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I have stumbled across the functions $$\mathrm{Gi}_s^{p,q}(x)=\sum_{n\geq0}\frac{x^{pn+q}}{(pn+q)^s}$$ And I would like to know where I can learn more about them.

These functions are interesting because they include certain other special functions as special cases.

For example, the polylogarithms: $$\mathrm{Li}_s(x)=\mathrm{Gi}_s^{1,1}(x)$$ and the inverse tangent integrals: $$\mathrm{Ti}_s(x)=-i\cdot\mathrm{Gi}_s^{2,1}(ix)$$ and the interesting relation $$\mathrm{Gi}_s^{p,p}(x)=\frac1{p^s}\mathrm{Li}_s(x^p)$$ As well as the Hurwitz zeta function: $$\mathrm{Gi}_s^{1,q}(1)=\zeta(s,q)$$ And similarily, a relation to the Lerch Transcendent: $$\Phi(z,s,\alpha)=\frac1{z^\alpha}\mathrm{Gi}_s^{1,\alpha}(z)$$

What I've found out so far is detailed below.

A hyper-geometric representation

We may note that $$\mathrm{Gi}_s^{p,q}(x)=x^q\sum_{n\geq0}\frac{\Gamma(n+1)}{(pn+q)^s}\frac{x^{pn}}{n!}$$ Setting $$t_n=\frac{\Gamma(n+1)}{(pn+q)^s}$$ We have that $$\frac{t_{n+1}}{t_n}=\frac{(n+1)(n+q/p)^s}{(n+q/p+1)^s}$$ so we have that $$\mathrm{Gi}_s^{p,q}(x)=x^q\,_{s+1}F_{s}\left(1,\frac{q}{p},...,\frac{q}{p};1+\frac{q}{p},...,1+\frac{q}{p};x^p\right)$$

A recurrence

We may notice that $$\begin{align} \frac{\partial}{\partial x}\mathrm{Gi}_s^{p,q}(x)&=\sum_{n\geq0}\frac{x^{pn+q-1}}{(pn+q)^{s-1}}\\ &=\frac1x\sum_{n\geq0}\frac{x^{pn+q}}{(pn+q)^{s-1}}\\ &=\frac1x\mathrm{Gi}_{s-1}^{p,q}(x)\\ \end{align}$$ So we of course have the $\mathrm{Li}$-style recurrence $$\mathrm{Gi}_s^{p,q}(x)=\int_0^x \frac{\mathrm{Gi}_{s-1}^{p,q}(t)}{t}\mathrm dt$$ With the easily shown base case of $$\mathrm{Gi}_0^{p,q}(x)=\frac{x^q}{1-x^p}$$ from which the recursive definitions of $\mathrm{Ti}$ and $\mathrm{Li}$ follow.

We may also consider the function $$\mathrm{Fi}_s^{p,q}(x)=\sum_{n\geq0}(-1)^n\frac{x^{pn+q}}{(pn+q)^s}$$ And by defining $\lambda_p=\exp\frac{i\pi}{p}$, we have $$\mathrm{Gi}_s^{p,q}(\lambda_p x)=\lambda_{p}^{q}\mathrm{Fi}_s^{p,q}(x)$$

clathratus
  • 17,161

2 Answers2

2

  • For $|z| < 1$ and $t \in (-1,\infty)$ $$g(s,t,z) = \sum_{n=1}^\infty z^n (n+t)^{-s}$$ For $|t|< 1$ from the binomial series $g(s,t,z) =\sum_{l=0}^\infty {-s \choose k} t^{-s-k} Li_{s+k}(z)$

  • For $a/q \in \Bbb{Q}$ $$q^{-s} z^{a/q} g(s,a/q,z) = \sum_{n=1}^\infty (z^{1/q})^{nq+a} (nq+a)^{-s}=\frac1q \sum_{k=1}^q e^{-2i \pi ak/q} Li_s(z^{1/q}e^{2i \pi k/q})$$

  • For $z \not \in [0,1)$ and $\Re(s) > 0$ $$\Gamma(s)g(s,t,z) = \int_0^\infty x^{s-1}f(x,t,z)dx, \qquad f(x,t,z)= \frac{e^{-xt}}{e^x/z-1}$$ Giving the contour integral $\int_C x^{s-1}f(x,t,z)dx = (1-e^{2i \pi s}) \Gamma(s) g(s,t,z)$ where $C$ encloses the branch cut $[0,\infty)$ to which you can try applying the residue theorem to obtain a functional equation.

  • For $\Re(s) > -K$

$$\Gamma(s)g(s,t,z) = \sum_{k=0}^K \frac{(-1)^k}{k!} \frac{\partial_x^k f(0,t,z)}{s+k}+ \int_0^\infty x^{s-1}(f(x,t,z)- 1_{x < 1} \sum_{k=0}^K \frac{\partial_x^k f(0,t,z)}{k!}) dx$$ obtaining the particular values $g(-k,t,z) = \frac{(-1)^k}{k!}\partial_x^k f(0,t,z)$

  • For $ \Re(x) > 0$ let the theta series $$\Theta(x,z,t) = \sum_n e^{2i \pi nz} e^{-\pi (n+t)^2 x} = x^{-1/2}\Theta(1/x,-t,-z)$$ (Poisson summation formula)

  • For $t \in (0,1), u \in \Bbb{R}$ $$\int_0^\infty x^{s/2-1} \Theta(x,u,t)dx = \pi^{-s/2}\Gamma(s/2)\sum_n |n+t|^{-s} e^{2i \pi nu} = g(s,t-1,e^{2i \pi u})+g(s,-t,e^{-2i \pi u})$$ $$ = \int_0^\infty x^{(1-s)/2-1}\Theta(x,-t,-u)dx=g(s,-u-1,e^{-2i \pi t})+g(s,u,e^{2i \pi t})$$

Then you can extend everything by analytic continuation

reuns
  • 77,999
1

I found a beautiful integral for $\mathrm{Gi}_s^{p,q}(z)$ when $q/p\in\Bbb Q^+$.

First we note that (see here) for $\text{Re }a>0$, $$\Phi(z,s,a)=\frac1{2a^s}+\frac{\log(1/z)}{z^a}\Gamma(1-s;a\log(1/z))\\ +\frac2{a^{s-1}}\int_0^\infty \frac{\sin\left[s\arctan(t)-ta\log(1/z)\right]}{(1+t^2)^{s/2}(e^{2\pi at}-1)}dt$$ Where $$\Gamma(s;x)=\int_x^\infty t^{s-1}e^{-t}dt$$ is the incomplete Gamma function.

Then we see that $$\mathrm{Gi}_s^{p,q}(z)=p^{-s}\mathrm{Gi}_s^{1,q/p}(z^p)$$ So we have that $$\mathrm{Gi}_s^{p,q}(z)=\frac{z^q}{p^s}\Phi(z^p,s,q/p)$$

And hence $$\mathrm{Gi}_s^{p,q}(z)=\frac{z^{q}}{2q^s}+p^{-s}\log(z^{-p})\Gamma[1-s;q\log(z^{-p})/p]\\ +\frac{2z^q}{q^{s-1}p}\int_0^\infty \frac{\sin\left[s\arctan(t)-tq\log(z^{-p})/p\right]}{(1+t^2)^{s/2}(e^{2\pi qt/p}-1)}dt$$

clathratus
  • 17,161