Harmonic zeta tail series

Let \mathcal{H}_n denote the n – th harmonic number. Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \mathcal{H}_n \left( \zeta(3) - \sum_{k=1}^{n} \frac{1}{k^3} \right)\]


Successively , we have:

    \begin{align*} \sum_{n=1}^{\infty} \mathcal{H}_n \left ( \zeta(3) - \sum_{k=1}^{n} \frac{1}{k^3} \right ) &= \sum_{n=1}^{\infty} \mathcal{H}_n \sum_{k=n+1}^{\infty} \frac{1}{k^3}\\ &= \sum_{k=2}^{\infty} \frac{1}{k^3} \sum_{n=1}^{k-1} \mathcal{H}_n \\ &= \sum_{k=2}^{\infty} \frac{1}{k^3} \sum_{n=1}^{k-1} \sum_{j=1}^{n} \frac{1}{j}\\ &=\sum_{k=2}^{\infty} \frac{1}{k^3} \sum_{j=1}^{n-1} \frac{1}{j} \sum_{n=j}^{k-1} 1 \\ &= \sum_{k=2}^{\infty} \frac{1}{k^3} \frac{1}{k^3} \sum_{j=1}^{k-1} \frac{k-j}{j} \\ &= \sum_{k=2}^{\infty} \frac{1}{k^2} \sum_{j=1}^{k-1} \frac{1}{j} - \sum_{k=2}^{\infty} \frac{k-1}{k^3} \\ &=\zeta(2, 1) -\zeta(2) + \zeta(3) \\ &= 2\zeta(3) - \zeta(2) \end{align*}

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Inverse zeta(3) limit

Evaluate the limit

    \[\ell= \lim_{T \rightarrow +\infty} \frac{1}{2T} \int \limits_{-T}^{T} \frac{\zeta(\frac{3}{2}+it)}{\zeta(\frac{3}{2}-it)} \, {\rm d}t\]

We are proving that the limit is \frac{1}{\zeta(3)}. Indeed , one has:

    \[\frac{\zeta(3/2+it)}{\zeta(3/2-it)} = \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) (d^2/n)^{it}\]

If x \neq 1 then

    \[\lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T x^{it}dt = \lim_{T \to +\infty} \frac{1}{2T} \frac{x^{iT}-x^{-iT}}{i\ln x} = 0\]

whereas if x=1 then the above limit is 1. Thus:

\begin{aligned} \lim_{T \to +\infty} \frac{1}{2T}\int \limits_{-T}^T \frac{\zeta(3/2+it)}{\zeta(3/2-it)} dt &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) \lim_{T \to +\infty} \frac{1}{2T}\int_{-T}^T (d^2/n)^{it}dt \\ &= \sum_{n=1}^\infty n^{-3/2} \sum_{d | n} \mu(d) 1_{n = d^2} \\ &= \frac{1}{\zeta(3)} \end{aligned}

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Conservative field

(i) Let f \in \mathcal{C}^2(\mathbb{R}) such that {\rm div(grad \;f)}=0 . Let \mathbb{D} \subseteq \mathbb{R}^2 be a \mathcal{C}^1 normal area. Prove that

    \[\oint \limits_{\partial \mathbb{D}} \left ( \frac{\partial f}{\partial y} , -\frac{\partial f}{\partial x} \right ) \cdot {\rm d}(x, y) =0\]

(ii) Examine if \bar{f}(x, y)=(2x \cos y , -x^2 \sin y) is a conservative field. If so, fiend a force of it.

(i) Using Green’s theorem we have that

    \begin{align*} \oint \limits_{\partial \mathbb{D}} \left ( \frac{\partial f}{\partial y} , -\frac{\partial f}{\partial x} \right ) \cdot {\rm d}(x, y) &= \iint \limits_{\mathbb{D}} \left ( -\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} \right )\, {\rm d}(x, y) \\ &= \iint \limits_{\mathbb{D}} \left ( -\nabla^2 f \right ) \, {\rm d}(x, y)\\ &= 0 \end{align*}

(ii) Yes, it is. We note that

    \[\nabla \left ( x^2 \cos y \right ) = 2x\cos y - x^2 \sin y\]

and of course a force is g(x, y)=x^2\cos y.

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An extraordinary sin integral

I was surfing the net today and I fell on this cute integral

    \[\mathcal{J}= \int_{-\infty}^{\infty} \sin \left( x^2 + \frac{1}{x^2} \right) \, {\rm d}x\]

I have seen integrals of such kind before like for instance this \displaystyle \int_{0}^{\infty} \sin \left( x^2 - \frac{1}{x^2} \right) \, {\rm d}x .In fact something  more general holds

    \[\int_0^{\infty}\sin \left( ax^2 - \frac{b}{x^2} \right) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{2a}} e^{-2 \sqrt{ab}}\]

where a, b>0.


The original integral does not fall into this category which is a real shame. Yet it does have a closed form and it does not contain an e in its final answer. Strange, huh? So similar but so different at the same time these two integrals. A sign changes everything.

We begin by exploring the integral

    \[\mathcal{J}'  = \int_0^\infty \exp \left( \frac{ia}{x^2} + ib x^2 \right) \, {\rm d}x\]

Manipulating the integral ( substitutions and known Gaussian results) reveals that

    \[\int_{0}^{\infty} \exp \left (\frac{ia}{x^2} + i b x^2 \right ) \, {\rm d}x =\frac{1}{2} \sqrt{\frac{\pi}{b}}\exp\left ( \frac{i \pi}{4} \right ) e^{2i \sqrt{ab}}\]

where a, b>0. Taking the imaginary part of the last expression we get that

    \[\int_{-\infty}^{\infty} \sin \left ( x^2 + \frac{1}{x^2} \right ) \, {\rm d}x = \sqrt{\frac{\pi}{2}} \left ( \sin 2 + \cos 2 \right )\]

and this is the final answer. See, no e!. Of course we can also extract the real part and calculate the corresponding integral involving \cos.

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On a nested sin sequence

Consider the sequence x_n defined recursively as

    \[x_1=1 \quad, \quad x_{n+1}=\sin x_n\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt{n} x_n = \sqrt{3}.


Lemma: If a_n is a sequence for which \displaystyle \lim_{n\to+\infty}(a_{n+1}-a_n)=a then

    \[\lim_{n\to + \infty}\frac{a_n}n=a.\]

Proof: In Stolz theorem we set x_{n}=a_{n+1} and y_n=n.

It is easy to see that x_n is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives

    \[\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}\]



Now, due to the lemma we have \lim\limits_{n\to+\infty} na_n^2 = 3 and the result follows.

Remark : The asymptotic now follows to be \displaystyle x_n \sim \sqrt{\frac{3}{n}}.

Problem: Find what inequality should \beta satisfy such that the series

    \[\mathcal{S}=\sum_{n=1}^{\infty} x_n^\beta\]


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