integral method gives an answer of zero
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Stashu Kozlowski
il 2 Apr 2023
Risposto: Walter Roberson
il 2 Apr 2023
Hello,
I have been trying compute
where β is a fixed constant. My code consists of the following:
k = 1.380649*10^(-23);
t = 5800;
beta = (k*t)^(-1);
fun = @(x) x.^3 ./ (exp(beta .* x) - 1);
q = integral(fun,0,Inf)
This results in the integral being evaluate as 0. However, I know that this integral is infact not zero over the domain
Am I making somekind of mistake?
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Walter Roberson
il 2 Apr 2023
The values are so small relative to your constants that they might as well be zero.
With your beta being 1/(k*t) you are ending up with some large exponents in the exp() -- large enough that you are going to overflow numerically.
syms x
Q = @(v) sym(v);
k = Q(1380649)*Q(10)^(-29);
t = Q(5800);
beta = (k*t)^(-1);
fun = x.^3 ./ (exp(beta .* x) - 1)
int(fun, 0, Inf)
vpa(ans)
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