All Questions
Tagged with closed-form gamma-function
109
questions
1
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0
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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 ...
- 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:
$$\, _4F_3\left(\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{2}{3};1,\frac{4}{3},\frac{4}{3};1\right)=\sum_{n=...
- 8,654
6
votes
3
answers
319
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Proving $\sum_{n=0}^\infty\frac{(-1)^n\Gamma(2n+a+1)}{\Gamma(2n+2)}=2^{-a/2}\Gamma(a)\sin(\frac{\pi}{4}a)$
Mathematica gives
$$\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$$
All I did is reindexing then using the series property $\sum_{n=...
- 25.8k
5
votes
3
answers
650
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Alternative approaches to showing that $\Gamma'(1/2)=-\sqrt\pi\left(\gamma+\log(4)\right)$
Starting from the definition of the Gamma function as expressed by
$$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\,dx\tag1$$
we can show that the derivative of $\Gamma(z)$ evaluated at $z=1/2$ is given by
$$\...
- 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?
- 37
4
votes
2
answers
394
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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 ...
- 11.8k
5
votes
1
answer
126
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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:
$$\int_{1}^{e^{\frac{1}{e}}} x^{x^{x^{....}}} = (e^{\frac{1}{e}})e - e - \frac{1}{2} - \sum_{k=1}^{\infty} \left( \frac{\gamma((k+2),(k))}{{k}^{(k+2)}\Gamma(...
3
votes
1
answer
149
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Closed form for $\Gamma (a+bi)\Gamma(a-bi)$ [duplicate]
I noticed that
$$\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}}.$$
Is there a closed form for $\Gamma (a+bi)\Gamma (a-...
- 311
1
vote
0
answers
57
views
Generating function containing Incomplete gamma function
Consider the following generating function :
$$\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}$$
Where , $\Gamma(p,q)$ is ...
- 916
1
vote
3
answers
212
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Evaluate $\int_0^\infty x^{n+\frac12}e^{-\frac x2}\log^2x\,dx$ and $\int_0^\infty x^ne^{-\frac x2}\log^2x\,dx$
Determine the closed forms of $$\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$$ where $s>0$ is an integer.
...
- 27.1k
2
votes
1
answer
152
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Closed form of $\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$
I was wondering if there is a closed form for
$$\prod_{i=0}^{N}\big(i!\big)^{{N}\choose{i}}$$
I know that for
$$\prod_{i=0}^{N}\big(i!\big)=G(N+2)$$
where we have expressed it as Barnes G-function. ...
- 493
2
votes
1
answer
118
views
Compute in a closed form the following sum : $\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$
Today Im going to find the closed form of :
$\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$
My attempt :
We know that : $\Gamma(z)=\int_0^{+\infty}t^{n-1}e^{-t}...
- 205
6
votes
3
answers
322
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Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$
How to show
$$\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}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{2},\...
- 8,654
15
votes
3
answers
1k
views
Closed-form of log gamma integral $\int_0^z\ln\Gamma(t)~dt$ for $z =1,\frac12, \frac13, \frac14, \frac16,$ using Catalan's and Gieseking's constant?
We have the known,
$$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) -(...
- 54.9k
5
votes
1
answer
205
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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 ...
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