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I recently found these Plouffe-like formulas using Mathematica's LatticeReduce. Has anybody seen/can prove these are indeed true?

$$\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}$$

And so on for other $\zeta(2n+1)$. The background for these are in my blog.

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up vote 7 down vote accepted

In this paper by Vepstas he builds a derivation for Plouffe's identities, and generalizes them to some degree.

Your identities for $\zeta(4n-1)$ follow from the results immediately following his Corollary 2 $$ q^{k-1}\sum\frac{1}{n^k(e^{2\pi pn/q}-1)}+p^{k-1}\sum\frac{1}{n^k(e^{2\pi qn/p}-1)} = q^{k-1}I_k(2\pi p/q) $$ with $p=\sqrt{2},q=1$, and using the expression for $I_k(x)$ from the middle of page 7 (I confirmed for $k=3$ and expect it will also match for $k=7$).

He uses additional machinery to establish the identities in Plouffe's form for $k=4n+1$, and it doesn't immediately admit generalization to your form, but maybe it can be tweaked.

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Yes, these Plouffe-like identities usually fall into two classes: for $\zeta(4n-1)$ and $\zeta(4n+1)$. I had hoped the general family using $\sqrt{2}$ was new, but even if it is not, I'm glad I found them on my own. –  Tito Piezas III Jun 25 '12 at 18:58
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It seems to have escaped attention that these sums may be evaluated using harmonic summation techniques.

Introduce the sum $$S(x; \alpha, p) = \sum_{n\ge 1} \frac{1}{n^p(e^{\alpha n x}-1)}$$ with $p$ a positive odd integer and $\alpha>1$, so that we seek e.g. $2 S(1; \pi\sqrt{2}, 3)+S(1; 2\pi\sqrt{2}, 3).$

The sum term 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 $$\lambda_k = \frac{1}{k^p}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{e^{\alpha x}-1}.$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{e^{\alpha x}-1} x^{s-1} dx = \int_0^\infty \frac{e^{-\alpha x}}{1-e^{-\alpha x}} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 1} e^{-\alpha q x} x^{s-1} dx = \sum_{q\ge 1} \int_0^\infty e^{-\alpha q x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 1} \frac{1}{(\alpha q)^s} = \frac{1}{\alpha^s} \Gamma(s) \zeta(s).$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x;\alpha,p)$ is given by

$$Q(s) = \frac{1}{\alpha^s} \Gamma(s) \zeta(s) \zeta(s+p) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^p} \frac{1}{k^s} = \zeta(s+p)$$ for $\Re(s) > 1-p.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.

First formula.

We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(2 + \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+3).$$

We shift the Mellin inversion integral to the line $s=-1$, integrating right through the pole at $s=-1$ picking up the following residues:

$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{\pi^3\sqrt{2}}{72 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{3}{2}\zeta(3)$$ and $$\frac{1}{2}\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^3\sqrt{2}x}{9}.$$

This almost concludes the proof of the first formula if we can show that the integral on the line $\Re(s) = -1$ vanishes when $x=1.$ To accomplish this we must show that the integrand is odd on this line.

Put $s = -1 - it$ in the integrand to get $$\pi^{1+it} \sqrt{2}^{1+it} \left(2 + 2^{1+it}\right) \Gamma(-1-it) \zeta(-1-it) \zeta(2-it).$$

Now use the functional equation of the Riemann Zeta function in the following form: $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ to obtain (with $s=-1-it$) $$\pi^{1+it} \sqrt{2}^{1+it} \left(2 + 2^{1+it}\right) \zeta(2+it) 2^{-1-it} \pi^{-1-it} \frac{1}{2\cos\left(\frac{\pi (-1-it)}{2}\right)} \zeta(2-it)$$ which is $$ \sqrt{2}^{1+it} \left(2^{-it} + 1\right) \zeta(2+it) \frac{1}{2\cos\left(\frac{\pi (1+it)}{2}\right)} \zeta(2-it)$$ and finally yields $$-\frac{1}{\sin(\pi i t/2)} \left(\sqrt{2}^{-1-it}+\sqrt{2}^{-1+it}\right) \zeta(2+it)\zeta(2-it).$$

It is now possible to conclude by inspection: the zeta function terms and the powers of the square root are even in $t$ and the sine term is odd, so the whole term is odd and the integral vanishes.

Second formula.

We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(4 - \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+5).$$

We shift the Mellin inversion integral to the line $s=-2$ (no pole on the line this time) picking up the following residues:

$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{\pi^5\sqrt{2}}{540 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{3}{2}\zeta(5)$$ and $$\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^5\sqrt{2}x}{540}.$$

It remains to verify that the integrand on the line $\Re(s)=-2$ is odd when $x=1$. Put $s=-2-it$ in the integrand to get $$\pi^{2+it} \sqrt{2}^{2+it} \left(4 - 2^{2+it}\right) \Gamma(-2-it) \zeta(-2-it) \zeta(3-it).$$ Applying the functional equation once again with $s=-2-it$ we obtain $$\pi^{2+it} \sqrt{2}^{2+it} \left(4 - 2^{2+it}\right) \zeta(3+it) 2^{-2-it} \pi^{-2-it} \frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)} \zeta(3-it)$$ which is $$\sqrt{2}^{2+it} \left(2^{-it} - 1\right) \zeta(3+it) \frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)} \zeta(3-it)$$ which is in turn $$\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right) \frac{1}{2\cos\left(\frac{\pi(2+it)}{2}\right)} \zeta(3+it) \zeta(3-it)$$ which finally yields $$-\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right) \frac{1}{2\cos(\pi i t /2)} \zeta(3+it) \zeta(3-it)$$

The product of the zeta function terms is even, as is the cosine term. The term in front is odd, so the integrand is odd as claimed.

Third formula.

We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(8 + \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+7).$$

We shift the Mellin inversion integral to the line $s=-3$, integrating right through the pole at $s=-3$ picking up the following residues:

$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{17 \pi^7\sqrt{2}}{37800 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{9}{2}\zeta(7)$$ and $$\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^7\sqrt{2}x}{1134} \quad\text{and}\quad \frac{1}{2} \mathrm{Res}(Q(s)/x^s; s=-3) = -\frac{\pi^7\sqrt{2}x^3}{2025}.$$

This almost concludes the proof of this third formula if we can show that the integral on the line $\Re(s) = -3$ vanishes when $x=1.$ To accomplish this we must show once more that the integrand is odd on this line.

Put $s = -3 - it$ in the integrand to get $$\pi^{3+it} \sqrt{2}^{3+it} \left(8 + 2^{3+it}\right) \Gamma(-3-it) \zeta(-3-it) \zeta(4-it).$$

By the functional equation we obtain with $s = -3-it$ $$\pi^{3+it} \sqrt{2}^{3+it} \left(8 + 2^{3+it}\right) \zeta(4+it) 2^{-3-it} \pi^{-3-it} \frac{1}{2\cos\left(\pi(-3-it)/2\right)} \zeta(4-it)$$ which is $$\sqrt{2}^{3+it} \left(2^{-it} + 1\right) \zeta(4+it) \frac{1}{2\cos\left(\pi(3+it)/2\right)} \zeta(4-it)$$ which finally yields $$\frac{1}{2\sin(\pi it/2)} \left(\sqrt{2}^{3-it} + \sqrt{2}^{3+it}\right) \zeta(4+it) \zeta(4-it).$$

This concludes it since the two zeta function terms together are even as is the square root term while the sine term is odd, so their product is odd.

A similar yet not quite the same computation can be found at this MSE link.

Another computation in the same spirit is at this MSE link II.

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