\right)\right)$. $\endgroup$ Dec 18, 2019 at 20:38

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