All Questions
6
questions
2
votes
0
answers
238
views
Showing $\int_{0}^{1}\frac{E(\tfrac{x}{\sqrt{x^2+8}})}{\sqrt{8-7x^2-x^4}}dx=\frac{1}{3}K(\frac{1}{\sqrt{2}})E(\frac{1}{\sqrt{2}})$
Context
$\begin{align}
K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}\tag{1}
\end{align}$
and
$\begin{align}
E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2t}dt\tag{2}
\end{align}$
the complete elliptic ...
- 323
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:
$$\, _4F_3\left(\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{2}{3};1,\frac{4}{3},\frac{4}{3};1\right)=\sum_{n=...
- 8,654
3
votes
2
answers
397
views
How to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}$
I want to Evaluate
$\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}.$
I tried from arcsin(x) series and got $\frac{1-z^4}{(1+z^4)^{\frac{2}{3}}}= 1-5(\frac{1}{2})z^4+9(\frac{(1)(3)}{(2)(4)})...
- 347
13
votes
2
answers
368
views
Prove known closed form for $\int_0^\infty e^{-x}I_0\left(\frac{x}{3}\right)^3\;dx$
I know that the following identity is correct, but I would love to see a derivation:
$$\int_0^\infty e^{-x}I_0\left(\frac{x}{3}\right)^3\;dx=\frac{\sqrt{6}}{32\pi^3}\Gamma\left(\frac{1}{24}\right)\...
- 3,343
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:
$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^{\frac{1}{12}}\pi\...
- 3,343
15
votes
1
answer
655
views
Closed form for $\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}$
Consider the following integral:
$$\mathcal{I}=\int_1^\infty\int_0^1\frac{\mathrm dy\,\mathrm dx}{\sqrt{x^2-1}\sqrt{1-y^2}\sqrt{1-y^2+4\,x^2y^2}}.$$
It can be represented as
$$\mathcal{I}=\int_1^\...
- 47.3k