All Questions
14
questions
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 ...
- 11
2
votes
1
answer
76
views
How to compute: $I_x =\int_0^1 t^x e^{-\frac{1}{1-t^2}} dt~~~x>0$?
$$I_x =\int_0^1 t^x e^{-\frac{1}{1-t^2}} dt~~~x>0$$
I first tried to see if I could solve the integer case.
$$I_n =\int_0^1 t^ne^{-\frac{1}{1-t^2}} dt $$
I have tried to find a possible ...
- 24.2k
28
votes
3
answers
1k
views
Interesting closed form for $\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta$
Some time ago I used a formal approach to derive the following identity:
$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^{\frac{1}{12}}\pi\...
- 3,343
23
votes
3
answers
1k
views
Integral $\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)}\,dx$
I would like to evaluate this integral:
$$\mathcal F(a)=\int_{-\infty}^\infty\frac{\Gamma(x)\,\sin(\pi x)}{\Gamma\left(x+a\right)}\,dx,\quad a>0.\tag1$$
For all $a>0$ the integrand is a smooth ...
- 1,271
12
votes
2
answers
688
views
Multiple integrals involving product of gamma functions
The following integral was posted a few days back on Integrals and Series forum:
$$\int_0^{2\pi} \int_0^{2\pi} \int_0^{2\pi} \frac{dk_1\,dk_2\,dk_3}{1-\frac{1}{3}\left(\cos k_1+\cos k_2+ \cos k_3\...
- 11k
15
votes
1
answer
599
views
Closed-form of $\int_0^1 \left(\ln \Gamma(x)\right)^3\,dx$
From the amazing result by Raabe we know that
$$LG_1=\int_0^1 \ln \Gamma(x)\,dx = \frac{1}{2}\ln(2\pi) = -\zeta'(0).$$
We also know that
$$LG_2 = \int_0^1 \left(\ln \Gamma(x)\right)^2\,dx = \frac{\...
- 12.4k
4
votes
0
answers
281
views
Non-recursive closed-form of the coefficients of Taylor series of the reciprocal gamma function
The reciprocal gamma function has the following Taylor series.
$$\frac{1}{\Gamma(z)}=\sum_{k=1}^{\infty}a_kz^k,$$
where the $a_k$ coefficient are given by the followint recursion.
$a_1=1$, $a_2=\...
- 12.4k
9
votes
1
answer
1k
views
Closed form for $\int_1^\infty\frac{\operatorname dx}{\operatorname \Gamma(x)}$
Is a closed form for $$\int\limits_1^{+\infty}\frac{\operatorname dx}{\operatorname \Gamma(x)}$$known?
I tried to find it, but all well-known integrals involving gamma-function (such as of $\log\...
- 4,301
7
votes
3
answers
518
views
A closed-form of product the gamma functions containing $\pi$ and $\phi$
Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result
\begin{equation}
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\...
- 19.4k
5
votes
1
answer
137
views
Expressing $\int_{-\infty}^\infty dx/(x^2+1)^n$ in terms of Gamma function
How to prove this identity for $n>1/2$?
$$\int_{-\infty}^{\infty}\frac{dx}{(x^2+1)^n}=\frac{\sqrt{\pi}\cdot \Gamma(n-\frac{1}{2}) }{\Gamma (n)}$$
- 51
25
votes
1
answer
1k
views
Strange closed forms for hypergeometric functions
So in the process of trying to find a derivation for this answer, the following interesting equalities arose (one can check with Wolfram Alpha/Mathematica):
$$\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(...
- 1,827
60
votes
1
answer
6k
views
Is it possible to simplify $\frac{\Gamma\left(\frac{1}{10}\right)}{\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$?
Is it possible to simplify this expression?
$$\frac{\displaystyle\Gamma\left(\frac{1}{10}\right)}{\displaystyle\Gamma\left(\frac{2}{15}\right)\ \Gamma\left(\frac{7}{15}\right)}$$
Is there a systematic ...
- 2,602
26
votes
1
answer
939
views
Is $K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi}$
Working on this conjecture, I found its corollary, which is also supported by numeric calculations up to at least $10^5$ decimal digits:
$$K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\...
- 47.3k
50
votes
5
answers
1k
views
Closed form of $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form of the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
- 13.2k