In this question the notation $\tilde{f}(x)$ refers to an analytic representation of the summatory function

$$f(x)=\sum\limits_{n=1}^x a(n)\tag{1}$$

that converges to

$$\underset{\epsilon\to 0}{\text{lim}}\left(\frac{f(x-\epsilon)+f(x+\epsilon)}{2}\right)\tag{2}$$

whereas the notation $\hat{f}(x)$ refers to a closed form expression for $\tilde{f}(x)$ which is undefined at the discontinuities of $f(x)$.

Consider the following three summatory functions

$$h(x)=\sum\limits_{n=1}^x \frac{1}{n}\tag{3}$$

$$s(x)=\sum\limits_{n=1}^x 1\tag{4}$$

$$g(x)=\sum\limits_{n=1}^x n\tag{5}$$

and their corresponding analytic representations

$$\tilde{h}(x)=\log(x)+\gamma+\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K (\text{Ei}(2 \pi i k x)+\text{Ei}(-2 \pi i k x))\right),\quad x>0\tag{6}$$

$$\tilde{s}(x)=x-\frac{1}{2}+\underset{K\to\infty}{\text{lim}}\left(\sum_{k=1}^K \frac{\sin(2 \pi k x)}{\pi k}\right),\quad x>0\tag{7}$$

$$\tilde{g}(x)=\frac{x^2}{2}-\frac{1}{12}+\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \frac{2 \pi k x \sin(2 \pi k x)+\cos(2 \pi k x)}{2 \pi^2 k^2}\right),\quad x\ge 0\tag{8}$$

Whereas formulas (6) and (7) above are both only valid for $x>0$, formula (8) above is valid for $x\ge 0$ and converges to an even function of $x$.

The primary objective of this question is to find a closed form expression for $\hat{h}(x)$ analogous to the closed form expressions for $\hat{s}(x)$ and $\hat{g}(x)$ defined below.

$$\hat{s}(x)=x-\frac{1}{2}+\frac{i \left(\log\left(1-e^{2 i \pi x}\right)-\log\left(1-e^{-2 i \pi x}\right)\right)}{2 \pi}\tag{9}$$

$$\hat{g}(x)=\frac{x^2}{2}-\frac{1}{12}+\frac{\text{Li}_2\left(e^{-2 i \pi x}\right)+\text{Li}_2\left(e^{2 i \pi x}\right)+2 i \pi x \left(\log\left(1-e^{2 i \pi x}\right)-\log\left(1-e^{-2 i \pi x}\right)\right)}{4 \pi^2}\tag{10}$$

Question: Is there a closed-form expression for

$$f_1(x)=\sum\limits_{k=1}^\infty (\text{Ei}(i k x)+\text{Ei}(-i k x))\,,\quad x\in\mathbb{R}\tag{11}$$

where $\text{Ei}(z)$ is the exponential integral function analogous to the closed form representations defined in formulas (12) and (13) below?

$$f_2(x)=\sum_{k=1}^{\infty} \frac{\sin(k x)}{\pi k}=\frac{i}{2 \pi} \left(\log\left(1-e^{i x}\right)-\log\left(1-e^{-i x}\right)\right)\tag{12}$$

$$f_3(x)=\sum\limits_{k=1}^{\infty} \frac{k x \sin(k x)+\cos(k x)}{2 \pi^2 k^2}=\frac{\text{Li}_2\left(e^{-i x}\right)+\text{Li}_2\left(e^{i x}\right)+i x \left(\log \left(1-e^{i x}\right)-\log \left(1-e^{-i x}\right)\right)}{4 \pi ^2}\tag{13}$$

The related Laplace transform

$$F_1(s)=s\, \mathcal{L}_x[f_1(x)](s)=s \int\limits_0^\infty f_1(x)\,e^{-s x}\,dx=-\sum\limits_{k=1}^\infty \log\left(1+\frac{s^2}{k^2}\right)\tag{14}$$

has the closed form representation

$$F_1(s)=-\log\left(\frac{\sinh(\pi s)}{\pi s}\right)\tag{15}$$

which is based on this answer to one of my related questions.

Mathematica doesn't seem to be capable of simplifying the sum in formula (11) above or deriving the inverse Laplace transform of $\log\left(\frac{\sinh(\pi s)}{\pi s}\right) \frac{1}{s}$ related to formula (15) above.

With respect to my question, the best I've been able to do so far is a finite sum where the number of terms are dependent on the magnitude of $x$, but that's not the objective solution. I suspect there's a closed form expression for $f_1(x)$ that's more analogous in complexity to formula (13) above than the simpler formula (12) above.

