All Questions
13
questions
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)}{\...
- 5,370
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:
$$\...
- 12.5k
15
votes
3
answers
1k
views
Prove that $_4F_3\left(\frac13,\frac13,\frac23,\frac23;1,\frac43,\frac43;1\right)=\frac{\Gamma \left(\frac13\right)^6}{36 \pi ^2}$
I found an interesting problem about generalized hypergeometric series in MO, stating that:
$$\, _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
1
vote
0
answers
89
views
Yet another bizarre identity involving hypergoemetric functions and gamma functions.
Let $d=4$, $T\ge d$ and $p\ge 0$ be integers.
By solving Spectral densities of finite dimensional sample covariance matrices we stumbled on a following identity.
\begin{eqnarray}
&&\sum\...
- 11.5k
14
votes
1
answer
1k
views
Prove $\int_0^1 \frac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left(9\Gamma(3/4)^2{}_4F_3(\cdots)+\Gamma(5/4)^2{}_4F_3(\cdots)\right)$
Mathematica gives the following. But how?!
$$\small{\int_0^1 \dfrac{4\cos^{-1}x}{\sqrt{2x-x^2}}\,dx=\frac{8}{9\sqrt{\pi}}\left(
9\Gamma\left(\tfrac{3}{4}\right)^2{}_4F_3\left( \begin{array}{c}\...
- 27k
9
votes
1
answer
448
views
Simple closed form for $\int_0^\infty\frac{1}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}}\;dx$
Some time ago, I used a fairly formal method (in the second sense of this answer) to derive the following integral, and am wondering whether it is correct or not:
$$\int_0^\infty\frac{1}{\sqrt{x^2+...
- 3,343
3
votes
0
answers
134
views
Is there a known transformation between $_2F_1\big(\tfrac12,\tfrac12;1;z\big)$ and $_2F_1\big(\tfrac12,\tfrac12;1;z^2\big)$?
In this post, the OP seeks a closed-form for,
$$A=\,_2F_1\big(\tfrac12,\tfrac12;1;\tfrac19\big)=1.02966\dots$$
Using the transformation,
$$\,_2F_1\big(\tfrac12,\tfrac12;1;z\big) = \tfrac2{1+\sqrt{1-z}}...
- 54.9k
14
votes
2
answers
400
views
Integral ${\large\int}_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}$
Mathematica claims that
$${\large\int}_0^1\!\!\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\sqrt\pi}{2^{\sqrt2}\sqrt2}\cdot\frac{\Gamma\left(\frac1{\sqrt2}\right)}{\Gamma\left(\frac12+\frac1{\sqrt2}\right)...
- 47.3k
2
votes
0
answers
238
views
Closed form expression for a sum
I want to calculate a sum of the form $$\sum_{k=0}^m \frac{\Gamma[m+1+\alpha-k]^2}{\Gamma[m+1-k]^2}\frac{\Gamma[x+k]}{\Gamma[x]k!}$$ where $m>0$ and belongs to integers and $\alpha$ takes half ...
- 21
5
votes
3
answers
1k
views
Hypergeometric 2F1 with negative c
I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is ...
- 159
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):
$$\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(...
- 1,827
33
votes
3
answers
2k
views
How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$?
I am interested in finding a general formula for the following integral:
$$\int_0^\infty J_\nu(x)^3dx,\tag1$$
where $J_\nu(x)$ is the Bessel function of the first kind:
$$J_\nu(x)=\sum _{n=0}^\infty\...
- 47.3k
34
votes
2
answers
2k
views
Closed form for $\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$
I need to find a closed form for these nested definite integrals:
$$I=\int_0^\infty\left(\int_0^1\frac1{\sqrt{1-y^2}\sqrt{1+x^2\,y^2}}\mathrm dy\right)^3\mathrm dx.$$
The inner integral can be ...
- 47.3k