Numerical Integration in Matlab

1 次查看(过去 30 天)
Hi guys,
I am trying to numerical integrate a function, but I am not able to do it correctly. Can anyone help me here? I think I am not writing the syntax correctly. I have attached the code below.
kb = 1.38 .* 10.^-23;
h = 1.05 .* 10.^-34;
n_puc = 1.81 .* 10.^19;
g = 5.2 .* 10.^-7;
B1 = 4.5 .* 10.^-20;
B2 = 410;
k_max = sqrt(4.* pi .* n_puc);
w_max = g .* k_max .* k_max;
T = 300;
c = (h.^2 ./ (2 .* pi .* kb .* T.^3 .* B1 .* exp(-B2 ./ T)));
% Integral:
fun = @(x) ((x .* exp((h .* x) ./ (kb .* T))) ./ (((exp((h .* x) ./ (kb .* T)) - 1).^2)));
K = c .* integral(fun,0,w_max);
Thanks in advance guys,
Raj.

采纳的回答

Jim Riggs
Jim Riggs 2020-12-1
编辑:Jim Riggs 2020-12-1
There seem to be two problems:
1) your function "fun" returns "NaN" at zero, so you cannot integrate starting from zero. You need to start from some non-zero value
2) I still get an error due to the integral function not being able to satify the tolerance. use the 'Relto' argument to reduce the tolerance, e.g.
xstart = 0.1;
D = integral(fun,xstart,w_max,'Relto',1e-3);
The problem I am seeing now is that the function is basically infinite at zero, therefore the integral value is highly dependent on the starting value. But the smaller you make the starting value, the larger you must make the Reltol in order to get a solution, so the numerical process is inherently limited.

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Mathematics 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by