All Questions
20
questions
4
votes
2
answers
238
views
Closed form for this generalisation of the gamma function. $f(x+1)=f(x)g(x+1) $
Just for curiosity I want to generalise the Pi function i.e $f(x+1) = f(x)g(x+1)$ for some differentiable function, I know this function probably has no closed form for general functions $g$ as I ...
- 6,620
1
vote
0
answers
128
views
Conjectured closed form for ${\it {Li_2}} \left( 1-{\frac {\sqrt {2}}{2}}-i \left( 1-{\frac {\sqrt { 2}}{2}} \right) \right)$
With Maple i find this closed form:
${\it {Li_2}} \left( 1-{\frac {\sqrt {2}}{2}}-i \left( 1-{\frac {\sqrt {
2}}{2}} \right) \right)$=$-{\frac {{\pi}^{2}}{64}}-{\frac { \left( \ln \left( 1+\sqrt {2}
...
- 303
16
votes
2
answers
599
views
On $\int_0^1\arctan\,_6F_5\left(\frac17,\frac27,\frac37,\frac47,\frac57,\frac67;\,\frac26,\frac36,\frac46,\frac56,\frac76;\frac{n}{6^6}\,x\right)\,dx$
Reshetnikov gave the remarkable evaluation,
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac24,\frac34,\frac54;\frac{1}{64}\,x\right)\,dx \\
&=\frac{3125}{...
- 54.9k
16
votes
2
answers
584
views
Conjecture $\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}=\frac{1}{12\pi^2}$
Wolfram Alpha numerical calculation shows that the quantity
$$
\sum_{n=1}^\infty\frac{12\pi^2n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}
$$
is 1 with high accuracy. Can anybody prove the resulting conjecture:...
- 1,695
20
votes
3
answers
908
views
Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$
I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!...
- 47.3k
57
votes
3
answers
2k
views
Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$
There is a known asymptotic expansion of harmonic numbers $H_n$ for $n\to\infty$:
$$\begin{align}H_n&=\gamma+\ln n+\sum_{k=1}^\infty\left(-\frac{B_k}{k\cdot n^k}\right)\\
&=\gamma+\ln n+\frac1{...
- 47.3k
22
votes
4
answers
1k
views
Closed form for $\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx$
I'm trying to find a closed form for the following integral:
$$\mathcal{J}(n)=\int_{-1}^1\frac{\ln\left(2+x\,\sqrt3\right)}{\sqrt{1-x^2}\,\left(2+x\,\sqrt3\right)^n}dx\tag1$$
I have conjectured values ...
- 47.3k
73
votes
2
answers
2k
views
Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$
Using a numerical search on my computer I discovered the following inequality:
$$\left|\,{_2F_1}\left(\frac14,\frac34;\,\frac23;\,\frac13\right)-\rho\,\right|<10^{-20000},\tag1$$
where $\rho$ is ...
- 1,063
28
votes
1
answer
924
views
Conjectured closed form for $\int_0^1x^{2\,q-1}\,K(x)^2dx$ where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind
I am interested in a general closed-form formula for integrals of the following form:
$$\mathcal{J}_q=\int_0^1x^{2\,q-1}\,K(x)^2dx,\tag0$$
where $K(x)$ is the complete elliptic integral of the 1ˢᵗ ...
- 47.3k
27
votes
2
answers
865
views
Conjecture: $\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\big(\!\frac{2x}{1+x}\!\big)\,dx\stackrel ?=\frac\pi{5\sqrt2}$
$$\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\left(\frac{2x}{1+x}\right)\,dx\stackrel ?=\frac\pi{5\sqrt2}$$
The integral above comes from the evaluation of the integral $A=\int_0^{\pi/2}\frac{f(\theta)}\...
- 27.7k
96
votes
2
answers
7k
views
Conjecture $\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi$
$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
The equality numerically holds up to at least $10^4$ decimal digits.
Can ...
- 47.3k
91
votes
2
answers
6k
views
Conjecture $\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})$
Let
$$\alpha=\sqrt{6}\ \sqrt{12+7\,\sqrt3}-3\,\sqrt3-6\,=\,\big(2+\sqrt{3}\big) \big(\sqrt{2} \sqrt[4]{27}-3\big)\,=\,\frac{3\sqrt{3}}{3+\sqrt2\ \sqrt[4]{27}}.\tag1$$
Note that $\alpha$ is the unique ...
- 47.3k
15
votes
2
answers
787
views
Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$
How do I prove
$$_4F_3\left(\frac18,\frac38,\frac58,\frac78;\ \frac14,\frac12,\frac34;\ \frac12\right)\stackrel?=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2\phantom{|}}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2\phantom{|...
- 47.3k
15
votes
2
answers
469
views
Conjecture $\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$
$$\int_0^1\frac{\ln\left(\ln^2x+\arccos^2x\right)}{\sqrt{1-x^2}}dx\stackrel?=\pi\,\ln\ln2$$
Is it possible to prove this?
- 47.3k
26
votes
1
answer
731
views
Simplify $\frac{_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2};\frac{3}{4}\right)}{\Pi\left(\frac{1}{4}\big|\frac{1}{\sqrt{3}}\right)}$
Is it possible to simplify the ratio
$$\mathcal{E}=\frac{_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\ 1,\frac{3}{2};\ \frac{3}{4}\right)}{\Pi\left(\frac{1}{4}\Big|\frac{1}{\sqrt{3}}\right)},$$
...
- 13.2k