Does the closed form of $f(t) = \int \frac{e^{2 \pi i \alpha t}}{e^{2 \pi i \beta t} - 1} dt$ exist?

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I have been working on finding close forms of various Fourier series.

The general approach is:

  1. From the series find the (not necessarily homogeneous) ordinary differential equation for which the series is the solution.
  2. Solve the differential equation.
  3. Solve for the arbitrary constants introduced as part of the solution to the diff. eq.

For several the series under investigation the "solution" provided by Maple to the differential equation is that function, $f(t)$, which has a derivative of: $\frac{e^{2 \pi i \alpha t}}{e^{2 \pi i \beta t} - 1}$ which maple writes as: $\int \frac{e^{2 \pi i \alpha t}}{e^{2 \pi i \beta t} - 1}$

Does anyone know if this integral has a closed form?

The parameters: $\alpha$ and $\beta$, are two real, non-zero constants such that: $\frac{\alpha}{\beta} \not\in \mathbb{Z}$

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The closed form of the integrals of this kind involves hypergeometric functions (or Incomplete Beta function in particular) $$\int \frac{e^{cx}}{e^x - 1} dt=-\frac{1}{c} e^{cx} {_2F_1}\left(1,c;c+1;e^x \right)+constant$$ With $x=bt$ and $bc=a$ $$b\int \frac{e^{at}}{e^{bt} - 1} dt=-\frac{b^2}{a} e^{at} {_2F_1}\left(1,c;c+1;e^{bt} \right)+constant$$ Formaly : $$\int \frac{e^{2 \pi i \alpha t}}{e^{2 \pi i \beta t} - 1} dt=\frac{\beta}{\alpha} e^{2 \pi i \alpha t} {_2F_1}\left(1,\frac{\alpha}{\beta};\frac{\alpha}{\beta}+1;e^{2 \pi i \beta t} \right)+constant$$