Simpsons Rule to Numerical integrate a function (Lorentzian Function)

Question

AnonAstroZ 2021-1-24

0
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评论： AnonAstroZ 2021-1-24

Hi all, i'm trying to prove the Lorentzian Profile is Unit Normalized (i.e = 1) VIA Simpsons Rule of Integration. The constants are the given parameters

Here is my functon/code:

function s=simprl(f,a,b,M)
h=(b-a)/(2*M);
s1=0;
s2=0;
for k=1:M
   x=a+h*(2*k-1);
   s1=s1+f(x);
end
for k=1:(M-1)
   x=a+h*2*k;
   s2=s2+f(x);
end
s=h*(f(a)+f(b)+4*s1+2*s2)/3;

here is my function that i am calling with the given parameters entered in, it should equal to 1

simprl(@(x) (1./pi).*((5e8)./2)./(x-4.5667e14).^2 + ((4.5667e14)./2).^2,-Inf,Inf,2)

It should equal to 1, but instead it is giving me a "NaN" answer. What is wrong?

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Answer 1

Alan Stevens 2021-1-24

1
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With an Inf limit you divide by Inf in simprl, resulting in a NaN. Use large, but finite limits.
You havem't writtn your Lorentzian correctly. The denominator should have the second term as (5e8/2)^2 not (4.5667e14/2)^2.
Simpson's rule with 4 panels is just too coarse here. Compare with Matlab's inbuilt integral function, which gets the right result.

The initial coding is clearer as:

Gmma = 5e8;
x0 = 4.5667e14;
lo = x0 - 100*x0; hi = x0 + 100*x0;
f = @(x) 1/pi.*(Gmma/2)./((x-x0).^2 + (Gmma/2).^2);
a=simprl(f,lo,hi,2);
disp(a)
% Compare with Matlab's inbuilt integral
mlab = integral(f,lo,hi);
disp(mlab)