Linked Questions
13 questions linked to/from Inverse Trigonometric Integrals
42
votes
7
answers
6k
views
Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?
Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\...
- 48k
26
votes
6
answers
3k
views
Evaluating $\int_0^{\large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $
It's my first post here and I was wondering if someone could help me with evaluating the
definite integral
$$ \int_0^{\Large\frac{\pi}{4}} \log\left( \cos x\right) \, \mathrm{d}x $$
Thanks in ...
- 269
22
votes
3
answers
1k
views
Prove $\int_{0}^{\pi/2} x\csc^2(x)\arctan \left(\alpha \tan x\right)\, dx = \frac{\pi}{2}\left[\ln\frac{(1+\alpha)^{1+\alpha}}{\alpha^\alpha}\right]$
When I showed to my brother how I proved
\begin{equation}
\int_{0}^{\!\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, \mathrm{d}x=\pi\ln\ln2
\end{equation}
using the following theorem by ...
- 19.2k
30
votes
2
answers
1k
views
Evaluate $\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}$
Question: How can we prove $$I=\int_0^{\pi/4}{(4\cot x\ln\sec x-x)\ln^2\tan xdx}=\frac5{2304}\pi^4?$$
I confirmed, numerically, that it holds for 1000 decimal places. This integral came up when I was ...
- 8,571
25
votes
2
answers
1k
views
How to prove $\int_0^1 \frac{\arctan^2(x)\ln\left(\frac{x}{(1-x)^2}\right)}x \, \mathrm{d}x=G^2$?
A while back I made a post asking for examples of integrals which evaluated to famous irrational constants (or constants that were very likely irrational but yet unproven to be). The top answer in ...
- 6,807
11
votes
4
answers
501
views
Prove that $\int_{0}^{\infty}\frac{(\arctan x)^3}{x^3}dx=\frac{3π}{2}\ln2-\frac{π^3}{16}$
Question: How to prove $$\int_{0}^{\infty}\frac{(\arctan x)^3}{x^3}dx=\frac{3π}{2}\ln2-\frac{π^3}{16}\: ?$$
I was able to prove $$\int_{0}^{\infty}\frac{(\arctan x)^2}{x^2}dx=π\ln2$$
using, $$f(x,y)=...
- 1,412
8
votes
6
answers
669
views
Integral: $\int_{0}^{1}\frac{\arctan^{2}\left(x\right)}{x}dx$
Context: While working on a contour integral for fun, I stumbled upon the following integral:
$$\int_{0}^{1}\frac{\arctan^{2}\left(x\right)}{x}dx.$$
I typed it into WolframAlpha and got that it equals
...
- 4,668
15
votes
2
answers
858
views
Compute polylog of order $3$ at $\frac{1}{2}$
How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{1}{2^nn^3}$$
I am aware this equals polylog of order $3$ at $\frac{1}{2}$ or $\operatorname{Li}_3\left(\frac{1}{2}\right)$, but how ...
- 10.8k
17
votes
3
answers
776
views
A conjectured value for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$
In evaluating the integral given here it would seem that:
$$\operatorname{Re} \operatorname{Li}_4 (1 + i) \stackrel{?}{=} -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{...
- 11.6k
16
votes
1
answer
978
views
Closed-form of $\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx$
Inspired by this question, is there a closed-form of
$$\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx\,?$$
Here $n \in \mathbb{N_+}$. In the answers to the question above we could find proofs of ...
- 12.1k
6
votes
3
answers
611
views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
- 24.6k
5
votes
1
answer
369
views
The identity $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx = 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin(anx) \, \mathrm dx$
Let $p(x)$ be a polynomial, and assume that $ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \, \mathrm dx $ converges.
How do you prove that $$ \int_{a}^{b} p(x) \cot \left(\frac{ax}{2} \right) \...
- 41k
7
votes
2
answers
365
views
How to calculate $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}h_n}{n^3}$ where $h_n=\sum_{k=1}^{n}\frac{1}{2k-1}$?
Well known is the relationship $$\displaystyle\sum_{n=1}^{\infty}\frac{h_n}{n^3}=\operatorname{Li}_4(\frac{1}{2})+\frac{1}{24}\ln^42-\frac{\pi^2}{24}\ln^22+\frac{7}{8}\zeta(3)\ln2-\frac{53\pi^4}{5760}$...
- 5,427