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2 votes
0 answers
52 views

How to define double factorial for non positive integers?

I studied double factorial which known for natural number $$ n!!=n(n-2)!! , 1!!=0!!=1$$ So we have for $n\in N$ $$ (2n)!!=2^n n! , (2n+1)!!=\frac{(2n+1)!}{2^n n!}$$ but I found on Math-World formula ...
Faoler's user avatar
  • 1,637
0 votes
0 answers
55 views

how to use Gauss Multiplication Formula for Gamma function?

I studied Gauss Multiplication Formula which known for $n\in Z^+ \wedge nx\notin Z^-\cup\{0\}$ $$\Gamma(nz)=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right)$$ but I didn't ...
Faoler's user avatar
  • 1,637
3 votes
2 answers
114 views

Is it possible to find the $n$th derivative of Gamma function?

By repeatedly differentiating $\Gamma(x)$, I noticed that $$\frac{d^{n}}{{dx}^{n}}\Gamma(x)=\sum_{k=0}^{n-1}\binom{n-1}{k}\psi^{(n-k-1)}(x)\,\frac{d^{k}}{{dx}^{k}}\Gamma(x),$$ where $\psi^{(a)}(x)$ is ...
Ali Shadhar's user avatar
  • 25.8k
5 votes
1 answer
231 views

Closed form for $\Gamma(a-x)$ where $a \in (0,1]$.

Now asked on MO here. I wonder if there is a closed form for $ \Gamma(a-x)$. And by closed form here I mean a finite combinations of elementary functions, powers of $\Gamma(a)$ and powers of $\Gamma(...
pie's user avatar
  • 6,620
2 votes
0 answers
238 views

Showing $\int_{0}^{1}\frac{E(\tfrac{x}{\sqrt{x^2+8}})}{\sqrt{8-7x^2-x^4}}dx=\frac{1}{3}K(\frac{1}{\sqrt{2}})E(\frac{1}{\sqrt{2}})$

Context $\begin{align} K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}\tag{1} \end{align}$ and $\begin{align} E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2t}dt\tag{2} \end{align}$ the complete elliptic ...
User's user avatar
  • 323
3 votes
0 answers
53 views

Find $\prod_{k=1}^n \frac{\Gamma (a_k/m)}{\Gamma (b_k/m)}$ algorithmically

It sometimes happens that $$\prod_{k=1}^n \frac{\Gamma (a_k/m)}{\Gamma (b_k/m)}$$ is algebraic for positive integers $m,n,a_k,b_k$. For example, $$\frac{\Gamma\left(\frac{1}{24}\right)\Gamma\left(\...
Nomas2's user avatar
  • 667
0 votes
0 answers
74 views

Is it possible to find a closed form for $i!$? [duplicate]

I am curious is there a closed form for $i!$? I tried to search for any closed form for this but I didn't find any. $$z! := \lim_{n \to \infty } n^z \prod_{k=1}^n \frac {k}{z+k}$$ $$i! =\lim_{n \to \...
Mathematics enjoyer's user avatar
2 votes
1 answer
164 views

Another weird limit involving gamma and digamma function via continued fraction

Context : I want to find a closed form to : $$\lim_{x\to 0}\left(\frac{f(x)}{f(0)}\right)^{\frac{1}{x}}=L,f(x)=\left(\frac{1}{1+x}\right)!×\left(\frac{1}{1+\frac{1}{1+x}}\right)!\cdots$$ Some ...
Ranger-of-trente-deux-glands's user avatar
4 votes
2 answers
188 views

A closed form for integrals of the type $\Gamma\left(\sigma+it\right)$?

Numerical evidence strongly suggests that: $$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, \sigma > 0$$ and $$\int_{-...
Agno's user avatar
  • 3,191
3 votes
1 answer
163 views

Does a closed-form expression exist for $ \int_0^\infty \ln(x) \operatorname{sech}(x)^n dx $?

