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Yes, there exists. The basic idea is to construct a train of picks. Picks are constructed to satisfy the following heuristics: Each pick at $x=n$ is of an approximate height $n$ and of an approximate width $n^{-3}$, thus the area is of an approximate order $n^{-2}$, whereas squared its height gets s...

Hmm.. we can make a somewhat generalization for this integral. Let us consider $\mathbb{N} \ni n \geq 2$ as well as the family of integrals: $$\mathcal{J}(n)=\int_0^{\pi/2} \frac{\log^n \tan x }{\sin^2 \left( x - \frac{\pi}{4} \right)} \, {\rm d}x$$ Then following the exact steps of Tolaso we get th...

Here is a solution: \begin{align*} \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{2n+1} &= \sum_{n=-\infty}^{-1} \frac{(-1)^n}{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\\ &= \frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{(-1)^n}{1-2n}\\ &= \frac{\pi}{4} - \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1...

Of course we are dealing with a Mellin Transform here. In general it holds that: $$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} \, {\rm d}x = \Gamma(s) \zeta(s) , \; s \in \mathbb{C} \mid s>1$$ And for future reference $$\int_{0}^{\infty} \frac{x^{s-1}}{e^x+1} \, {\rm d}x = \left\{\begin{matrix} \Gamma(s...

Hello Vaggelis. Here is a solution: The second condition tells us that there exists an $M>0$ such that for every $ x> M$ to hold $f(x)<2$. Since $f$ is continuous on $[0, M]$ a maximum is attained and let that be at $f(x_0)$. Then: $$f(x_0) \geq f(x) \;\; \forall x \in [0, M] \quad \quad \text{and} ...

The answer would be no due to Parsheval's identity. Indeed: $$\sum_{n=1}^{\infty}a_n^2 =\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x \Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{\ln^2 (n+1)}=\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x$$ which is a contradiction, since the series $\sum \limits_{n...

Another way would be using digamma. Here is how: \begin{align*} \sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )\left ( 4n+3 \right )} &= \frac{1}{2}\sum_{n=0}^{\infty}\left [ \frac{1}{4n+1}- \frac{1}{4n+3} \right ]\\ &= \frac{1}{8}\sum_{n=0}^{\infty}\left [ \frac{1}{n+1/4}- \frac{1}{n+3/4}...

In more general if $k \in \mathbb{N}$ then:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log^{k}n}{n}=\frac{(-1)^{n}\left(\log2\right)^{k+1}}{k+1}+(-1)^{k-1}\sum_{j=0}^{k-1}\gamma_{j}\binom{k}{j}\log^{k-j}2$$

where $\gamma_i$ are the Stieljes constants.

We recall that the eta function is defined as $$\eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \left(1- 2^{1-s} \right) \zeta(s)$$ Hence differentiating once we have that: \begin{align*} \eta'(s) &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \log n}{n^s} \\ &= \sum_{n=2}^{\infty} \frac{(-1)^...

Evaluate \(\displaystyle\sum_{n\geq1}\frac{1}{(2n-1)(3n-1)(4n-1)}\). A somewhat different approach would be using the digamma function. As Demetres shows we have that: $$\frac{1}{(2n-1)(3n-1)(4n-1)} = \frac{2}{2n-1} - \frac{9}{3n-1} + \frac{8}{4n-1}$$ Thus: \begin{align*} \sum_{n \geq 1} \frac{1}{(...