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23 votes

Separating Gamma in two independent functions

$\newcommand\Ga\Gamma$If this factorization were true, we would get $$1=\frac{f(3)g(1)\,f(4)g(2)}{f(4)g(1)\,f(3)g(2)} =\frac{\Ga(3,1)\Ga(4,2)}{\Ga(4,1)\Ga(3,2)}=\frac{19}{16},$$ a contradiction. So, ...
Iosif Pinelis's user avatar
14 votes
Accepted

Series involving factorials

The sum $$\sum_{k=0}^\infty \frac{(a+k)!\,(b+k)!}{k!\,(a+b+c+k+1)!}z^k.$$ is not only a generalized hypergeometric series; it's the original ungeneralized Gauss hypergeometric series, $$\frac{\Gamma(a+...
Ira Gessel's user avatar
  • 16.5k
12 votes

A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture

A conceptual way to tackle this question is to look at universal distributions on $\mathbf{Q}/\mathbf{Z}$, studied by Kubert and Lang among others. Distributions arise naturally in number theory, see ...
François Brunault's user avatar
11 votes
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A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture

It turns out that triplication is not needed here: the recursion $\Gamma(z+1) = z \Gamma(z)$, the reflection formula $$ \Gamma(z) \Gamma(1-z) = \frac\pi{\sin \pi z}, $$ and the duplication formula $$ ...
Noam D. Elkies's user avatar
10 votes

"unexpected" residue formula for $\Gamma^3(s)/(\Gamma(3s)(e^{2\pi is}-1)) $

$\def\Res{\operatorname*{Res}} \def\G{\Gamma} \def\e{\varepsilon} \def\p{\pi} \def\ZZ{\mathbb{Z}} \def\QQ{\mathbb{Q}} \def\NN{\mathbb{N}} \def\j{\psi} \def\z{\zeta} \def\To{\rightarrow} \def\f{\...
user26872's user avatar
  • 416
10 votes
Accepted

Reference request: proofs of integrals presented in Erdélyi's *Table of Integral Transforms*

Titchmarsh’s Fourier integrals (1937, 7.6.4) has proof and attribution to Ramanujan.
Francois Ziegler's user avatar
10 votes

An interesting infinite product involving the factorial function with connection to the K and gamma function

I do not know if there is any closed form for this product, but you can rewrite it as follows. First, consider the logarithm of your product, so that you get: $$ L:=\log \left ( \prod_{n=2}^{\infty} (...
Manuel Norman's user avatar
10 votes
Accepted

Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)

Let's rewrite the given problem $$\sum _{n=0}^{\infty } \frac{\Gamma \left(n+\frac{1}{2}\right)^2 \Gamma \left(n+\frac{s}{2}\right)}{\Gamma (n+1)^2 \Gamma (n+s)}=\frac{\pi ^2 2^{1-s} \Gamma \left(\...
T. Amdeberhan's user avatar
9 votes
Accepted

An integral identity evaluating the gamma function

Yes, there is a trick which generalizes to analogous integrals on the classical cones, using the Gamma functions attached to these cones. In this, the simplest case, the starting point is the ...
paul garrett's user avatar
  • 22.6k
9 votes

Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)

This is a special case of Watson's Theorem $$\def\h{\frac{1}{2}} \def\g#1{\Gamma(#1)\,} {}_3F_2\left({a,\ b,\ c\atop\h a+\h b+\h, 2c }\biggm| 1 \right) =\frac{\g\h\g{c+\h}\g{\h a+\h b +\h}\g{c+\h -\...
Ira Gessel's user avatar
  • 16.5k
8 votes

An integral identity evaluating the gamma function

It is nothing but beta-function. Consider only positive $x$ and denote $1/(x^2+1)=t$. You get $$\int_0^\infty (1+x^2)^{-z/2-1}dx=\frac12 \int_0^1 t^{z/2-1/2}(1-t)^{-1/2}dt=\frac12 B((z+1)/2,1/2)=\\=\...
Fedor Petrov's user avatar
8 votes

Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)

FWIW, Maple (which gets the same result) says this comes from "definite summation using hypergeometric functions". Hmmm: it looks like this comes from $$ {}_{3}^{}F_{2}^{} \left(\frac{1}{2},\...
Robert Israel's user avatar
8 votes

Hankel determinant of incomplete gamma functions

Your quantity is $$ P(n,\alpha)=\det_r A_{i,j}(n,\alpha),$$ with $$ A_{i,j}(n,\alpha)=\int_0^\alpha t^{n+r-i-j}e^{-t}dt.$$ By the Andreief identity, this is $$ P(n,\alpha)=\frac{1}{r!}\int_0^\alpha e^{...
Marcel's user avatar
  • 2,542
8 votes
Accepted

Evaluating the series $\sum_{n=0}^{\infty} n! x^n$ and inverse variable-fractional-derivatives

$$ G(x) = \sum_{n=0}^\infty n!x^n \tag1$$ Another approach is to observe that the series $G(x)$ formally satisfies the differential equation $$ x^2 G'(x) + (x-1) G(x) + 1 = 0 . \tag2$$ The unique ...
Gerald Edgar's user avatar
  • 40.4k
7 votes
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Intuition behind the Riemann $\zeta$ functional equation

