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To add a derivation of the identity mentioned in galactus's post: We might begin by considering the function $f(s) = \dfrac{(-x)^s\Gamma(3s+1)}{\Gamma(2s+1)\Gamma(s+1)}$, then appying the residue theorem over the function $\pi\csc (\pi s) f(s)$ in the rectangular contour with vertices $\gamma_T = (0...

Cool C! :clap2: I didn't notice the connection between the integral and Dirichlet's Lambda Function. \begin{align*}\int_0^{\infty} \left(\frac{x}{e^{x}-e^{-x}} - \frac{1}{2}\right)\frac{\,dx}{x^2}&= \int_0^{\infty}\int_0^{\infty} \left(\frac{x}{e^{x}-e^{-x}} - \frac{1}{2}\right)se^{-sx}\,ds\,dx\...

Calculate the integral: $$\int_{0}^{\infty}\left(\frac{x}{e^{x}-e^{-x}}-\frac{1}{2}\right)\frac{\,dx}{x^2}$$

\begin{align*}\sum\limits_{k=1}^{\infty} \frac{1}{k^2+k+1} &= \frac{1}{1+2\alpha}\sum\limits_{k=1}^{\infty} \frac{1}{(k-\alpha)}+\frac{1}{(-k-1-\alpha)}\\&= \frac{1}{1+2\alpha}\left(\frac{1}{1+\alpha}+\frac{1}{\alpha}\right)-\frac{1}{1+2\alpha}\sum\limits_{k=-\infty}^{\infty} \frac{1}{k-\alp...

Nice solutions!!

Calculate the limit (if it exists): $$\lim\limits_{x \to 1^{+}} \int_x^{x^2} \frac{\,dt}{\log t}$$

This in principle is not too far from my first attempt, but I am adding it for the sake of variety: \begin{align*}&\int_0^{\infty} \frac{(1-\sin ax)(1-\cos bx)}{x^2}\,dx\\&= \int_0^{\infty} (1-\sin ax)(1-\cos bx) \int_0^{\infty} ye^{-yx}\,dy\,dx\\&= \int_0^{\infty} \int_0^{\infty} ye^{-y...

Compute the limit (if it exists): $$\lim\limits_{n \to \infty} \sum\limits_{k=0}^{n-1} (-1)^k\frac{(n-k)^k}{k!}$$

\begin{align*}&\int_0^{\infty} \frac{\sin^2 \tan x}{x^2}\,dx\\&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin^2 \tan x}{x^2}\,dx\\&= \frac{1}{2}\sum\limits_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan (x+n\pi)}{(x+n\pi)^2}\,dx\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \si...

We assume $a > 0$ (is a positive real number). We'll use $\displaystyle \int_0^{2\pi} f\left(a+re^{\pm ix}\right)\,dx = 2\pi f(a)$ for $f$ analytic in $|z-a| < R$ where, $ 0< r < R$ to simplify the integral. \begin{align*}I(a)&=\int_0^{\pi} \log^2 (1-2a\cos x + a^2)\,dx \\&= \frac{1}{2}\int_...