# Search Results

Results tagged with Search options user 16314
9 results

Questions related to the calculus and analysis branches of Mathematica, including, but not limited to, limits, derivatives, integrals, series, and residues.

Another approach if you want to avoid the use of AsymptoticIntegrate (whose very presence I learnt today, thanks @Roman:-)!). Timing[ FullSimplify[ Integrate[(y (1 - x)^2 (1 + x))/(x + (1 - x^2) y …
answered May 1 by Dimitris
I think this is not a answer but can serve as a summary of the useful workarounds. Still we do not have a rigorous explanation of the buggy (?) behavior. Since Integrate[BesselJ[0, x], {x, 0, \[Infi …
answered Oct 16 '15 by Dimitris
I don't see any any particular reason why the integral should be real. chi = 62/27; phi[x_] := x Log[x] + (1 - x) Log[1 - x] + chi x (1 - x) integrand = phi[x] Log[phi[x]] + (1 - phi[x]) Log[1 - …
answered Feb 27 '17 by Dimitris
Just a workaround. In:= Clear["Global*"] In:= f[x_, t_] := Abs[Re[Exp[I*x]/(1 - t*Exp[I*x])]] In:= Timing[ resAn[t_] = Integrate[f[x, EulerGamma], {x, 0, 2*Pi}] /. EulerGamma -> t] O …
answered Nov 18 '14 by Dimitris
(Mathematica 10) Numerically: Needs["NumericalCalculus"] NLimit[Power[Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2, (n)^-1], n -> ∞] // Chop // Rationalize[#, .000001] & (* 64/27 *) Symbolically Lim …
answered Apr 8 '17 by Dimitris
If you are sure about your integral's behavior you can try Integrate[Exp[I z] 1/z, {z, -Infinity, Infinity}, GenerateConditions -> False] (* I π *)
answered Feb 27 by Dimitris
I guess this is a problem of version 9.0. In Mathematica 10 I get the correct result: expr = -((E^(2 n x μ) (-1 + Gamma[2, n x μ]) (λR^2))/((λR^2 + (E^(n x μ) ((n μ - λR^2)))))^2) (n μ E^( …
answered Feb 21 '17 by Dimitris