Your intuition breaks down because $\zeta(1-2k)$ has a closed form in terms of Bernoulli numbers, but no powers of $\pi$ at all. This was known to Euler (via Abel summation as the series is of course ...
Stopple's user avatar
  • 10.8k
7 votes
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Asymptotic behavior of integral with gamma functions

If I just insert the large-$z$ asymptotics of $\Gamma(z)\rightarrow \sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}}$, and take $z>1/2$ real for simplicity, I find $$5^z\,F(z)\rightarrow \int_0^\infty \left(...
Carlo Beenakker's user avatar
7 votes

One-line proof of the Euler's reflection formula

It can be shown (from the Beta function) that $$ \Gamma(1-x) \Gamma(x) = \mathrm{B}(x, 1 -x) = \int_0^{\infty} \frac{s^{x-1} d s}{s+1} \label{1}\tag{1} $$ Now we show that $$ \int_0^{\infty} \...
Herman Jaramillo's user avatar
7 votes
Accepted

New method to compute square roots

Let me "unclutter" the basic formula $S(x,a)=\sqrt{x}$, starting from the definition in the OP, $$S(x,a) =\sum_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)...
Carlo Beenakker's user avatar
7 votes
Accepted

Representing $\Gamma(a-x)$ in terms of $\Gamma(kx)$ and $\Gamma(a)$ and elementary functions

It is highly unlikely that further useful identities of this form exist. This can be seen by inspecting the poles and zeroes of the functions in the identities that do exist (noting from the Hadamard ...
Terry Tao's user avatar
  • 110k
6 votes

One-line proof of the Euler's reflection formula

This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral: $$ \Gamma(x)\Gamma(1-x) =\int^\...
Steve's user avatar
  • 61
6 votes
Accepted

On the integral $\int_0^1\log(x!)dx$ revisited

Details of the simple integration by series for $\int_0^1\log(x!)dx$ mentioned above (hopefully yours may be treated analogously, if you wish to try it). Start from the series of the logarithm of ...
Pietro Majer's user avatar
  • 56.9k
6 votes
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Converse of a result of Koblitz and Ogus on algebraic products of gamma values

A convenient way to formulate this kind of questions is to use the language of distributions introduced by Kubert and Lang. I gave a short account in this answer to a previous MO question. Your ...
François Brunault's user avatar
6 votes

Reference request: proofs of integrals presented in Erdélyi's *Table of Integral Transforms*

Another approach appears as a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways to Titchmarsh and this to prove it $ \int_{\mathbb{R}}\...
Dabed's user avatar
  • 256
6 votes

Analytic continuation of convergent integral

Technically, your integral is not well-defined because the path goes through $z=1$; the remedy I see presently (unless you have a definition for the contour going through $z=1$), is to move the ...
Jack L.'s user avatar
  • 1,443
6 votes
Accepted

$\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} z^{2k}$ is an elementary function

As Carlo noted, for $n$ an even integer, $S_{n,m}(z)$ is an elementary function of $z$. What about $n$ odd? When $n,m$ are both odd, I get something in terms of arcsinh, also elementary. But for $n$ ...
Gerald Edgar's user avatar
  • 40.4k
6 votes
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Extended binomial coefficients and the gamma function

There's nothing special about the gamma function; the failure of the limit to exist when $a$, $b$, and $n$ are negative is exactly the same as the failure of $$\lim_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ ...
Ira Gessel's user avatar
  • 16.5k
6 votes
Accepted

Integral calculus with Gamma function

You fix $\alpha$ and denote your integral to the left by $I(\beta )$. Then $I$ is convergent and analytic on the semi-plane $H=\{\beta\in{\mathbb C}\mid\Re (\beta )>0\}$. The right hand side too is ...
Constantin-Nicolae Beli's user avatar
6 votes

How to evaluate inverse Laplace transform of $e^{- \sqrt{s}} $ using series?

Let me first look at a simpler example, instead of the square root consider the inverse Laplace transform of $e^{-s}$. If you write the series expansion and invert term by term you obtain $$L^{-1}_s\...
Carlo Beenakker's user avatar
6 votes
Accepted

Eisenstein $E_2$ at imaginary quadratic arguments

Exactly the same Chowla--Selberg formula is valid, but you must apply it to the modified (non-holomorphic) $$E_2^*(\tau)=E_2(\tau)-3/(\pi\Im(\tau))$$ In other words, $E_2^*(\tau)/\eta^4(\tau)$ is an ...
Henri Cohen's user avatar
  • 11.9k
5 votes
Accepted

A generalized logarithmic function

Let us calculate $f_{\epsilon}(x) - \frac{1}{\epsilon}\log(1 + x)$ using your formula for $\log(1 + x)$ as $f_1(x)$(I didn't checked it but believe that it is correct): $f_\epsilon(x) - \frac{1}{\...
Aleksei Kulikov's user avatar

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