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21 votes
3 answers
1k views

On the general form of the family $\sum_{n=1}^\infty \frac{n^{k}}{e^{2n\pi}-1} $

I. $k=4n+3.\;$ From this post, one knows that $$\sum_{n=1}^\infty \frac{n^{3}}{e^{2n\pi}-1} = \frac{\Gamma\big(\tfrac{1}{4}\big)^8}{2^{10}\cdot5\,\pi^6}-\frac{1}{240}$$ and a Mathematica session ...
Tito Piezas III's user avatar
26 votes
3 answers
1k views

Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$

Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
Piotr Shatalin's user avatar
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
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 ...
X.C.'s user avatar
  • 2,602
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\...
Vladimir Reshetnikov's user avatar
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\...
Anon's user avatar
  • 3,343
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
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 ...
Vladimir Reshetnikov's user avatar
22 votes
2 answers
2k views

Closed form (or an ODE) for the integral $\int_0^\infty \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz$

Is there a closed form for: $$I(p)=\int_0^\infty \frac{1+z^2}{1+z^4} \frac{z^p}{1+z^{2p}} dz$$ The integral has a number of nice properties: $$I(p)=I(-p)$$ $$I(p)=2\int_0^1 \frac{1+z^2}{1+z^4} \...
Yuriy S's user avatar
  • 31.7k
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\...
CuriousGuest's user avatar
  • 4,301
33 votes
1 answer
1k views

Closed form for $\sum_{n=1}^\infty\frac{\psi(n+\frac{5}{4})}{(1+2n)(1+4n)^2}$

This question came up in the process of finding solution to another problem. Eventually, the problem was solved avoiding calculation of this sum, but it looks quite interesting on its own. Is there a ...
Laila Podlesny's user avatar
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 ...
july's user avatar
  • 1,271
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}\...
Chris Culter's user avatar
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 ...
balping's user avatar
  • 159
3 votes
0 answers
160 views

Two complementary continued fractions that are algebraic numbers

Define the two similar continued fractions, $$x=\cfrac{1}{km\color{blue}+\cfrac{(m-1)(m+1)} {3km\color{blue}+\cfrac{(2m-1)(2m+1)}{5km\color{blue}+\cfrac{(3m-1)(3m+1)}{7km\color{blue}+\ddots}}}}\tag1$$...
Tito Piezas III's user avatar

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