I am trying to find a closed-form expression for the following integral $$ \int_0^\infty \ln(x) \operatorname{sech}(x)^n dx $$ There are specific values that I would like to generate (Table of ...
Gabriel Demirdag's user avatar
7 votes
0 answers
194 views

The closed-form of $1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot3}{2\cdot4}\right)^k-13\left(\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\right)^k+\dots$?

(A related MSE question by P. Singh.) First define, $$F_k = 1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^k-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^k+17\left(\...
Tito Piezas III's user avatar
1 vote
0 answers
52 views

Does the rest of this family of continued fractions have closed forms?

The pattern for the continued fractions below is quite straightforward. $F_1$ has numerators with all the integers but, $F_2\; \text{is missing}\; 2m+1 = 3,5,7,\dots\\ F_3\; \text{is missing}\; 3m+1 = ...
Tito Piezas III's user avatar
3 votes
2 answers
191 views

Proving $\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=\frac{\Gamma (1/4)}{4\sqrt{2}\pi^{7/4}}$

I conjecture that $$\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=\frac{\Gamma (1/4)}{4\sqrt{2}\pi^{7/4}}$$ because the left-hand side and right-hand side agree to at least $50$ decimal places. Is the ...
Nomas2's user avatar
  • 667
9 votes
2 answers
341 views

On the cubic counterpart of Ramanujan's $\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots$?

We have Ramanujan's well-known, $$\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots\color{blue}+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\...
Tito Piezas III's user avatar
0 votes
0 answers
34 views

Does the following expression have a closed form?

I have a ratio that has upper incomplete Gamma functions. $$ r = \frac{\Gamma(n, (a+c) x) - \Gamma(n, (a+c) y)}{\Gamma(n, a x) - \Gamma(n, a y)} $$ Here, $a, c, x, y$, are positive real and $y > x$....
CfourPiO's user avatar
  • 109
4 votes
0 answers
96 views

Evaluate $\sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(2b)_n} \frac{\psi^{(0)}\left (n+ \frac{a}2+1 \right ) }{\Gamma\left (n+ \frac{a}2+1 \right ) }$

Define $$S(a,b)=\sum_{n=0}^{\infty} \frac{(a)_n (b)_n}{(2b)_n} \frac{\psi^{(0)}\left (n+ \frac{a}2+1 \right ) }{\Gamma\left (n+ \frac{a}2+1 \right ) },$$ where $\psi^{(0)}(x)=\frac{\Gamma^\prime(x)}{\...
Setness Ramesory's user avatar
2 votes
1 answer
115 views

Generalization of Gauss multiplication formula for $\Gamma(jm+kn+a);j,k\in\Bbb N$?

A hypergeometric single sum, like a Mittag Leffler function uses the Pochhammer symbol $(a)_n$ multiplication formula to easily have a univariate hypergeometric function $_p\text F_q$ closed form: $$\...
Тyma Gaidash's user avatar
1 vote
1 answer
61 views

Closed form for $\int_0^{\pi/2}(\cos^2{\theta})^{1/n}d\theta$

Is there a closed form for $f(n)=\int_0^{\pi/2}(\cos^2{\theta})^{1/n}d\theta$ for every $n\in\mathbb{N}$ ? I suspect there may be, because of the following apparent equalities. $f(1)=\frac{\pi}{4}$ $f(...
Dan's user avatar
  • 25.8k
1 vote
1 answer
133 views

Does $\lim_{x\to 0} \left(2^{1-x!}3^{1-x!!}4^{1-x!!!}5^{1-x!!!!}6^{1-x!!!!!}\cdot\cdot\cdot\right)^{\frac{1}{x}}=L$ admits a closed form?