The function $f_1(x)$ defined in formula (11) above can also be expressed in terms of the exponential integral function $E_1(z)$ as

$$f_1(x)=-\sum\limits_{k=1}^\infty (E_1(i k x)+E_1(-i k x))\,,\quad x\in\mathbb{R}\tag{16}$$

or in terms of the "upper" incomplete gamma function $\Gamma(0,z)$ as

$$f_1(x)=-\sum\limits_{k=1}^\infty (\Gamma (0,i k x)+\Gamma (0,-i k x))\,,\quad x\in\mathbb{R}\tag{17}$$

or assuming $x\ge 0$, in terms of the cosine integral function $\text{Ci}(x)$ as

$$f_1(x)=2\sum\limits_{k=1}^\infty \text{Ci}(k\,x)\,,\quad x\ge 0\,.\tag{18}$$

I'm looking for a result for $x\in\mathbb{R}$ which is useful in the evaluation of other related formulas under investigation, but I suppose a result for $x\ge 0$ is better than nothing.

Figure (1) below illustrates the function $(\text{Ei}(i x)+\text{Ei}(-i x))$, related to the term of formula (11) above for $f_1(x)$, has a branch point at $x=0$, is an even function of $x$ with a decaying oscillation around the real axis as $|x|\to\infty$, and $\underset{x\to\infty}{\text{lim}}(\text{Ei}(i x)+\text{Ei}(-i x))=0$.

Illustration of (Ei(i x)+Ei(-i x))

Figure (1): Illustration of $(\text{Ei}(i x)+\text{Ei}(-i x))$

Figure (2) below illustrates the function $\text{Ci}(x)$, related to the term of formula (18) above for $f_1(x)$, has an undesirable branch cut along the negative real axis.

Illustration of $Im(Ci(x))

Figure (2): Illustration of $\Im(\text{Ci}(x))$

Figure (3) below illustrates formula (6) for $\tilde{h}(x)$ above in orange overlaid on the blue reference function $h(x)$ defined in formula (3) above where formula (6) is evaluated at $K=20$. The red discrete evaluation points in Figure (3) below illustrate the evaluation of formula (6) for $\tilde{h}(x)$ at integer values of $x$.

Illustration of formula (6) for}h(x)

Figure (3): Illustration of formula (6) for $\tilde{h}(x)$

Formulas (11), (12), and (13) above lead to the following formulas for $\zeta(s+1)$, $\zeta(s)$, and $\zeta(s-1)$.

$$\zeta(s+1)=s\, \mathcal{M}_x[f_1(2 \pi x)](-s)=s \int\limits_0^\infty f_1(2 \pi x)\,x^{-s-1}\,dx$$ $$=2^{s+1}\, \pi^s\, \cos\left(\frac{\pi s}{2}\right)\, \Gamma(-s)\, \underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N n^s\right),\quad\Re(s)<-1\tag{19}$$

$$\zeta(s)=s\, \mathcal{M}_x[f_2(2 \pi x)](-s)=s \int\limits_0^\infty f_2(2 \pi x)\,x^{-s-1}\,dx$$ $$=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma (1-s) \underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N n^{s-1}\right),\quad\Re(s)<0\tag{20}$$

$$\zeta(s-1)=s\, \mathcal{M}_x[f_3(2 \pi x)](-s)=s \int\limits_0^\infty f_3(2 \pi x)\,x^{-s-1}\,dx$$ $$=-2^{s-1} \pi^{s-2} \cos\left(\frac{\pi s}{2}\right) \Gamma(2-s) \sum\limits_{n=1}^N n^{s-2},\quad\Re(s)<1\tag{21}$$

  • 3
    $\begingroup$ ComplexYetTrivial gave a closed form in a comment in the answer to this similar question $\endgroup$ May 19, 2022 at 0:44
  • $\begingroup$ @TymaGaidash Thanks, but that closed form isn't useful to me as it's analogous to the derivation of $\sum\limits_{k=1}^\infty\frac{\sin(2 \pi k x)}{k}=\pi \left(\lfloor x\rfloor -x+\frac{1}{2}\right)$ from $\lfloor x\rfloor =x-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^\infty \frac{\sin (2 \pi k x)}{k}\right)$ whereas what I'm looking for is analogous to $\sum\limits_{k=1}^{\infty } \frac{\sin(2 \pi k x)}{k}=\frac{1}{2} i \left(\log \left(1-e^{2 i \pi x}\right)-\log \left(1-e^{-2 i \pi x}\right)\right)$. $\endgroup$ May 19, 2022 at 2:33
  • $\begingroup$ Also posted to MathOverflow, mathoverflow.net/questions/440928/… $\endgroup$ Feb 17, 2023 at 1:26


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