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$$=\frac{1}{(n+1)!}\int_{0}^{\infty}\frac{w^{n+1}}{e^{w}-1}dw\int_{0}^{1}t^{n}e^{-tw}dt$$ We can rewrite the inner integral as: \begin{align*}\int_0^{1} t^ne^{-wt}\,dt &= (-1)^n\frac{d^n}{dw^n}\left(\int_0^1 e^{-wt}\,dt\right) \\&= (-1)^n \left(\frac{1-e^{-w}}{w}\right)^{(n)} \\&= (-1)^...

We begin by observing that: \begin{align} \int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx&= \frac{1}{i}\int\limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\&= \frac{1}{i}\int\limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\&= \frac{1}{i}\int\limits_{\gamma} ...

Let $\displaystyle \{a_n\}_{ n\ge 1}$ be a real valued sequence, such that for every real convergent sequence $\{r_n\}$, the series $\displaystyle \sum\limits_{n=1}^{\infty} r_na_n$ converges. Does it follow that $\displaystyle \sum\limits_{n=1}^{\infty} |a_n| < \infty$ (converges absolutely)?

Hmm.. I am pretty confident that a closed form of the general case: $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\binom{2n}{n}x^n $$ exists. Is anyone aware of? We may start with $\displaystyle \sum\limits_{n=0}^{\infty} \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}}$ dividing by $x$ on both sides and inte...

I suppose I could add the contour way of dealing with it (although it's not so elegant as T's solution). Integrating by parts we have: \begin{align*}I&=\int_0^{\infty} \frac{\sin^2 x}{x^2}e^{-2ax}\,dx \\&= \int_0^{\infty} \frac{1}{x}\left(e^{-2ax}\sin 2x - ae^{-2ax}(1-\cos 2x)\right)\,dx\\&a...

We might connect the given integrand to the spherical bessel function v.i.a. Rayleigh Formula: $$j_n(x) = (-1)^nx^n \left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^n \frac{\sin x}{x}$$ and note that our case is $n = 2$, $\displaystyle j_2(x) = \left(\frac{3}{x^2} - 1\right)\frac{\sin x}{x} - \f...

\begin{align*}&\frac{1}{n^2}\log \left(\prod\limits_{j=1}^{n}\binom{n+j}{n}\right) \\&= \frac{1}{n^2}\log \left(\prod\limits_{j=1}^{n}\frac{(n+j)_{n}}{j!}\right)\\&= \frac{1}{n^2}\sum\limits_{j=1}^{n}(n+1-j)\log(n+j) - \frac{1}{n^2}\sum\limits_{j=1}^{n}j\log (n+1-j)\\&= \frac{1}{n}\s...

Wow T! That's neat.

Awesome result!

Humor me if the response seems odd :mrgreen: Assuming $C$ is the best constant for the inequality, $|f(x) - f(y)| \le C|x-y|^2$ for $x,y \in \mathbb{R}$ (we know $0 \le C \le 1$ and $C = 0$ implies $f$ is constant), Then, \begin{align*}|f(y) - f(x)| &= \left|\sum\limits_{k=1}^{\infty}\left( f\le...