Justify an approximation of $\sum_{k=1}^\infty\operatorname{Ei}(-k)$, where $\operatorname{Ei}(x)$ denotes the exponential integral function

Question

When I've considered an integral involving the fractional part function I got two terms in the result, one of which was $$\sum_{k=1}^\infty\operatorname{Ei}(-k),\tag{1}$$ where $\operatorname{Ei}(x)$ denotes the exponential integral function, see if you want this MathWorld.

Question. I would like to know how to justify an approximation (what method I need to justify an approximation, isn't required a very good approximation, of this series of integrals). Also if you can convince to me that there is no a closed-form of $(1)$ as we can suspect (that is the sum of previous series, in terms of constants or particular values of some special functions) I appreciate it. Many thanks.

Of course if you need refer the literatue feel free to do it, that I can to search such books or papers and read those interesting facts from those. Also I don't know if this question or similar were asked in this site MSE.

Answer 1

In simpler terms, you are looking for an approximation of $$ -\sum_{k\geq 1}\int_{k}^{+\infty}\frac{du}{u e^u} = -\int_{1}^{+\infty}\frac{\lfloor u \rfloor }{u e^u}\,du = -\frac{1}{e}+\int_{1}^{+\infty}\frac{\{u\}}{ue^u}\,du=-1+\int_{0}^{+\infty}\frac{\{u\}}{ue^u}\,du$$ where the inverse Laplace transform of $\frac{1}{u e^u}$ is the indicator function of $(1,+\infty)$ and the Laplace transform of $\{u\}$ is $\frac{1}{s^2}\left(1-\frac{s}{e^s-1}\right)$. It follows that $$ -\sum_{k\geq 1}\int_{k}^{+\infty}\frac{du}{u e^u} = \color{red}{-\int_{1}^{+\infty}\frac{ds}{s(e^s-1)}}$$ which is simple to approximate numerically. For instance, by replacing the integrand function with $\frac{e^{3/2}}{s(e-1)e^{3s/2}}$ we get that the RHS is not far from $\frac{e^{3/2}}{e-1}\text{Ei}\left(-\frac{3}{2}\right)\approx -0.26$.

Answer 2

For what concerns the closed form, there is no "elegant" close form, that is, no cool expression.

Yet there is a numerical close form which is:

$$-0.2867924309492973(...)$$

Playing a bit with numbers I found a very rough and poor approximation, which actually consist in the first $4$ digits, which is:

$$-\frac{41}{143} \approx -0.2867$$

An approximation is justified when no exact formula exists, and when it brings final results which are pretty close to the true ones. This all depends upon the used technique of approximation, the order of approximation (like when you use Taylor Series), and the final error.

Answer 3

See my new result here. I show that your sum is equal to

$$\frac{1}{2}\left(\ln\left(2\pi\right)-\gamma\right)-1+\sum_{n=2}^{\infty}\frac{B_n}{n!\left(n-1\right)}$$

See also this post.