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1h
awarded  Nice Answer
7h
reviewed Approve Contour integral of $\frac{1}{\sqrt z}$ with branch cut
7h
answered Contour integral of $\frac{1}{\sqrt z}$ with branch cut
10h
comment About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
@Chris'ssistheartist: I'm surprised I don't see it more. I use it with respect to Fourier transforms because I know a lot of FTs off the top of my head. Parseval is not limited to FTs and so it is an incredibly versatile tool for evaluating integrals and sums.
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
revised About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
deleted 59 characters in body
17h
answered About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
1d
answered How to show that this interesting difference of products is $O \left( \frac{1}{n^2} \right) $
1d
comment Integral from minus infinity to plus infinity
If you're going to derive a result that has already been derived here and probably in about 457 other places within this site, at least have the decency to format the content properly.
1d
comment Integral from minus infinity to plus infinity
Here is an interesting way to show a similar result using the residue theorem: math.stackexchange.com/questions/1266856/…
1d
comment Integral from minus infinity to plus infinity
And further, the "solution" involving the Gamma function is circular. We prove that Gamma function results by using the gaussian integral whose result you are allegedly deriving.
1d
answered Improper rational/trig integral comes out to $\pi/e$
1d
comment Improper rational/trig integral comes out to $\pi/e$
If you are familiar and know what I am talking about, I will be happy to outline a solution using that method. If not, then it will be of no use to you.
1d
revised Improper rational/trig integral comes out to $\pi/e$
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1d
comment Improper rational/trig integral comes out to $\pi/e$
This is easy using complex analysis and the residue theorem. If that is beyond you, then it is a tough one.
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comment Not the toughest integral, not the easiest one
@Zach466920: debates like this are old and useless. If you think Mathematica has rendered these problems an anachronism, then you're not alone; downvote and move on. Otherwise, your (ill-informed and tiresome) commentary merely serves to deface a decent problem. Moreover, I will add that Mathematica and its ilk, while quite useful, can't do many of these integrals and even once in a while get them wrong. And many of us do appreciate the art of integration and rather enjoy these problems. Thus, there is disagreement. A rarity on this site.
2d
comment Question about asymptotic behaviour of argument of complex number
@BasicUser: ceiling
2d
answered Question about asymptotic behaviour of argument of complex number
2d
comment Volume of partially filled spherical cap?
@Rahul: then how on earth is it a duplicate?
Jul
25
comment Sum of the series $\sum\limits_{n=0}^\infty \frac{1}{(3n+1)^3}$
@JackD'Aurizio: This argument is as old as Math.SE. There will always be people that mistake Mathematica/Maple output for actual mathematics ability. The answers speak for themselves. I find that arguing about this in the comments is not worth my time. Just downvote and move on.