On $\mathrm{\sum\limits_{x=1}^\infty Ci(x)}$.

I have held out on asking this question as it seems a bit simple, but I have also asked some similar summation questions. This brought about the idea of adding it to the collection. The problem uses properties of the Cosine Integral function with related functions.

Here is a related question:

Prove $$\sum_{n=1}^\infty \text{Ci}(\pi n)=\frac{\ln(2)-\gamma}{2}$$

Also see the hypergeometric definitions. The expression with the Confluent-U hypergeometric function looks like some type of sum transform. Note that Wolfram Research defines Ci(x) and related functions with an additional logarithm term which will be ignored. Without this term, the Wolfram version just has an extra imaginary part, but the sums will all be over the real part of the summand:

$$\mathrm{Ci(1)+Ci(2)+Ci(3)…=\sum_{x=1}^\infty Ci(x)= \sum_{x=1}^\infty Chi(ix)= \sum_{x=1}^\infty\frac{Ei(-ix)+Ei(ix)}{2}=-\sum_{x=1}^\infty\frac{Γ(0,ix)+Γ(0,-ix)}{2}=\sum_{x=1}^\infty G_{2,3}^{1,2}\left(_{1,0,0}^{\ \ 1,1} \big|x\right)=\sum_{x=1}^\infty (Ei(ix)-i Si(x))=-\sum_{x=1}^\infty e^x U(1,1,-x)=\sum_{x=1}^\infty x\,_2F_1(1,1,2,2,x)=\sum_{x=1}^\infty \sum_{y=1}^\infty \frac{x^y}{yy!}=-\sum_{x=1}^\infty\sum_{y=1}^\infty\sum_{z=1}^y\frac{e^x(-x)^y}{zy!}=ln\prod_{x=1}^\infty e^{Ci(x)}=0.630…}$$

The Abel-Plana formula seems to work, so here is the constant using it. Note that you can split the integral:

$$\mathrm{\sum_{x=0}^\infty Ci(x+1)= sin(1)-\frac{Ci(1)}{2}+\frac i2\int_0^\infty [Ci(1+ix)-Ci(1-ix))][coth(\pi x)-1]dx}$$

What is a way to evaluate the summation in Exact form? Even though there is no explicit $(-1)^x$, this sum converges by the Alternating Series test as it is decreasing with a limit to 0 of the summand which is just Ci(x) on $\text x\ge 1$. Please correct me and give me feedback!

Note:

See this question on

Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$?

and

On $\mathrm{\sum\limits_{n=0}^\infty \left(C(n)-\frac{\sqrt\pi}{2\sqrt2}\right)+ \sum\limits_{n=0}^\infty \left(S(n)-\frac{\sqrt\pi}{2\sqrt2}\right)}$

Jus for fun, here is another nice identity with credit from @ComplexYetTrivial and @Metamorphy. This will use the principal branch for simplicity. Also, please remember the Pi-Product notation. This works for $x>0$ and for $x=\frac{1}{2\pi}$, the formula works for all $b\ne 0$. It includes the Harmonic numbers and an indicator function of natural numbers:

$$\mathrm{\prod_{n=1}^\infty b^{Ci(2\pi n x)}= b^{\sum\limits_{n=1}^\infty Ci(2\pi n x)}=b^{-\frac γ2}\sqrt{\frac{b^{H_{\lceil x\rceil}}b^{\frac{I_{\Bbb N}(x)}{2\lceil x\rceil}}}{\sqrt[\lceil x\rceil]{b} b^{ln(x)}}}\mathop\implies ^{x=\frac{y}{2\pi}} \prod_{n=1}^\infty b^{Ci(n y)}= b^{-\frac {γ+ln(2)+ln(\pi)}2}\sqrt{\frac{b^{H_{\left\lceil \frac{y}{2\pi}\right\rceil}}b^{\frac{I_{\Bbb N}\left(\frac {y}{2\pi}\right)}{2 \left\lceil \frac{y}{2\pi}\right\rceil}}}{\sqrt[\left\lceil \frac{y}{2\pi}\right\rceil]{b} b^{ln y}}}}$$

Another fascinating special case is the following. You can derive similar ones like it yourself if you like, so I will only post one. I will also expand it using Euler’s formula. Note I will use the principal branch only:

$$\mathrm{log_{-1}\prod_1^\infty (-1)^{Ci(x)}=\sum_1^\infty Ci(x)\implies \prod_1^\infty (-1)^{Ci(x)} =(-1)^{\sum_1^\infty Ci(x)}=(-1)^\frac{ln(2\pi)-γ}{2}=i^{ln(2\pi)-γ}=i^{-γ}(2\pi)^\frac{i\pi}{2}=cos\left(\frac{\pi(γ-ln(2\pi))}{2}\right)+i\,sin \left(\frac{\pi(γ-ln(2\pi))}{2}\right)=-\sqrt{\frac{1+cos(\pi(γ-ln(2\pi))}{2}}+ \sqrt{\frac{1-cos(\pi(γ-ln(2\pi))}{2}} =-0.39810115624278019936718245779... + 0.91734152277009760782638804362... i}$$

$$\newcommand{\Ci}{\operatorname{Ci}}\Ci(x):=\int_\infty^x\frac{\cos t}{t}\,dt=\frac{\sin x}{x}-\frac{\cos x}{x^2}+2\int_x^\infty\frac{\cos t}{t^3}\,dt=\frac{\sin x}{x}+\mathcal{O}(x^{-2})$$ implies that $S(x):=\sum_{n=1}^\infty\Ci(nx)$ converges for $x>0$. Now we use well-known $$s(x):=\sum_{n=1}^\infty\frac{\sin nx}{n},\qquad c(x):=\sum_{n=1}^\infty\frac{\cos nx}{n^2},$$ namely $s(x)=(\pi-x)/2$ for $0<x<2\pi$ and $c'(x)=-s(x)$ for $x\notin 2\pi\mathbb{Z}$. Then $$S(x)=\frac{s(x)}{x}-\frac{c(x)}{x^2}+2\int_x^\infty\frac{c(t)}{t^3}\,dt=\frac{s(x)}{x}-\int_x^\infty\frac{s(t)}{t^2}\,dt$$ after integration by parts. Further, for $0\leqslant\delta\leqslant 1$ we compute $$\int_{2(n-\delta)\pi}^{2n\pi}\frac{s(t)}{t^2}\,dt=\frac14\left(\frac{2\delta-1}{n-\delta}+\frac1n-2\log\frac{n}{n-\delta}\right),$$ which gives, using the asymptotics of the harmonic numbers, $$\int_{2\pi}^\infty\frac{s(t)}{t^2}\,dt=\frac14\sum_{n=1}^\infty\left(\frac1n+\frac1{n+1}-2\log\frac{n+1}{n}\right)=\frac{2\gamma-1}{4}.$$

This leads to an expression for $S(x)$ as a finite sum (of $\approx x/2\pi$ terms). In particular, $$S(1)=\frac{\log(2\pi)-\gamma}{2}\approx 0.6303307007539\cdots$$