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:

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

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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

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

How can I find an alternate representation for “S” without writing an infinite summation? Please do not just write out the sum and say that is the answer as I am looking for any “convenient” evaluation of the summation. You can find an integral expression of “S”. Please correct me and give me feedback!

  • $\begingroup$ What makes you so sure there is a representation of S other than as the infinite sum? $\endgroup$ – Moko19 Jul 15 at 14:06
  • $\begingroup$ @Tyma_Gaidash I read it, but I don't see how the fact that $\sum \frac{Q(n,n)}{n^n}$ has a convenient compact integral form means that S needs to have one too $\endgroup$ – Moko19 Jul 15 at 14:17
  • 1
    $\begingroup$ @Tyma_Gaidash I'm not saying it doesn't exist. I have no clue if it exists. My point is that your question makes the assumption that it exists; I want to know what is behind that assumption $\endgroup$ – Moko19 Jul 15 at 14:25
  • $\begingroup$ @TymaGaidash $\Gamma(-n,z)$ has a logarithmic branch point at $z=0$. What do you mean by $\Gamma(-n,-n)$? $\endgroup$ – Gary Jul 16 at 15:30
  • $\begingroup$ As I said there is a logarithmic singularity at $z=0$. There is no analytic continuation to $\mathbb C$ or $\mathbb C$ minus the origin. The value of the integral you are proposing will depend on the path of integration. Computer algebra softwares are often bad with multivalued functions. $\endgroup$ – Gary Jul 16 at 15:39

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