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Introduction. The purpose here is to motivate the reader to study Mellin transforms and their applications in the evaluation of harmonic sums as well as verify and perhaps improve a certain application of the method. Mellin transform manipulation is a very useful skill to have especially in number theory or computer science.

Problem statement. Inspired by the computations at this MSE link I and this MSE link II we introduce the harmonic sum $$ S(x) = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{\cosh((2k+1)x)}.$$

Use the Mellin transform $$\mathfrak{M}\left(\frac{1}{\cosh(x)}; s\right)$$ and the functional equation of the Hurwitz Zeta function to establish a functional equation for $S(x).$ Evaluate $$S(\pi/2)$$ making use of a special property of the functional equation, which is simpler than the ones at this MSE link III but requires proof just the same.

Remark. We do ask that the details of the calculation be included as they are relevant to an understanding of the process and in the present case include a remarkable eureka moment.

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Suppose we seek to show that $$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{\exp((2k+1)\pi/2)+\exp(-(2k+1)\pi/2)} = \frac{\pi}{16}.$$

The sum term
$$S(x) = \sum_{k\ge 1} \frac{(-1)^{k+1}}{2k-1} \frac{1}{\exp(x(2k-1))+\exp(-x(2k-1))}$$

is harmonic and may be evaluated by inverting its Mellin transform. We are interested in $S(\pi/2).$

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 $g^*(s)= \mathfrak{M}(g(x);s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{(-1)^{k+1}}{2k-1}, \quad \mu_k = 2k-1 \quad \text{and} \quad g(x) = \frac{1}{\exp(x)+\exp(-x)}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is computed as follows: $$g^*(s) = \int_0^\infty \frac{1}{\exp(x)+\exp(-x)} x^{s-1} dx = \int_0^\infty \frac{\exp(-x)}{1+\exp(-2x)} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 0} (-1)^q \exp(-(2q+1)x) x^{s-1} dx = \Gamma(s) \sum_{q\ge 0} \frac{(-1)^q}{(2q+1)^s} = \Gamma(s) \beta(s)$$

where $$\beta(s) = 4^{-s} \left(\zeta\left(s,\frac{1}{4}\right) -\zeta\left(s,\frac{3}{4}\right)\right).$$

Note that $\beta(s)$ does not have a pole at $s=1$ and $\beta(1) = \frac{\pi}{4}.$

Hence the Mellin transform $Q(s)$ of $S(x)$ is given by $$ Q(s) = \Gamma(s) \beta(s) \beta(s+1) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \beta(s+1)$$ where $\Re(s) > 0$.

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 two beta function terms cancel the poles of the gamma function term and we are left with just

$$\mathrm{Res}(Q(s)/x^s; s=0) = \frac{\pi}{8}.$$

This shows that $$S(x) = \frac{\pi}{8} + \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds.$$

To treat the integral recall the duplication formula of the gamma function: $$\Gamma(s) = \frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right).$$

which yields for $Q(s)$

$$\frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \beta(s) \beta(s+1)$$

Furthermore observe the following variant of the functional equation of the Riemann zeta function adapted to the beta function: $$\Gamma\left(\frac{s+1}{2}\right)\beta(s) = 2^{1-2s} \pi^{s-1/2} \Gamma\left(1-\frac{s}{2}\right) \beta(1-s)$$

which gives for $Q(s)$ $$\frac{1}{\sqrt\pi} 2^{s-1} \Gamma\left(\frac{s}{2}\right) 2^{1-2s} \pi^{s-1/2} \Gamma\left(1-\frac{s}{2}\right) \beta(1-s)\beta(s+1) \\ = 2^{-s} \pi^{s-1} \frac{\pi}{\sin(\pi s/2)} \beta(1-s)\beta(s+1) \\ = 2^{-s} \frac{\pi^s}{\sin(\pi s/2)} \beta(1-s)\beta(s+1).$$

Now put $s=-u$ in the remainder integral to get

$$\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} 2^u \frac{\pi^{-u}}{\sin(-\pi u/2)} \zeta(1+u)\zeta(1-u) x^u du \\ = -\frac{1}{2\pi i} \int_{1/2-i\infty}^{1/2+i\infty} 2^{-u} \frac{\pi^{u}}{\sin(\pi u/2)} \zeta(1+u)\zeta(1-u) (4 x / \pi^2)^u du.$$

We have shown that $$S(x) = \frac{\pi}{8} - S(\pi^2/4/x).$$

In particular we get $$S(\pi/2) = \frac{\pi}{8} - S(\pi/2) \quad\text{or}\quad S(\pi/2) = \frac{\pi}{16}$$ as claimed.

