Solve $\sum_{n=1}^\inftyΓ(-n,-n)\mathop= \pi\left(\frac1e-1\right)i+ \sum_{n=1}^\infty \frac{(-1)^n \text{Ei}(n)}{n!}+\sum_{n=1}^\infty a_n･(-e)^n $

Motivation:

It may be possible to find an integral representation using:

Integral form(s) of a general tetration/power tower integral solution: $$\sum\limits_{n=0}^\infty \frac{(pn+q)^{rn+s}Γ(An+B,Cn+D)}{Γ(an+b)}$$

After seeing

this related question

with the use of gamma functions, I found that $$\mathrm{\int_0^\frac1e x^{-x} dx=\mathop \sum_{n≥1}\frac{Q(n,n)}{n^n}=\mathop \sum_{n≥0}\frac{E_{−n}(n+1)}{n!}=.48689…}$$ source 1 source 2

Here is a graph of the real side of the second to last series.

This result made me wonder how to evaluate the following. The definition of the exponential integral function still works as Re(n)<0:$$\mathrm {S\mathop=^{def}\sum_{n\in \Bbb Z^-} Γ(n,n)= \sum_{n=1}^\infty Γ(-n,-n) = \sum_{n=1}^\infty \frac{E_{n+1}(-n)}{(-n)^{n}}=\sum_{n=1}^\infty\int_{-n}^\infty \frac{dt}{t^{n+1}e^t}= \int_1^\infty\sum_{n=1}^\infty\frac{e^{nt}}{(-n)^nt^{n+1}}dt=\sum_{n=1}^\infty Q(n,n)Γ(n)=-0.5948551252563932027611440348… + 1.98586530379887152055255019996… i}$$

Here is an attempt at an integral representation. If only I could find a summation that I could evaluate because this “$n^n$” in the denominator makes it hard to find a closed form for the sum: $$\mathrm{S=\sum_{n=1}^\infty \int \frac d{dx} Γ(-n,-n\,x)\big|_0^1dx=\int_0^1\sum_{n=1}^\infty \frac{e^{nx}}{n^n x^{n+1}}dx\implies S=\sum_{n=1}^\infty \int_a^b \frac d{dx} Γ(-n,-n\,f(x)) dx =\int_a^b \sum_{n=1}^\infty -\frac{e^{n\,f(x)}f’(x)}{(-n\,f(x))^nf(x)}dx}$$

S converges because of this graph and is almost an alternating series with the absolute value of the summand decreasing to 0. Here are the partial sums for the negative bounds of summation:

Finally, here is the graph of the summand Γ(n,n) plotted at the index bounds:

A possible hint is understanding how to derive the $\int_0^\frac1e x^{-x}dx $ integral sum. You can “reverse” substitute and “reverse” integrate as I showed above.

The following uses an exponential integral function. Here is a result showing that you can easily find an alternate expression for “S” using @Steven Clark’s great answer in:

Nice result

$$\mathrm{\int_1^\infty \frac{dx}{xe^x-1}=\sum_{n\ge1} E_{n}(n)=.269292…}$$

Here is a third example to support that this “S” constant should have another expression. This example uses the lower regularized incomplete gamma function:

Related problem

$$\mathrm{\int_{-1}^0 e^{te^t}dt=1+\sum_{n=1}^\infty \frac{(-1)^nP(n+1,n)}{n^{n+1}}=.772158…}$$

Here is a possible form using the Abel-Plana formula. It may theoretically converge, but Wolfram Alpha has trouble finding it numerically. There are many possibilities for substitution, so I will leave it here:

$$\mathrm{\sum_{n=0}^\infty Γ(-n-1,-n-1)=\frac{Ei(1)-e+i\pi}{2}+\int_0^\infty Γ(-x-1,-x-1)+i\frac{Γ(-ix-1,-ix-1)-Γ(ix-1,ix-1)}{e^{2\pi x}-1}dx}$$

Here is theoretically working integral representation. We can integrate by each term:

$$\sum_1^\infty Γ(-n,-n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_C\frac{Γ(x-n)Γ(x)}{Γ(x+1)(-n)^x}dx=\frac{1}{2\pi i}\int_{k-i\infty}^{k+i\infty}\frac{1}{ (-1)^{x}x}\sum_{n=1}^\infty \frac{Γ(x-n)}{n^x}dx,\text{max}(\text{Re}(n),0)<k$$

How can I find an alternate representation for “S”? You can find a better integral expression of “S”. If not, then please find a closed form, optional, or a series representation which converges faster. Please correct me and give me feedback!

Solution 1:

It is pretty easy to find a closed form for the imaginary part of the sum. Here is some motivation for finding the rest of the answer by expanding and putting corresponding terms into an infinite sum:

$$\sum_{-\infty}^{-1} \Gamma(n,n)=\text{Ei(1)}+i\pi-\frac{\text{Ei(2)}+i\pi }2+ \frac{\text{Ei(3)}+i\pi }{6}-\frac{\text{Ei(4)}+i\pi }{24}+…-e+\frac{3e^2}8-\frac{7e^3}{81}+\frac{47 e^4}{3072}+… =\sum_{n=1}^\infty \frac{(-1)^n\pi i}{n!}+\sum_{n=1}^\infty \frac{(-1)^n \text{Ei}(n)}{n!}+\sum_{n=1}^\infty a_n(-e)^n, a_n=\left\{1,\frac38,\frac7{81},\frac{47}{3072},\frac{427}{187500},\frac{13}{43740},\frac{36013}{1037664180},\frac{155011}{42278584320},…\right\}$$

Therefore we can use the series expansion for $e^x$ and simplify: $$\sum_{-\infty}^{-1} \Gamma(n,n)= -0.5948551252563932027611440348 +\pi\left(\frac1e-1\right)i= \pi\left(\frac1e-1\right)i+ \sum_{n=1}^\infty \frac{(-1)^n \text{Ei}(n)}{n!}+\sum_{n=1}^\infty a_n(-e)^n $$

I tried using the following identify with the Generalized Exponential integral function:

$$\text{Ei}(x)+i\pi=-\text E_1(-x),\text E_v(z)=\int_1^\infty \frac{e^{-zt}}{t^v}dt,\text{Re}(z)>0$$

which did not work with the restriction.

We could probably do something similar to:

Prove $$\displaystyle\sum_{-\infty}^0\text H(x)=\frac12\sum_0^\infty\frac{\text H_\frac x2-\text H_\frac{x-1}2}{x!}=\frac{\text{Ei}(2)-\text{Ei}(1)}e$$

which implies the Gompertz Constant G which is very close to the real part of the sum:

$$-\text G=e\text{Ei}(-1)= -0.59634736232319407434107849936927 ≈\text{Re}\sum_{-\infty}^{-1} \Gamma(n,n)= -0.5948551252563932027611440348 $$

How can I find the rest as an integral or in closed form? Please correct me and give me feedback!