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Oct
10
comment A 1D numerical integral Mathematica cannot compute, from physics
Usually for what I need $h (\omega)$ is the result of a numeric integral over other variables, so does not have an analytic form. The problem is more complicated (two numeric integrations + an infinite sum to end with...). Modifying the contour choice is great advice, thanks! I'm looking into it right now...
Oct
10
comment A 1D numerical integral Mathematica cannot compute, from physics
J.M. is back: care to elaborate? The usual way in physics textbook to calculate that sum is to convert it to a contour integral. I'm no expert in this field, which other options do I have to numerically calculate an infinite sum? Note that here I chose a very simple $h(\omega)$ usually it is more complicated...
Oct
10
comment A 1D numerical integral Mathematica cannot compute, from physics
Many thanks! I modified the question to add it!
Oct
10
comment A 1D numerical integral Mathematica cannot compute, from physics
I just tried $\beta=10$ and $\beta=20$, the convergence problems are unchanged. They disappear in the low-temperature limit, $\beta=100$, for example.
Oct
10
comment A 1D numerical integral Mathematica cannot compute, from physics
Sorry, I forgot to mention it, for the plots I am using $\beta=0.2$. Physically speaking it is the inverse temperature, so $\beta=0.2$ corresponds to high temperatures.
May
14
comment Converting HeavisideTheta[]s and Sign[]s functions to a single Piecewise[]
Done. Actually it's not a quartic strictly speaking, but while solving the equation in order to find the zeroes, you get a quartic and so you get 4 solutions in the end.
May
14
comment Converting HeavisideTheta[]s and Sign[]s functions to a single Piecewise[]
No, they are not. Some of them are quartic, however it's possibile to find the zeroes of each one analytically.
May
14
comment Is Mathematica really getting this limit wrong?
Thank you very much Daniel. Just one additional question: why is everyone saying that the limit should go unevaluated? Isn't Sign[A] a correct result?
Mar
1
comment Non-linear integral equation
@george2079: it should converge, at least it is eq. 5 in this paper: journals.aps.org/prl/pdf/10.1103/PhysRevLett.74.1633 The authors refer to an unpublished paper for the numerical part, commenting as follows: letting $x=\tan(\beta)$, we set up a Gaussian-quadrature grid for $\beta$ and convert the above equations into a matrix form which can be solved iteratively. The logarithmic singularities are treated separately.
Feb
28
comment Non-linear integral equation
I edited the question to include some more details! Thank you for your links!
Jul
4
comment MathLink and a function of a real variable
Thank you! Accepted!
Jul
4
comment MathLink and a function of a real variable
That's what I need, I thought that N[integer] would remain an integer, but this is not the case. If you want to convert your comment into an answer I will accept it. Thank you!
Nov
5
comment Interpolating a function of two variables
chris: works like a charm! Too bad it's only a comment and I can't accept it... :-)
May
22
comment Finding ranges of a parameter for which a function is always positive
That's exactly what I was looking for! Thank you!
Apr
17
comment Coulomb potential as a Fourier transform
I can't reproduce the result, if I copy your code and paste it into a new Mathematica workbook I get: -(Sinh[m r]/(4 [Pi] r)) however, this is clearly the way to go! :-) Answer accepted!
Apr
17
comment Coulomb potential as a Fourier transform
I'm fine with the logs, but the EulerGamma is completely unexpected!
Apr
17
comment Coulomb potential as a Fourier transform
@F'x In 2D it fails to evaluate the output with the mass term. In 3D it gets stuck for ages thinking and then gives the input as output (not able to solve it, I suppose). I remember that in 2D the mass is not needed for convergence, so I've tried and the output is: 1/2 (-HeavisideTheta[-x] (2 EulerGamma + Log[-x - I y] + Log[-x + I y]) - HeavisideTheta[x] (2 EulerGamma + Log[x - I y] + Log[x + I y])) while I was expecting: $k \ln(r)$