Questions tagged [catalans-constant]
For questions about special identities and problems involving Catalan's constant as well as general questions about the constant itself.
94
questions
2
votes
2
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Is $\sum _{n=0} ^{\infty} \frac{1}{(2n+1)^2(2n+3) \binom{n-1/2}{n}} = G - \frac{1}{2}$?
Wolfram Mathematica claims that
$$\sum _{n=0} ^{\infty} \frac{1}{(2n+1)^2(2n+3) \binom{n-1/2}{n}} = G - 1/2$$
where $G$ is the Catalan constant.
I can't figure out how to prove that. I found a similar ...
3
votes
2
answers
146
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Show that $\int_0^1K^2(k)dk=\frac12\int_0^1K'^2(k)dk$
By switching integrals in double integral, I showed that
$$\int_0^1K(k)dk=\int_0^1K'(k)dk=2G$$
where $K(k)$ is complete elliptic integral of the first kind and $K'(k)=K(\sqrt{1-k^2})$ is its ...
- 11.9k
2
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4
answers
200
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Show that $\int_0^{\frac\pi 2}\frac{\theta-\cos\theta\sin\theta}{2\sin^3\theta}d\theta=\frac{2C+1}4$
While trying to evaluate $\int_0^1 k^2K(k)dk$ related to elliptic integral of the first kind, by integral switching method, I reached the trigonometric integral
$$\int_0^{\frac\pi 2}\frac{\theta-\cos\...
- 11.9k
1
vote
1
answer
105
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$\int_0^1E(k)dk$ without switching integrals
Let $E=E(k), K=K(k)$ be the complete elliptic integrals of the second and the first kind respectively. It is well-known that $\frac12\int_0^1K\,dk=G$ is the Catalan constant. We can also find, by ...
- 11.9k
3
votes
1
answer
146
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closed form for $\psi^{(-2)}\left(\frac{1}{4} \right)$
the polygamma function of negative order exist in a lot of complicated integrals like this integral:
$$\int_0^{\frac{1}{4}} x \psi(x) dx=\left(x \psi^{(-1)}(x) \right)^{\frac{1}{4}}_0-\int_0^{\frac{1}{...
- 1,647
6
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$I = \int_0^{\pi/2} \left( \int_0^{\pi/2} \frac{\log(\cos(x/2)) - \log(\cos(y/2))}{\cos(x) - \cos(y)} dx\right) \ dy $
\begin{align*}
I = \int_0^{\pi/2} \left( \int_0^{\pi/2} \frac{\log(\cos(x/2)) - \log(\cos(y/2))}{\cos(x) - \cos(y)} dx\right) \ dy
\end{align*}
What I do so far
Let $u = \cos(x/2)$ and $v = \cos(y/2)$....
11
votes
0
answers
255
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Solve the integral $\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx$
I tried to solve this integral and got it, I showed firstly
$$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$
and for other integral
$$\int_0^...
- 1,647
26
votes
4
answers
816
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Is there an expression for $I(n)=\int_{0}^{\frac{\pi}{4}}x\tan^{n}x dx$?
I've played around a little with this integral, and I can straightforwardly evaluate it with a substitution $\tan x\mapsto x$ in terms of the Beta function if the bounds were $(0,\pi /2)$. But for the ...
- 586
8
votes
3
answers
248
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Struggling to figure out why $\int\limits_{0}^{1}\frac{\ln \left(x^{2}\right)}{\left(1+x^{2}\right)^{2}}dx = -G-\frac{\pi}{4}$.
Struggling to figure out why $$\int\limits_{0}^{1}\frac{\ln \left(x^{2}\right)}{\left(1+x^{2}\right)^{2}}dx = -G-\frac{\pi}{4}$$
I've taken note of the exponent in the logarithm, tried substituting $...
- 586
9
votes
2
answers
782
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Another integral representation of Catalan's Constant?
The integral in question:
$$I=\int_{0}^{1} {\frac{\arctan\left(\frac{x-1}{\sqrt{x^2-1}}\right)}{\sqrt{x^2-1}}}dx.$$
So I have a solution and it's rather straightforward, but I don't much like it. I'm ...
0
votes
2
answers
107
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Ways to solve $\int_0^{\frac{\pi}{2}} \frac{x}{\sin{x}}dx$ [duplicate]
$$\int_0^{\frac{\pi}{2}} \frac{x}{\sin{x}}dx$$
I tried using
$$\int_0^{\frac{\pi}{2}} \frac{2ix}{e^{ix}-e^{-ix}}dx$$
And then factoring and making use of the Geometric Series
$$\int_0^{\frac{\pi}{2}} \...
- 478
4
votes
2
answers
205
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Evaluation of tricky Gamma infinite sum
I want to prove that:
$$\sum_{n=0}^\infty\frac{n}{(n+1)^2}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}=\frac{4G}{\sqrt{\pi}}+\sqrt{\pi}\log2$$
and
$$\sum_{n=0}^\infty(-1)^n\...
- 3,402
5
votes
0
answers
184
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Is there a simple way to evaluate $\int_0^1 \arctan x \ln \left(\frac{1-x}{1+x}\right) d x?$
After trying many methods but fail, I try the following substitution.
Letting $t=\frac{1-x}{1+x}$ preserves the interval and transforms the integral into
$$I=\int_0^1 \arctan x \ln \left(\frac{1-x}{1+...
- 22.3k
5
votes
2
answers
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Calculating a limit of a sequence of integrals that approaches the catalan constant
Let $f\colon[0, 1] \to \mathbb{R}$, $f(x)=\frac{\arctan(x)}{x}, x \in (0, 1], f(0)=1$. Prove that $$\lim_{n \to \infty} n \left(\frac{\pi}{4}-n\int_0^1\frac{x^n}{1+x^{2n}}dx \right)=\int_0^1f(x)dx$$
...
- 1,043
3
votes
2
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286
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For $n\ge m\ge 1$, how far can we walk with $ \int_0^{\frac{\pi}{2}} \frac{x^n}{\sin^m x} d x$?
In the post, I tackled the integral by power series and integration by parts and obtained that
$$
\int_0^{\frac{\pi}{2}} \frac{x^2}{\sin x} d x=2\pi G-\frac{7}{2}\zeta(3)
$$
where $G$ is the Catalan’s ...
- 22.3k