I try to simplify this limit : $$\lim_{x\to 0} \left(2^{1-x!}3^{1-x!!}4^{1-x!!!}5^{1-x!!!!}6^{1-x!!!!!}\cdots\right)^{\frac{1}{x}}=L$$ Where we compose the Gamma function with itself . From the past ...
Ranger-of-trente-deux-glands's user avatar
1 vote
0 answers
43 views

Computing $\frac{d^{n-1}}{dz^{n-1}}\frac{z}{\ln\left(z!\right)}^{n}$

I am trying to compute $$\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z}{\ln\left(z!\right)}\right)^{n}$$ The problem arises when dealing with inversion formulae. My question is, can this expression be ...
user2549157's user avatar
0 votes
1 answer
54 views

Find $a$ such that the limit is zero

Problem : Let $x>0$ then define : $$f(x)=\left(\left(\frac{1}{x}\right)!\left(x!\right)\right)^{\frac{1}{x+\frac{1}{x}}}$$ Then find $a$ such that : $$\lim_{x\to\infty}f(x)-\frac{1}{2}\left(\frac{1}...
Ranger-of-trente-deux-glands's user avatar
3 votes
1 answer
125 views

Neat function $I(x,y)= \sum\limits_{n=1}^{\infty}\bigg|\int_0^1 \frac{1}{t~(\log t)^y}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg| $. Closed form?

Consider the following function: $$ I(x,y)=\sum_{n=1}^{\infty}\bigg|\int_0^1 \frac{1}{t~(\log t)^y}\exp\bigg(\frac{n^x}{\log t}\bigg) ~dt~\bigg|$$ For $x=0$ and letting $y$ vary we get the Gamma ...
zeta space's user avatar
3 votes
0 answers
173 views

Question on a closed-form expression related to the harmonic number $H_n$

In this question the notation $\tilde{f}(x)$ refers to an analytic representation of the summatory function $$f(x)=\sum\limits_{n=1}^x a(n)\tag{1}$$ that converges to $$\underset{\epsilon\to 0}{\text{...
Steven Clark's user avatar
  • 7,631
0 votes
2 answers
145 views

Closed from for the series involving gamma function

Is there a closed form for the fallowing series, $$\sum_{n=1}^\infty \Gamma\left(n+\frac12+\frac12k\right)\Gamma\left(n+\frac12-\frac12k\right)\frac{x^{n}}{(2n+1)!}$$ where $k\notin\mathbb{Z}$. I ...
user avatar
17 votes
1 answer
1k views

What is $\mathcal{R}$?

First of all, I am asking this question entirely out of curiosity. It basically randomly popped out of my mind. So I am asking for the value of an infinite series. Let's call it, $\mathcal{R}=\sum_{n=...
Rounak Sarkar's user avatar
6 votes
1 answer
307 views

Closed form of $\sum_{n=1}^\infty \frac{1}{\sinh n\pi}$ in terms of $\Gamma (a)$, $a\in\mathbb{Q}$

This question and this question are about $$\sum_{n=1}^\infty \frac{1}{\cosh n\pi}=\frac{1}{2}\left(\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}-1\right)$$ and $$\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}=\frac{1}{...
Poder Rac's user avatar
  • 966
1 vote
1 answer
123 views

Proof about the power series of reciprocal multifactorials $m_x(k)=\sum_{n=0}^\infty \frac{x^n}{n\underbrace{!\cdots!}_{\text{k times}}}$

The proof I've attempted mimics very closely the answer on this question. How to prove the formula for the Reciprocal Multifactorial constant? Pre-requisite definitions: A multifactorial of order $k \...
Bhoris Dhanjal's user avatar
0 votes
0 answers
41 views

Incomplete upper gamma for a non-integer number of degrees of freedom

I can't seem to nail the closed form of the incomplete upper gamma function for the number of degrees of freedom $s$ being a fraction type $n/2$ where $n$ is integer. For the case when $s$ is integer ...
Emile Zäkiev's user avatar
48 votes
1 answer
1k views

How to evaluate double limit of multifactorial $\lim\limits_{k\to\infty}\lim\limits_{n\to 0} \sqrt[n]{n\underbrace{!!!!\cdots!}_{k\,\text{times}}}$

Define the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ where the product extends to the least positive integer of $n$ modulo $k$. In this answer, I derived one of several analytic ...
TheSimpliFire's user avatar
  • 27.1k
6 votes
1 answer
86 views

If $r>0$ and $r\notin \mathbb{N}$, is there a simple method to evaluate $ \sum_{n=\lceil r \rceil}^{\infty} {\binom{n}{r}^{-1}}?$

Let $r>0,r\in \mathbb{R}\setminus\mathbb{N}$. Empirically, I have noticed the following relation: $$ \sum_{n=0}^{\lfloor r \rfloor} \frac{1}{\binom{n}{r}} = - \sum_{n=\lceil r \rceil}^{\infty} \...
Integrand's user avatar
  • 8,369
1 vote
0 answers
95 views

Is there a closed form for the binomial expression $\binom{-1/m}{k} $?

