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15h
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
@Did it's not only about supporting the classical ideas for approaching a certain class of integrals, series and limits, but it's about a revolution of ideas, I mean we don't need to be stuck on the same classical style, but happily greet the new attacking ideas even if they are not in a complete form as expected.
15h
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
@Did Still, the answer that was previously chosen has its incontestable value, and I don't try to defend its weak points of the proofs, perhaps many proofs here have such points, but what is important is that it produces new ideas, it makes you think of new ways of approaching the problem. I won't choose a mistake on MSE, but even a mistake has its incontestable values, it might simply push you on a fruitful direction, and the experience may confirm that. In this case it's not about mistakes, but about a deeper approach that most of the users won't easily understand without further support.
15h
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
Cute approach (+1)
15h
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
Nice way to go (+1)
1d
accepted Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
1d
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
@Did Indeed, some steps need explanations but this fact doesn't mean it cannot be an accepted answer. Anyway, I'll choose another answer to avoid any possible misleading (for others).
1d
awarded  Nice Question
1d
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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1d
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
By the way, I don't think I saw anyone on this site showing the exceptional power of Parseval tool more than you. Perhaps it's one of your favourite tools. :-)
1d
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
(+1) for the powerful tool that is good to know and that leads to an elegant solution. However, the interesting thing is getting the answer by real analysis exclusively, not using not even complex numbers. I'm curious if in the math literature there is any tool like that mentioned.
1d
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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2d
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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2d
comment Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $
Great! A big (+1)! Looking forward to the final form of solution! :-)
2d
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
edited tags
2d
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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2d
asked About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
2d
comment Not the toughest integral, not the easiest one
Great approach! A very simple and powerful substitution (+1).
2d
accepted Not the toughest integral, not the easiest one
2d
comment Not the toughest integral, not the easiest one
Very nice attack! (+1)
Jul
28
comment Not the toughest integral, not the easiest one
Can we derive now from the answer that $I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$ $=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$ ?