Finding the function that corresponds to a Fourier series

Question

Im trying to reverse engineer a Fourier series that came about in my work by happenstance. The series in question: $$ \sum^{\infty}_{n=0} \frac{c}{c^{2}+(2n+1)^{2}}\sin((2n+1)x)-\frac{2n+1}{c^{2}+(2n+1)^{2}}\cos((2n+1)x) $$ The sum in question only needs to be valid on $x\in [0 , t] , t < \pi$, so it may correspond to not necessarily periodic function. The series is very close to a kind of triangle wave if it weren't for c. You may also notice that the coefficients correspond to the Laplace transform's: $\mathscr{L}\{ \sin(ct)\}(2n+1) , \mathscr{L}\{ \cos(ct)\}(2n+1)$ (for the respective function). This seems like it could be leveraged but I have been unsuccessful in doing so. Another avenue I have persued is expressing it in terms of the Lerch-Zeta fuction as follows. $$ -\frac{1}{2}\operatorname{Re}\bigg(e^{ix}\zeta\bigg(\frac{x}{\pi},1,\frac{1}{2}(1-ic)\bigg) \bigg) $$ (Edited because it was incorrectly the imaginary part) But this seems like a very heavy piece of mathematical machinery for a series such as this. I have tried to find lists of known Fourier series but haven't seen it apear in those lists. Any help you can provide on any potential leads would be great, Thank You!

Addition:

After working the problem for a bit one addition I can make is the following expression is equivalent to the series above.

$$ \frac{e^{-cx}}{2}\operatorname{Re}\bigg(\psi\big( \frac{1-ic}{2} \big) \bigg)-e^{-cx}\operatorname{Im}\bigg(\int^{x}_{0} te^{ct}\mathscr{L}^{-1}\{ \ln\big(\Gamma\big( \frac{1+is}{2} \big)\big) \}(t) dt \bigg) $$

This was found by taking the Laplace transform of the Lerch-Zeta function, which then produced a series that could be represented with Digamma functions. I then took the inverse Laplace transform of this expression using known properties of the Laplace transform. However, now the problem can be characterized by trying to find the inverse Laplace transform of either the Digamma function or the Log-Gamma function on the $\operatorname{Re}(z)=1/2$ line of the complex plane. I have been unable yet to find such an expression. I have an inkling that it may involve the Jacobi-Theta function and Modular Forms by looking at different expansions of the Digamma/Log-Gamma functions, but haven't been able to over-come some domain conditions and nuances regarding Laplace transforms and complex logarithm's. Again any help you can provide would be greatly appreciated.

Answer 1

With respect to

$$f_c(x)=\sum_{n=0}^\infty \left(\frac{c}{c^2+(2 n+1)^2} \sin((2 n+1) x)-\frac{2 n+1}{c^2+(2 n+1)^2} \cos((2 n+1) x)\right)\tag{1}$$

Mathematica simplfies

$$\sum\limits_{n=0}^\infty \frac{c}{c^2+(2 n+1)^2} \sin((2 n+1) x)\tag{2a}$$

to

$$g_c(x)=$$ $$\frac{1}{8} \left(e^{-i x} \left(\Phi\left(e^{-2 i x},1,\frac{1}{2}-\frac{i c}{2}\right)-\Phi\left(e^{-2 i x},1,\frac{1}{2} (1+i c)\right)\right)+e^{i x} \left(+\Phi\left(e^{2 i x},1,\frac{1}{2} (1+i c)\right)-\Phi\left(e^{2 i x},1,\frac{1}{2}-\frac{i c}{2}\right)\right)\right)\tag{2b}$$

where $\Phi(z,s,a)$ is the Hurwitz–Lerch transcendent and

Mathematica simplifies

$$\sum\limits_{n=0}^\infty -\frac{(2 n+1) \cos ((2 n+1) x)}{c^2+(2 n+1)^2}\tag{3a}$$

to

$$h_c(x)=$$ $$\frac{1}{8 c} i e^{-i x} \left(\left(e^{2 i x}\right)^{\frac{1}{2}-\frac{i c}{2}} \left(\left(e^{2 i x}\right)^{i c} \left(B_{e^{2 i x}}\left(\frac{1}{2}-\frac{i c}{2},0\right)+2 B_{e^{2 i x}}\left(\frac{3}{2}-\frac{i c}{2},-1\right)\right)-B_{e^{2 i x}}\left(\frac{1}{2}+\frac{i c}{2},0\right)-2 B_{e^{2 i x}}\left(\frac{3}{2}+\frac{i c}{2},-1\right)\right)+\left(e^{-2 i x}\right)^{-\frac{1}{2}-\frac{i c}{2}} \left(\left(e^{-2 i x}\right)^{i c} \left(B_{e^{-2 i x}}\left(\frac{1}{2}-\frac{i c}{2},0\right)+2 B_{e^{-2 i x}}\left(\frac{3}{2}-\frac{i c}{2},-1\right)\right)-B_{e^{-2 i x}}\left(\frac{1}{2}+\frac{i c}{2},0\right)-2 B_{e^{-2 i x}}\left(\frac{3}{2}+\frac{i c}{2},-1\right)\right)\right)\tag{3b}$$

where $B_z(a,b)$ is the Euler beta function.

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Therefore $f_c(x)$ defined in formula (1) above can be evaluated as

$$f_c(x)=g_c(x)+h_c(x)\tag{4}.$$

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I believe $f_c(x)$ can also be represented as the inverse Laplace transform

$$f_c(x)=\mathcal{L}_s^{-1}\left[\frac{H_{\frac{1}{2} i (c+i)}+H_{-\frac{1}{2} (1+i c)}-H_{\frac{1}{2} i (i+s)}-H_{-\frac{1}{2} (i s+1)}}{4 (c+s)}\right](x)$$ $$=\frac{1}{2 \pi i} \int\limits_{-\infty }^{\infty} \frac{H_{\frac{1}{2} i (c+i)}+H_{-\frac{1}{2} (1+i c)}-H_{\frac{1}{2} i (i+s)}-H_{-\frac{1}{2} (i s+1)}}{4 (c+s)}\, e^{x s} \, ds\tag{5}$$

where $H_s$ is the analytical Harmonic number function.

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Formula (5) above follows from simplification of the term-wise Laplace transform of formula (1) for $f_c(x)$ above, and formula (5) above leads to the alternate representation

$$f_c(x)=$$ $$\frac{i \left((-c+i)\, e^{i x} \, _2F_1\left(1,-\frac{1}{2} i (c+i);-\frac{1}{2} i (c+3 i);e^{2 i x}\right)+(c+i)\, e^{-i x} \, _2F_1\left(1,\frac{1}{2} (1+i c);\frac{1}{2} (3+i c);e^{-2 i x}\right)\right)}{2 \left(c^2+1\right)}\tag{6}.$$

where $_2F_1(a,b;c;z)$ is the Hypergeometric2F1 function