Why can't I get this integration to work?
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Kendra Hamilton
il 6 Ott 2023
Commentato: Kendra Hamilton
il 6 Ott 2023
Hi, I'm trying to integrate the function I(z) from -infinity to infinity. This is my code:
syms P w0 r n lamda z
Pi = sym(pi);
zR = Pi*w0^2*n/lamda;
I(z) = 2*P*zR^2/(Pi*w0^2)*1/(zR^2+z^2)*exp(-2*r^2*zR^2/(w0^2*zR^2+w0^2*z^2));
I1 = int(I, z, -inf, inf)
but it just prints
I1 = int((2*P*n^2*w0^2*pi*exp(-(2*n^2*r^2*w0^4*pi^2)/(lamda^2*(w0^2*z^2 + (n^2*w0^6*pi^2)/lamda^2))))/(lamda^2*(z^2 + (n^2*w0^4*pi^2)/lamda^2)), z, -Inf, Inf)
Can someone tell me what I'm doing wrong?
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Walter Roberson
il 6 Ott 2023
Spostato: Walter Roberson
il 6 Ott 2023
syms P w0 r n lamda z real
Pi = sym(pi);
zR = Pi*w0^2*n/lamda;
I(z) = 2*P*zR^2/(Pi*w0^2)*1/(zR^2+z^2)*exp(-2*r^2*zR^2/(w0^2*zR^2+w0^2*z^2));
IE = expand(I)
I1 = int(IE, z, -inf, inf)
Now look at the form of that integral. Can we find a simpler integral that has the same general form but still cannot be integrated? Yes.
syms B C x
F = exp(-B/(x^2+C))
int(F, x, -inf, inf)
A key point for this is that if you hardcode B as 1 then MATLAB can proceed, and likewise if you hard-code C as 0. But the combination of the two is more than MATLAB can deal with.
Maple is not able to deal with this integral either
Your actual integral is worse than this.
Your basic problem, then, is that there is no closed form integral for that expression that can be found by MATLAB (or Maple either)
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