# Weighted sum of cosines

Consider $$f(x) = \sum_{k=1}^\infty \cos(kx) k^\alpha.$$ The first question is: does this have a name (Mathematica gives it as a sum of polylogs of complex arguments, but this seems unnatural). Also, does the logarithmic derivative have a name and/or nice properties? (The exponent $\alpha$ in my applications is a negative real number, but I will take what I can get...)

• The sum can't start at $0$. – Lucian Oct 22 '14 at 14:25
• @Lucian truly you are wise in the ways of science. – Igor Rivin Oct 22 '14 at 14:43
• You may want to consult this MSE link to see how your sum can be evaluated for $\alpha<-1$ a negative integer. – Marko Riedel Oct 22 '14 at 19:31
• @MarkoRiedel Wow, that's very cool! – Igor Rivin Oct 22 '14 at 19:57
• @MarkoRiedel Is it important that $\alpha$ be integral? – Igor Rivin Oct 22 '14 at 19:58

We treat the case $\alpha = -4$ because even $\alpha$ yields maximal cancelation in the residue computation that follows. (Gamma function poles.)

The sum term $$S(x) = \sum_{n\ge 1} \frac{\cos(nx)}{n^4}$$ is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{1}{k^4}, \quad \mu_k = k \quad \text{and} \quad g(x) = \cos(x).$$

We need the Mellin transform $g^*(s)$ of $g(x) = \cos(x)$ which was computed at this MSE link and found to be $$g^*(s) = \Gamma(s) \cos(\pi s/2)$$ with fundamental strip $\langle 0,1 \rangle.$

Hence the Mellin transform $Q(s)$ of $S(x)$ is given by $$Q(s) = \Gamma(s) \cos(\pi s/2) \zeta(s+4) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s+4)$$ where $\Re(s+4) > 1$ or $\Re(s) > -5$.

Intersecting the fundamental strip and the half-plane from the zeta function term we find that the Mellin inversion integral for an expansion about zero is $$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds$$ which we evaluate in the left half-plane $\Re(s)<1/2.$

The cosine term cancels the poles of the gamma function term at odd negative integers and the zeta function term the poles at even negative integers. We are left with just four poles.

\begin{align} \mathrm{Res}(Q(s)/x^s; s=0) & = \frac{\pi^4}{90} \\ \mathrm{Res}(Q(s)/x^s; s=-2) & = - \frac{\pi^2}{12} x^2 \\ \mathrm{Res}(Q(s)/x^s; s=-3) & = \frac{\pi}{12} x^3 \quad\text{and}\\ \mathrm{Res}(Q(s)/x^s; s=-4) & = -\frac{1}{48} x^4 \end{align}

Hence in a neighborhood of zero, $$S(x) = \frac{\pi^4}{90} - \frac{\pi^2}{12} x^2 + \frac{\pi}{12} x^3 -\frac{1}{48} x^4.$$

We will now show that this is exact for $x\in(0,2\pi).$ Put $s= \sigma + it$ with $\sigma \le -9/2$ where we seek to evaluate $$\frac{1}{2\pi i} \int_{-9/2-i\infty}^{-9/2+i\infty} Q(s)/x^s ds.$$

Recall that with $\sigma > 1$ and for $|t|\to\infty$ we have $$|\zeta(\sigma+it)| \in \mathcal{O}(1).$$

Furthermore recall the functional equation of the Riemann Zeta function $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ which we re-parameterize like so $$\zeta(s+4) = 2\times (2\pi)^{s+3} \cos\left(-\frac{\pi (s+3)}{2}\right) \Gamma(-s-3) \zeta(-s-3)$$ which is $$\zeta(s+4) = 2\times (2\pi)^{s+3} \sin(\pi s/2) \frac{\Gamma(1-s)}{s(s+1)(s+2)(s+3)} \zeta(-s-3).$$

Substitute this into $Q(s)$ to obtain $$\Gamma(s) \cos(\pi s/2) \times 2 \times (2\pi)^{s+3} \sin(\pi s/2) \frac{\Gamma(1-s)}{s(s+1)(s+2)(s+3)} \zeta(-s-3).$$

Use the reflection formula for the Gamma function to obtain $$\cos(\pi s/2) \times 2 \times (2\pi)^{s+3} \sin(\pi s/2) \times \frac{\pi}{\sin(\pi s)} \frac{1}{s(s+1)(s+2)(s+3)} \zeta(-s-3),$$ in other words we have $$Q(s) = \pi(2\pi)^{s+3} \frac{\zeta(-s-3)}{s(s+1)(s+2)(s+3)}.$$

This finally implies (with $\sigma< -9/2$ we have $\Re(-s-1) > 7/2$) $$|Q(s)/x^s|\sim 8\pi^4 (2\pi)^{\sigma} x^{-\sigma} |t|^{-4}.$$ or $$|Q(s)/x^s|\sim 8\pi^4 (x/2/\pi)^{-\sigma} |t|^{-4}.$$

We see from the term in $|t|$ that the integral obviously converges. (This much we knew already.) Moreover, when $x\in(0,2\pi)$ we have $(x/2/\pi)^{-\sigma}\to 0$ as $\sigma\to -\infty.$ The term in $x$ does not depend on the variable $t$ of the integral and may be brought to the front. This means that the contribution from the left side of the rectangular contour that we employ as we shift to the left vanishes in the limit, proving the exactness of the formula for $S(x)$ in the interval $(0,2\pi)$ obtained earlier.

As I have mentioned elsewhere there is a theorem hiding here, namely that certain Fourier series can be evaluated by inverting their Mellin transforms which is not terribly surprising and which the reader is invited to state and prove.