I'm interested in binomial coefficients of the form $$\binom{-1/m}{k} ,$$ where $m$ is a positive integer. For $m=2$, it holds that \begin{align} \binom{-1/2}{k} &= (-4)^{-k} \binom{2k}{k} \qquad ...
Max Muller's user avatar
  • 7,148
15 votes
3 answers
1k views

Prove that $_4F_3\left(\frac13,\frac13,\frac23,\frac23;1,\frac43,\frac43;1\right)=\frac{\Gamma \left(\frac13\right)^6}{36 \pi ^2}$

I found an interesting problem about generalized hypergeometric series in MO, stating that: $$\, _4F_3\left(\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{2}{3};1,\frac{4}{3},\frac{4}{3};1\right)=\sum_{n=...
Infiniticism's user avatar
  • 8,654
6 votes
3 answers
319 views

Proving $\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a)$

Mathematica gives $$\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a),\quad 0<a<1$$ All I did is reindexing then using the series property $\sum_{n=...
Ali Shadhar's user avatar
  • 25.8k
5 votes
3 answers
650 views

Alternative approaches to showing that $\Gamma'(1/2)=-\sqrt\pi\left(\gamma+\log(4)\right)$

Starting from the definition of the Gamma function as expressed by $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx\tag1$$ we can show that the derivative of $\Gamma(z)$ evaluated at $z=1/2$ is given by $$\...
Mark Viola's user avatar
  • 181k
1 vote
0 answers
55 views

Closed form of $\sum_{n=1}^\infty (n+k)!(a/n)^n$

I got this equality: $$\sum_{n=1}^\infty (n+k)!\left(\frac{a}{n}\right)^n=a(k+1)!\int_{0}^{1}\frac{dx}{(1+ax\ln x)^{k+2}}$$ when $|a|<e$ then, does this series have a closed form?
Ckk's user avatar
  • 37
4 votes
2 answers
394 views

On a log-gamma definite integral

A very famous log-gamma integral due to Raabe is $$\int_0^1 \log \Gamma (x) \, dx = \frac{1}{2} \log (2\pi).$$ Several proofs of this result can be found here. I would like to know about the ...
omegadot's user avatar
  • 11.8k
5 votes
1 answer
126 views

Is there a way to simplify the solution to $\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{x^{...}}}} dx$

My result for this integral is as follows: $$\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{....}}} = (e^{\frac{1}{e}})e - e - \frac{1}{2} - \sum_{k=1}^{\infty} \left( \frac{\gamma((k+2),(k))}{{k}^{(k+2)}\Gamma(...
Nathan Bradley's user avatar
3 votes
1 answer
149 views

Closed form for $\Gamma (a+bi)\Gamma(a-bi)$ [duplicate]

I noticed that $$\Gamma (3+2i)\Gamma (3-2i)=\frac{160\pi}{e^{2\pi}-e^{-2\pi}}$$ and $$\Gamma (2+5i)\Gamma (2-5i)=\frac{260\pi}{e^{5\pi}-e^{-5\pi}}.$$ Is there a closed form for $\Gamma (a+bi)\Gamma (a-...
UraUra's user avatar
  • 311
1 vote
0 answers
57 views

Generating function containing Incomplete gamma function

Consider the following generating function : $$\sum_{k=0}^\infty\sum_{n=0}^\infty\sum_{m=0}^\infty \frac {n^{2m+4}(-1)^m\Gamma(2k+1,-(am+b))}{m!(am+b)^{2k+1}} x^{2k}$$ Where , $\Gamma(p,q)$ is ...
bambi's user avatar
  • 916
1 vote
3 answers
212 views

