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Jan
20
revised Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$
I added a necessary restriction.
Jan
17
revised How to find PV $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$
I reorganized my answer and corrected some typos.
Jan
16
revised Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
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Jan
11
comment Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
Thank you. And I wasn't accusing you of copying a part of my answer.
Jan
11
comment Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
I knew the reason for introducing the $-1+a$. I evaluated an integral on here doing something similar. I just wanted to show that you actually didn't need to change the function in this particular case to use that contour. My only gripe now is that the second half of the first section of your answer is now basically just a restatement of the first part of my answer (which was posted 7 hours earlier). I linked to a question in which I prove the lemma I stated. And Daniel Fischer proves it in the comment section of that question as well.
Jan
11
comment Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
But why is that relevant here? We're dealing with a specific function and a specific contour that satisfy certain conditions. That was the point of my answer, namely to explain that those conditions were satisfied and because of that changing the function unnecessary. Your answer states explicitly that we cannot consider $f(z)$ and that doughnut contour because the integral along the small semicircle about the origin blows up as the semicircle shrinks. But that's not true.
Jan
11
comment Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
I'm almost certain that the integral using $f(z)$ does not blow up along the small semicircle about the origin as the semicircle shrinks. It would if the Laurent expansion of $f(z)$ at the origin had $\frac{1}{z^{2n}}$ terms, but it doesn't. For geometrical reasons the integral remains finite. That was the point of my answer. Furthermore, $\text{PV} \int_{-\infty}^{\infty} \frac{1}{x^{3}} \ dx = 0$. But your statement implies that the Cauchy principal value of that integral is not finite.
Jan
11
revised Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
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Jan
11
revised Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
I edited the title to make it more clear, removed 2 tags, and added 2 tags.
Jan
11
answered Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
Jan
9
comment How to prove $\int_0^1 \frac{\arctan( a x)}{x \sqrt{1 - x^2}} dx = \frac{\pi}{2} \ln (a + \sqrt{1 + a^2})$?
Almost a duplicate: math.stackexchange.com/questions/834205/… Felix Marin's answer does address your question, though.
Jan
7
revised How to evaluate the following integral involving hyperbolic functions?
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Jan
7
revised How to evaluate the following integral involving hyperbolic functions?
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Jan
5
comment How to evaluate $\int_0^1 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}} dx$?
math.stackexchange.com/questions/805893/…
Jan
5
comment Evaluating $\int_{-\infty}^{\infty}\frac{\sin ax-a \sin x}{x^3(x^2+1)} \ dx$ using contour integration
What I stated previously was not correct. I was looking at just a particular value of $a$. In the first case, $\lim_{r \to 0} \int_{\pi}^{0} g(re^{it}) \ ir e^{it} dt = -\pi i \ \text{Res}[g(z),0]$ since $g(z)$ has a simple pole at the origin. But it can also be shown by expanding $f(z)$ in a Laurent series at the origin and integrating termwise that $\lim_{r \to 0} \int_{\pi}^{0} f(re^{it}) \ ir e^{it} dt = - \pi i \ \text{Res} [f(z),0] $ even though $f(z)$ has a pole of order $3$ at the origin. And $$\text{Im} \ g(x) = \text{Im} \ f(x) = \frac{\sin (ax) - a \sin(x)}{x^{3}(x^{2}+1)}.$$
Jan
4
comment Proving that $\sum_{n=2}^{\infty }\frac{\zeta (n)}{2^{n-1}}=\log(4)$
$\sum_{n=1}^{\infty} \zeta(n+1) x^{n} = -\gamma - \psi(1-x)$
Jan
2
revised Integral $I=\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx$
corrected typos and added an explanation about why the integral vanishes
Jan
1
revised Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
added links and better explained why the integral vanishes
Dec
31
revised How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$?
added an explanation
Dec
31
revised How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$?
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