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, ...

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

### "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{\...

10
votes

Accepted

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

Titchmarshâ€™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

### 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)...

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 ...

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\...

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 ...

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}{\...

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