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Basically it equals to

$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$

However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !

The Engels form of the Cauchy – Schwartz inequality gives us: \begin{align*} \sum \frac{\log_{x_1}^4 x_2}{x_1+x_2} & \geq \frac{\left (\sum \log_{x_1}^2 x_2 \right )^2}{\sum (x_1+x_2)} \\ &= \frac{\left ( \sum \log_{x_1}^2 x_2 \right )^2}{2\sum x_1} \\ &\!\!\!\!\!\!\overset{\text{AM-GM}}{\geq } \fra...

Evaluate the limit:

$$\ell= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m$$

It is closely related to another famous integral; namely $$\int_{0}^{\pi}\frac{\mathrm{d} \theta}{1-2a\cos \theta+a^2}\quad , \quad |a|<1$$ Evaluation of the integral: For $|a|<1$ we have successively: \begin{align*} \int_{0}^{\pi} \frac{{\rm d}x}{1-2a \cos x+a^2} &= \frac{1}{2} \int_{-\pi}^{\pi} \f...

Hi ,

I'm sorry but I do not understand what exactly you wrote down! Could you please elaborate?

Let $ABC$ be a triangle and denote $a, b, c$ the lengths of the sides $BC , CA$ and $AB$ respectively. If $abc \geq 1$ then prove that

$$\sqrt{\frac{\sin A}{a^3+b^6+c^6}} + \sqrt{\frac{\sin B}{b^3+c^6+a^6}} + \sqrt{\frac{\sin C}{c^3 + a^6+b^6}} \leq \sqrt[4]{\frac{27}{4}}$$

Oh, also how about adding an Equation Editor button next to the Preview Button so that it links to an equation editor to speed up typesetting?

Let $\mathcal{V}$ be a linear space over $\mathbb{R}$ such that $\dim_{\mathbb{R}} \mathcal{V} < \infty$ and $f:\mathcal{V} \rightarrow \mathcal{V}$ be a linear projection such that any non zero vector of $\mathcal{V}$ is an eigenvector of $f$. Prove that there exists $\lambda \in \mathbb{R}$ such t...

I really like the add of the topic tags. Now every topic can be sorted into categories and be found easier. Hey , what about a live topic preview? That would be fantastic!

Let $f$ be analytic in the disk $|z|<2$. Prove that: $$\frac{1}{2\pi i} \oint \limits_{\left | z \right |=1} \frac{\overline{f(z)}}{z-\alpha} \, \mathrm{d}z = \left\{\begin{matrix} \overline{f(0)} & , & \left | \alpha \right |<1 \\\\ \overline{f(0)} - \overline{f\left ( \frac{1}{\bar{\alpha}} \right...