Evaluate $\int_0^\infty x^{n+\frac12}e^{-\frac x2}\log^2x\,dx$ and $\int_0^\infty x^ne^{-\frac x2}\log^2x\,dx$

Determine the closed forms of $$\mathfrak I_1=\int_0^\infty x^{n+\frac12}e^{-\frac x2}\log^2x\,dx\quad\text{and}\quad\mathfrak I_2=\int_0^\infty x^ne^{-x/2}\log^2x\,dx$$ where $s>0$ is an integer. ...
TheSimpliFire's user avatar
  • 27.1k
2 votes
1 answer
152 views

Closed form of $\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$

I was wondering if there is a closed form for $$\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$$ I know that for $$\prod_{i=0}^{N}\big(i!\big)=G(N+2)$$ where we have expressed it as Barnes G-function. ...
Wiliam's user avatar
  • 493
2 votes
1 answer
118 views

Compute in a closed form the following sum : $\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$

Today Im going to find the closed form of : $\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$ My attempt : We know that : $\Gamma(z)=\int_0^{+\infty}t^{n-1}e^{-t}...
Dinize markob's user avatar
6 votes
3 answers
322 views

Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$

How to show $$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$ I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\...
Infiniticism's user avatar
  • 8,654
15 votes
3 answers
1k views

Closed-form of log gamma integral $\int_0^z\ln\Gamma(t)~dt$ for $z =1,\frac12, \frac13, \frac14, \frac16,$ using Catalan's and Gieseking's constant?

We have the known, $$I(z)=\int_0^z\ln\Gamma(t)~dt=\frac{z(1-z)}2+\frac z2\ln(2\pi)+z\ln\Gamma(z)-\ln G(z+1)$$ or alternatively, $$I(z)=\int_0^z\ln\Gamma(t)~dt= \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) -(...
Tito Piezas III's user avatar
5 votes
1 answer
205 views

Prove $\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$

Prove $$S=\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}.$$ I don't know how to evaluate this problem . At first I used partial fraction but I got ...
Kînan Jœd's user avatar
1 vote
1 answer
45 views

Expressing the coefficients of $(1-x)^{1/4}$ using factorials

From the fact that $1\times3\times5\times\ldots\times(2n-1)=\frac{(2n)!}{2^nn!}$, we can show that $$ (1-x)^{1/2}=\sum_{n=0}^\infty \frac{(2n-2)!}{(n-1)!n!2^{2n+1}}x^n. $$ However, can I do the same ...
Ma Joad's user avatar
  • 7,534
4 votes
1 answer
171 views

closed form of the following integral :$\int_{0}^{\infty}- \sqrt{x}+ \sqrt{x}\coth (x) dx$?

I have tried to evaluate this:$\int_{0}^{\infty}- \sqrt{x}+ \sqrt{x}\coth (x)$ using the the following formula $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\Big( \frac{x^{...
zeraoulia rafik's user avatar
1 vote
0 answers
37 views

Closed form for an integral involving an incomplete Gamma function?

I am trying to find a closed form for this integral: $$\int_{0}^{\infty}\int_{0}^{\infty}e^{-d_{p,s}^v\,x-d_{s,p}^v\, y+d_{p,p}^v\,\frac{\xi_1\,\sigma^2\,xy}{P_p\,\xi_2\,xy+\sigma^2}}\mathrm dy\...
X. Zhang's user avatar
3 votes
2 answers
398 views

How to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}$

I want to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}.$ I tried from arcsin(x) series and got $\frac{1-z^4}{(1+z^4)^{\frac{2}{3}}}= 1-5(\frac{1}{2})z^4+9(\frac{(1)(3)}{(2)(4)})...
Unik Sillavich's user avatar
1 vote
0 answers
68 views

Closed form for product over Gamma function

Is there a "closed form" (with which I mean an expression not involving an indexed sum or product) for any of these four products? $$\prod_{k=1}^{n} \Gamma(\frac{x}{k*2+1})$$ $$\prod_{k=1}^{n} \Gamma(...
SmallestUncomputableNumber's user avatar

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