Addendum. The functional equation for $\beta(s)$ can be derived from the functional equation of the Hurwitz Zeta function: $$\zeta\left(1-s, \frac{m}{n}\right) = \frac{2\Gamma(s)}{(2\pi n)^s} \sum_{k=1}^n \left[\cos\left(\frac{\pi s}{2} - \frac{2\pi k m}{n}\right) \zeta\left(s,\frac{k}{n}\right)\right].$$ where $1\le m\le n.$

This yields $$\zeta(1-s, 1/4) = \frac{2\Gamma(s)}{2^{3s}\pi^s} \left[ \cos(\pi s/2 - \pi/2) \zeta(s, 1/4) + \cos(\pi s/2 - \pi) \zeta(s, 1/2) \\ + \cos(\pi s/2 - 3\pi/2) \zeta(s, 3/4) + \cos(\pi s/2 - 2\pi) \zeta(s)\right] \\ = \frac{2\Gamma(s)}{2^{3s}\pi^s} \left[ \sin(\pi s/2) \zeta(s, 1/4) - \cos(\pi s/2) \zeta(s, 1/2) \\ -\sin(\pi s/2) \zeta(s, 3/4) + \cos(\pi s/2) \zeta(s)\right].$$

Similarly $$\zeta(1-s, 3/4) = \frac{2\Gamma(s)}{2^{3s}\pi^s} \left[ \cos(\pi s/2 - 3\pi/2) \zeta(s, 1/4) + \cos(\pi s/2 - 3\pi) \zeta(s, 1/2) \\ + \cos(\pi s/2 - 9\pi/2) \zeta(s, 3/4) + \cos(\pi s/2 - 6\pi) \zeta(s)\right] \\ = \frac{2\Gamma(s)}{2^{3s}\pi^s} \left[- \sin(\pi s/2) \zeta(s, 1/4) - \cos(\pi s/2) \zeta(s, 1/2) \\ +\sin(\pi s/2) \zeta(s, 3/4) + \cos(\pi s/2) \zeta(s)\right].$$

Subtract to obtain $$4^{1-s} \beta(1-s) = \frac{2\Gamma(s)}{2^{3s}\pi^s} (2\sin(\pi s/2) \zeta(s, 1/4) - 2\sin(\pi s/2) \zeta(s, 3/4))$$ or $$4^{1-s} \beta(1-s) = \frac{2\Gamma(s)}{2^{3s}\pi^s} 2\sin(\pi s/2) 4^s \beta(s)$$

which is $$\beta(1-s) = 2^s \frac{\Gamma(s)}{\pi^s} \sin(\pi s/2) \beta(s) \\ = \frac{1}{\pi^{s+1/2}} 2^{2s-1} \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{s+1}{2}\right) \sin(\pi s/2) \beta(s) \\ = \frac{1}{\pi^{s+1/2}} 2^{2s-1} \frac{\pi}{\sin(\pi s/2)} \Gamma\left(1-\frac{s}{2}\right)^{-1} \Gamma\left(\frac{s+1}{2}\right) \sin(\pi s/2) \beta(s)$$

which yields $$\beta(1-s)\Gamma\left(1-\frac{s}{2}\right) = \pi^{1/2-s} 2^{2s-1} \beta(s) \Gamma\left(\frac{s+1}{2}\right)$$ which is the desired result.

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  • $\begingroup$ Very informative indeed. The proof of functional equation for beta function at the end is kind of icing on the cake. +1 $\endgroup$ – Paramanand Singh Oct 17 '15 at 3:10
  • $\begingroup$ The functional equation is from the paper "Mellin transform and its applications" by Szpankowski. $\endgroup$ – Marko Riedel Oct 17 '15 at 3:15

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