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,
$$\frac{\Gamma(a+...
12
votes
A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture
A conceptual way to tackle this question is to look at universal distributions on $\mathbf{Q}/\mathbf{Z}$, studied by Kubert and Lang among others. Distributions arise naturally in number theory, see ...
11
votes
Accepted
A tantalizing Gamma quotient to challenge the Rohrlich-Lang Conjecture
It turns out that triplication is not needed here:
the recursion $\Gamma(z+1) = z \Gamma(z)$, the reflection formula
$$
\Gamma(z) \Gamma(1-z) = \frac\pi{\sin \pi z},
$$
and the duplication formula
$$
...
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.
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:
$$ L:=\log \left ( \prod_{n=2}^{\infty} (...
10
votes
Accepted
Method to evaluate an infinite sum of ratio of Gamma functions (how does Mathematica do it?)
Let's rewrite the given problem
$$\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(\...
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 ...
9
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{\...
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
$$\def\h{\frac{1}{2}}
\def\g#1{\Gamma(#1)\,}
{}_3F_2\left({a,\ b,\ c\atop\h a+\h b+\h, 2c }\biggm| 1 \right)
=\frac{\g\h\g{c+\h}\g{\h a+\h b +\h}\g{c+\h -\...
8
votes
An integral identity evaluating the gamma function
It is nothing but beta-function. Consider only positive $x$ and denote $1/(x^2+1)=t$. You get $$\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)=\\=\...
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
$$ {}_{3}^{}F_{2}^{} \left(\frac{1}{2},\...
8
votes
Hankel determinant of incomplete gamma functions
Your quantity is
$$ 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^{...
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 ...
7
votes
Accepted
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 ...
7
votes
Accepted
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
$$5^z\,F(z)\rightarrow \int_0^\infty \left(...
7
votes
One-line proof of the Euler's reflection formula
It can be shown (from the Beta function) that
$$
\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} \...
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,
$$S(x,a) =\sum_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)...
6
votes
One-line proof of the Euler's reflection formula
This can be done without invoking the Beta function explicitly. Directly from the integral definition of $\Gamma(x)$,the product in question is a double integral:
$$
\Gamma(x)\Gamma(1-x) =\int^\...
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 ...
6
votes
Accepted
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 ...
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}}\...
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 ...
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 $n$ ...
6
votes
Accepted
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
$$\lim_{(x,y,z)\to(0,0,0)} \frac{xy}{z}$$ ...
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 ...
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
$$L^{-1}_s\...
5
votes
Accepted
An integral identity relate to the Gamma function or the Beta function
To evaluate $\int_{-\infty}^\infty {dx\over (1+ix)^a\,(1-ix)^b}$, view this as $\int_{-\infty}^\infty \widehat{f_a}(x)\,\overline{\widehat{f_{\overline{b}}}(x)}\,dx$ where $f_c$ are functions whose ...
5
votes
Accepted
Integral involving the gamma function
I understand that the OP is after the imaginary part of the integral in the small-$\alpha$ limit. To evaluate this limit, I note that Wilson's theorem implies that, for real positive $x$, the function
...
5
votes
Accepted
How to compute the following integral $I_{\alpha,\beta}$
Assume first $\beta>1$ so that the integral converges and let
$$f(x)=x^{\alpha}(1-x)^{-1}(-\log x)^{\beta}.$$
Then
$$0=\int_{0}^{1}df\\=\alpha\int_{0}^{1}x^{\alpha-1}(1-x)^{-1}(-\log x)^{\beta}dx
+ ...
5
votes
Why is the Gamma function shifted from the factorial by 1?
I'm going to elaborate on Pietro Majer's answer a bit.
Suppose $S_1,S_2$ are independent random variables for which for all Borel sets $A\subseteq [0,\infty),$
\begin{align}
\Pr(S_1\in A) & = \...
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