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### How to prove that $\big(sinc(x) + sinc(x-1) \big)^{*r} = \frac{\Gamma{(r+1)}}{\Gamma{(x+1)}\Gamma{(r-x+1)}}$

First of all, recall that the $sinc$-function is defined to be $$sinc(x) := \frac{\sin{(\pi x)}}{\pi x}$$ and is well-defined also at the origin. Next thing is to note that the $sinc$-function is ...
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### Legendre Duplication by Logarithmic Derivatives

The issue was I did not write out the indices on the summations. Alfors has the sums going from 0 to $\infty$ because he scooted the residual $\frac{1}{z^2}$ term inside since it is adding 0 to it. ...

### Integral of Choose functions

Just use the definition of the generalized binomial coefficient: $$\int_{0}^{4}\binom{x}{5}dx =\frac{1}{5!}\int_{0}^{4}x\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)dx=0$$

### Meromorphic continuation of the multifactorial

I will answer you briefly by summarizing only the formulas that I have already given in 3 answers and questions: Definition of multifactorial Error of Fourier series of the multifactorial $\infty$-...

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### Is Beta function essential to evaluating the integral $\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^n}$?

Any standard method (including scaling $x$ in the first display equation in the question statement) yields $$\int_0^\infty \frac{dx}{a + x^4} = \frac{\pi}{2 \sqrt 2} a^{-3 / 4} .$$ If $n$ is a ...
### $e^{-\mu } \sum_{k=m }^{\infty} \frac{\mu^k}{k!} \le (\frac{\mu}{m})^m e^{m-\mu}$ when $1\le \mu \le m$
Divide by $e^{-\mu}$, and also divide by $\mu^m$, you get an equivalent inequality $$\sum_{k=m}^\infty \frac{\mu^{k-m}}{k!}\leqslant \left(\frac{e}m\right)^m.$$ LHS is obviously an increasing function ...