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According to Hankel's formula $$ \frac{1}{\Gamma(z)}=\frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}dt, $$ where $C$ is Hankel contour. So $$ \frac{x^x}{\Gamma(x+1)}=\frac{i}{2\pi}\int_C(-t)^{-x-1}e^{-xt}dt,\quad …

This is particular case of a classic integral studied by Ramanujan. See Chapter 11 in Hardy's book, "Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work", where it is shown that $$ \ …

Start from the well known formula \begin{equation} 2^{N-1} \prod _{k=1}^N \left[\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)\right]=\cos N(x-y)-\cos N(x+y), \end{equation} take logarithmic de …

A more general result is due to C. Störmer (Acta Mathematica December 1895, Volume 19, Issue 1, pp 341–350)

First of all this has nothing to do with the inflection point of $e^{-\alpha x^2}$. According to Poisson summation formula (see Whittaker, Watson, Modern analysis, chapter 21.51) $$ \sum_{n=-\infty}^\ …

The generalization looks like this $$ \int_{-\infty}^{\infty} \binom{n}{\alpha x}^l dx =\sum_{k=-\infty}^\infty\binom{n}{\alpha k}^l,\quad 0<\alpha\le 2/l,~l\in\mathbb{N}\tag{1} $$ where $n$ need not …

This problem can be reduced at least formally to a compact double integral, which might be easier to solve. Starting with the integral representation for the Gamma function, we write the double sum a …

Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\c …

We will use the following well known fact (e.g., see sections 1.1 and 1.2 in this article): Given $f(x)$ with period $1$, its Fourier series $$ f(x)=\sum_{j=0}^\infty a_j\cos(2\pi jx), $$ and a positi …

UPDATE The integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 A dozen integrals: Russell …

According to the representation for Jacobi polynomials https://en.wikipedia.org/wiki/Jacobi_polynomials#Alternate_expression_for_real_argument $$ P^{(0,n-m)}_m(x)=\sum_{j=0}^m \binom{m}{j}\binom{n}{j} …