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15h
comment Feynman technique of integration for $\int^\infty_0 \exp\left(\frac{-x^2}{y^2}-y^2\right) dx$
Related to Cortizol's comment and Dr. MV's answer see this thread.
16h
comment Is there any relationship between the Riemann z function and strange attractors?
Two more things you may enjoy : the universality of zeta and the related paper by Woon "Riemann zeta function is a fractal". Fine reading,
1d
awarded  Nice Answer
2d
comment Value of polylogarithms $\mathrm{Li}_s(1)$ for $s<1$.
Yes @gammatester. From the equality for $\Re s >1$ and since $\zeta$ admits a unique analytic continuation over the whole complex plane except $s=1$ we must have the equality for $s\neq 1$ (else $\operatorname{Li}_s(1)$ would not be analytic in $s$ !).
May
20
comment About Mertens' first theorem
Nice to meet you here @draks! An answer would require more work (at least for me) to find something simple, compare the solutions and why not provide a proof... :-) (but I am too busy !). Cheers,
May
20
comment About Mertens' first theorem
Glad it helped @wiskundeliefhebber! I don't know a simple proof of this (which may involve a goot part of the material needed for PNT). Fine continuation,
May
20
comment About Mertens' first theorem
See $(17)$ here and the links concerning $B_3$ for example at OEIS.
May
20
revised Value of polylogarithms $\mathrm{Li}_s(1)$ for $s<1$.
added 80 characters in body
May
20
answered Value of polylogarithms $\mathrm{Li}_s(1)$ for $s<1$.
May
18
comment Transforming 2d coordinate system while keeping points placement
More simply replace $y\to 1-y$.
May
17
comment Is there any relationship between the Riemann z function and strange attractors?
Concerning iterations and fractals you may enjoy this (probably not an answer to your question but neat!)
May
16
awarded  Nice Answer
May
16
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Ant: You may try with this paper by Bailey and Borwein (experimental mathematics is fun !).
May
16
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Ant: anyway I am trying another experimental path and, should you have a generating function for $5/2,17/54,257/2250,8933/154350,309817/8930250,18242179/792410850,\cdots$, then I could give you a (still not proved) formula for $I(a,1)$. Cheers,
May
16
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Ant: Of course without a proof you can't be sure but I could challenge you to disprove them! :-) This should be easy by increasing the precision but the neat thing about the integer relation algorithms is that you will get the same constants by increasing the precision once you crossed the minimal threshold while in my link the constants will change rather randomly with the number of digits (the total count of digits staying near the fixed precision). Of course this could be an illusion with a second threshold much higher and so on (like when playing with $\theta$ functions).
May
15
comment Lambert-W Function Integral
Wikipedia's contribution...
May
15
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Lucian: unfortunately $I(3/4,1)$ has coefficients outside of my table (unknown) so that more work would be required in this case ($I(1/5,1)$ on the other side works but I'll have to let place for more general answers!). Cheers,
May
15
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Lucian: since you propose... Yes! you may update it ! :-) (I'm not sure it works in this case but...)
May
15
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@Ant: that's pslq (or LLL) at work! Compute a value to high precision (say $400$ digits), conjecture a vector of expressions like $[\ln(3)^3, \ln(3)\ln(2)^2,\zeta(3)]$ and so on and the software will use an integer relation algorithm to return you the wished integer coefficients like here.
May
15
comment Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$
@VladimirReshetnikov: I updated my answer. Concerning the more general $I(a,1)$ a generating function for the fractions $5/2,17/54,257/2250,8933/154350,309817/8930250,18242179/792410850$ should allow to conclude (I think...)