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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 ...
A. Kristian Winstén's user avatar
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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. ...
ADS_Fibonacci's user avatar
3 votes

Integral of Choose functions

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

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 $\infty$-...
Math Attack's user avatar
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Completeness of Gamma family

You have to assume that the random variables are independent. Hence the Likelihood has the following expression $$L_\theta(x_1,\ldots,x_n) = \frac{\alpha^{\alpha n}}{\Gamma(\alpha)^n} \prod_{k=1}^n ...
Christophe Leuridan's user avatar
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Evaluating $\sum\limits_{x=2}^\infty \frac{1}{!x}$ in exact form.

I cannot see this one included above because it uses a much simpler function than those above, but it might be equivalent to one of them in the end. We start from the exponential generating function ...
alex.peter's user avatar
1 vote

Calculation of a derivative of a function related to the Euler Gamma function

Too long for a comment. I am grateful to Inbo Gottlieb-Fenves for the nice answer. I am now realizing that I should have said that $ F(x)=\frac2π\int_0^{π/2}(\cos \theta)^x d\theta, $ so that $$ \frac ...
Bazin's user avatar
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1 vote

Calculation of a derivative of a function related to the Euler Gamma function

Set $f(x) = \Gamma(\frac{1+x}{2})$, $g(x) = \sqrt{\pi}\Gamma(1+\frac{x}{2})$. We compute the derivatives of $f$ and $g$ to get $$f'(x) = \frac{1}{2}\int_0^\infty t^{\frac{x-1}{2}}e^{-t}\ln(t)\, dt \...
Inbo Gottlieb-Fenves's user avatar
1 vote

Mistake with Integration with Beta, Gamma, Digamma Fuctions

One-line solution using steps in More (Almost) Impossible Integrals, Sums, and Series (2023), page 193, the sequel of (Almost) Impossible Integrals, Sums, and Series (2019), for a similar integral. ...
user97357329's user avatar
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0 votes
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$\frac{\Gamma\left(x+\frac{1}{x}+1\right)}{2}\geq \left(\Gamma(x+1)\Gamma\left(\frac{1}{x}+1\right)\right)x^{-\frac{1}{2\pi}\ln\left(x\right)},x>0$

If $$f(x)=\text{lhs - rhs}\qquad \qquad f\left(\frac{1}{x}\right)=f(x)$$ the first derivative cancels at $x=1$ and $$f''(1)=3+\frac{1}{\pi }-\frac{\pi ^2}{3}\quad > \quad 0$$ Using a Taylor ...
Claude Leibovici's user avatar
0 votes

$\frac{\Gamma\left(x+\frac{1}{x}+1\right)}{2}\geq \left(\Gamma(x+1)\Gamma\left(\frac{1}{x}+1\right)\right)x^{-\frac{1}{2\pi}\ln\left(x\right)},x>0$

You can bring the given inequality in the form $$\frac{x^{c\ln x}}{2(x+\frac1x+1)}\geq B(x+1,\frac1x+1)$$ where $c>0$ or specially $c=\frac1{2\pi}$ and where $B(.,.)$ is the beta function. It is ...
Bob Dobbs's user avatar
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0 votes

$\frac{\Gamma\left(x+\frac{1}{x}+1\right)}{2}\geq \left(\Gamma(x+1)\Gamma\left(\frac{1}{x}+1\right)\right)x^{-\frac{1}{2\pi}\ln\left(x\right)},x>0$

$$\frac{(x+1/x)!}{2} \ge (x)!\cdot(1/x)!\cdot x^{\frac{-1}{2\pi}\log(x)}$$ say $(x)! = e^{f(x)}$ $$e^{f(x+1/x)}\cdot (1/2) \ge e^{(x)!}\cdot e^{(1/x)!}\cdot x^{\frac{-1}{2\pi}\log(x)}$$ $$f(x+1/x)+\...
Aderinsola Joshua's user avatar
0 votes

Limit of sum of ratios of gamma function like products

Using first Pochhammer symbols and converting to the gamma function $$ \prod_{j=1}^{n}\frac{kj-1}{kj}=\frac{\left(i-\frac{1}{k}\right)_{n-i+1}}{(i)_{n-i+1}}=\frac{\Gamma (i)\, \Gamma \left(n+1-\frac{1}...
Claude Leibovici's user avatar
1 vote
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Question about derivative of the falling factorial

You have $f(n, n) = \Gamma(n + 1)$ but this by no means implies that $\frac d{dx}f(x, n)\vert_{x = n} = \Gamma'(n + 1)$. Even if you can extend the arguments of $f$ to a differentiable function $f(x, ...
WhatsUp's user avatar
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2 votes
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Values for which $\frac{K(k')}{K(k)}=a+i$ where $a$ is an algebraic number.

The ideas in your question related to complex values of $K'/K$ have been studied in past. In particular the theory of complex multiplication says that if $\tau$ is an algebraic number of degree $2$ ...
Paramanand Singh's user avatar
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0 votes

Limit of sum of ratios of gamma function like products

Let's first consider the infinite product$\displaystyle\lim_{n\to\infty} \prod_{j=1}^{n}\frac{kj-1}{kj}$ We know that if $1+a_j=1-(1-\frac{kj-1}{kj})>0$ then $\displaystyle\prod_{j=1}^{\infty}1+a_j$...
Conor_Meise's user avatar
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pochammer symbol and hypergeometric function

$$(a)_b=\frac{\Gamma (a+b)}{\Gamma (a)}$$ $$\frac{(a)_b}{b!}=\frac{\Gamma (a+b)}{\Gamma (a)\, \Gamma (b+1)}$$ So, in you case, the numerator is a constant $\Gamma(?)$. Find it and work the ...
Claude Leibovici's user avatar
3 votes

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

We are glad to have 3 nice and inspiring alternative solutions. I now want to give one more by introducing a reduction formula : $$ I_{n+1}=\frac{4 n-1}{4 n} I_n \text { for any } n\in \mathbb{N}. $$ ...
Lai's user avatar
  • 18k
6 votes

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

\begin{align}n\geq 1,J_n=\int_0^\infty \frac{1}{(1+x^4)^n}dx\end{align} For $0\leq t< 1$, Define, \begin{align}F(t)&=\sum_{n=1}^\infty J_nt^n\\ &=\sum_{n=1}^\infty\left(\int_0^\infty \frac{...
FDP's user avatar
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4 votes

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

The antiderivative $$I=\int \frac {dx}{ \left(1+x^a\right)^n}=x \,\, _2F_1\left(\frac{1}{a},n;1+\frac{1}{a};-x^a\right)$$ So, the definite integral is $$J=\int_0^\infty \frac {dx}{ \left(1+x^a\right)^...
Claude Leibovici's user avatar
8 votes
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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 ...
Travis Willse's user avatar
2 votes
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Integral relating gamma function's linear subspaces

As @MariuszIwaniuk notes, these integrals are computable. In fact, not only by Mathematica [sic], but by human beings: replace the cosine by its expression in terms of exponentials. Grouping things ...
paul garrett's user avatar
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9 votes
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$ 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 ...
Fedor Petrov's user avatar
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