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2h
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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7h
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. :-)
7h
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.
7h
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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14h
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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15h
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! :-)
16h
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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17h
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
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17h
asked About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
1d
comment Not the toughest integral, not the easiest one
Great approach! A very simple and powerful substitution (+1).
1d
accepted Not the toughest integral, not the easiest one
1d
comment Not the toughest integral, not the easiest one
Very nice attack! (+1)
1d
comment Not the toughest integral, not the easiest one
Can we derive now from the answer that $I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$ $=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$ ?
1d
comment Not the toughest integral, not the easiest one
@Zach466920 well, I think it's about different visions of the same thing, math. Maybe I wasn't clear enough, but my point is that I'm interested in the art of the mathematics, that is I'm interested in the art of the way to the exactness. One can use Mathematica and easily finish the story of the exactness, but the clever way to the solution is what I'm interested in. It's also the fact that some particular cases of a generalization might allow clever solutions that might not be caught be the general solutions proposed.
1d
comment Not the toughest integral, not the easiest one
@Zach466920 I wouldn't debate the difficulty level of these integrals, but on MSE are posted also integrals at research level that is over the competition level. This post is more about the art of mathematics, that is we don't just come up with a solution for the sake of having a solution, but we want more than that (in this case, an elegant solution, as required).
1d
comment Not the toughest integral, not the easiest one
@Zach466920 I don't think that appealing to the collective brilliance of the MSE team for a neat solution is a wrong thing especially when you see that Mathematica shows you that $\displaystyle \int \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$ $=\frac{2 \sqrt{x} \sqrt{x+1} \left(\text{Li}_2\left(e^{-2 \sinh ^{-1}\left(\sqrt{x}\right)}\right)+\log (x) \log \left(\sqrt{x}+\sqrt{x+1}\right)-\sinh ^{-1}\left(\sqrt{x}\right)^2-2 \sinh ^{-1}\left(\sqrt{x}\right) \log \left(1-e^{-2 \sinh ^{-1}\left(\sqrt{x}\right)}\right)\right)}{\sqrt{x (x+1)}}$
1d
asked Not the toughest integral, not the easiest one
1d
comment Calculating in closed form $\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$
@VladimirReshetnikov Thanks! Indeed, it's a way that works.
2d
revised Calculating in closed form $\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$
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2d
comment Calculating in closed form $\int_0^{\infty} \frac{\text{PolyLog}^{(1,0)}(1,-x)}{1+x^2} \, dx$
@JacquesCarette Well, I could have used the notation $\operatorname{Li}_s^{(n,0)}(x)$ (where $n$ is the $n$th derivative with respect to $s$), and for the second part, it's about generalized Stieltjes constants (where for this case we consider the proper partial derivative), see here en.wikipedia.org/wiki/Stieltjes_constants.