All Questions
Tagged with closed-form calculus
835
questions
595
votes
14
answers
384k
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Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$
I need help with this integral:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
The integrand graph looks like this:
$\hspace{1in}$
The ...
- 13.2k
135
votes
2
answers
6k
views
How to prove $\int_0^1\tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$
How can one prove that
$$\int_0^1 \tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$$
- 1,499
122
votes
5
answers
41k
views
An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$
I need your help with this integral:
$$\mathcal{K}(p)=\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx,$$
where $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy ...
- 21.7k
118
votes
4
answers
12k
views
Compute $\int_0^{\pi/4}\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)} x\exp(\frac{x^2-1}{x^2+1}) dx$
Compute the following integral
\begin{equation}
\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, ...
- 19.4k
101
votes
1
answer
4k
views
Arithmetic-geometric mean of 3 numbers
The arithmetic-geometric mean$^{[1]}$$\!^{[2]}$ of 2 numbers $a$ and $b$ is denoted $\operatorname{AGM}(a,b)$ and defined as follows:
$$\text{Let}\quad a_0=a,\quad b_0=b,\quad a_{n+1}=\frac{a_n+b_n}2,...
- 47.3k
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
81
votes
1
answer
4k
views
The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$
What tools or ways would you propose for getting the closed form of this integral?
$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
EDIT: It took a while since I made this post. I'll ...
- 44.4k
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
71
votes
5
answers
4k
views
Show that $\int_{0}^{\pi/2}\frac {\log^2\sin x\log^2\cos x}{\cos x\sin x}\mathrm{d}x=\frac14\left( 2\zeta (5)-\zeta(2)\zeta (3)\right)$
Show that :
$$
\int_{0}^{\Large\frac\pi2}
{\ln^{2}\left(\vphantom{\large A}\cos\left(x\right)\right)
\ln^{2}\left(\vphantom{\large A}\sin\left(x\right)\right)
\over
\cos\left(x\right)\sin\left(x\...
- 3,955
70
votes
5
answers
9k
views
Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$
Is there a closed form for the integral
$$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$
I do not have a strong reason to be sure it exists, but I ...
- 1,249
69
votes
5
answers
4k
views
Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
Please help me to find a closed form for the following integral:
$$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$
I was told it could be calculated in a ...
- 13.2k
66
votes
2
answers
4k
views
Prove that $\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx =\frac{\pi e}{24} $
I've found here the following integral.
$$I = \int_{0}^{1}\sin{(\pi (1-x))}x^x(1-x)^{1-x}\,dx=\int_{0}^{1}\sin{(\pi x)}x^x(1-x)^{1-x}\,dx=\frac{\pi e}{24}$$
I've never seen it before and I also didn'...
- 12.4k
64
votes
7
answers
25k
views
What is the antiderivative of $e^{-x^2}$
I was wondering what the antiderivative of $e^{-x^2}$ was, and when I wolfram alpha'd it I got $$\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space \text{erf} (x) + C$$
So, I ...
- 2,259
62
votes
3
answers
4k
views
Closed Form for $~\int_0^1\frac{\text{arctanh }x}{\tan\left(\frac\pi2~x\right)}~dx$
Does $$~\displaystyle{\int}_0^1\frac{\text{arctanh }x}{\tan\left(\dfrac\pi2~x\right)}~dx~\simeq~0.4883854771179872995286585433480\ldots~$$ possess a closed form expression ?
This recent post, in ...
- 48.5k