Can't integrate function using Matlab

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David Harra
David Harra 2023-2-9
评论： David Harra 2023-2-9
I am trying to do a basic integration of a formula for 4 different values. After doing a bit of reading, apparently I need to create a function handle but I am not getting any success. I have put my attempt below. I might be missing more though.
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
% eta_r = P.*exp(-r./L);
% I want to integrate eta from zero to infinity where P has changing
% variables to give 4 different plots.
r = 0:100;
% I was just testing below to see if a loop would run before attempting to
% plot anything
P = zeros(1,length(L));
fun = @(r) P.*exp(-r./L);
for ii=1:numel(sigma_d)
P= 1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2)) ;
eta_r = integral(fun,0,inf);
end

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采纳的回答

Torsten
Torsten 2023-2-9
Ran in:
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
r = 0:100;
for ii=1:numel(sigma_d)
P= @(L)1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(L,r) P(L).*exp(-r./L);
eta_r(ii,:)=integral(@(L)fun(L,r),0,Inf,'ArrayValued',true);
end
plot(r,eta_r)
grid on
  4 个评论
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Dyuman Joshi
Dyuman Joshi 2023-2-9
Ran in:
You will have to apply that individually -
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
r = 0:100;
LineTypes={'-' '--' ':' '-.'};
for ii=1:numel(sigma_d)
P= @(L)1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(L) P(L).*exp(-r./L);
eta_r(ii,:)=integral(@(L)fun(L),0,Inf,'ArrayValued',true);
plot(r,eta_r(ii,:),'LineStyle', LineTypes{ii}, 'LineWidth', 1.5)
hold on
end
xlabel('r (\mu m)')
xticks([0 20 40 60 80 100])
ylabel('\eta(r)');
ylim([0 1])
title('Spatial Correlation Function LBAR=25')
set(gca, 'FontSize',10);
legend('\sigma_d =0.25', '\sigma_d= 0.5', '\sigma_d =0.75', '\sigma_d = 1')

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更多回答（1 个）

Dyuman Joshi
Dyuman Joshi 2023-2-9
Ran in:
You will have to explicitly change the function handle to change its definition
P=pi;
fun = @(x) P*x;
P=3;
%fun(3) is not equal to 3*3=9
fun(3)
ans = 9.4248
L=0:100;
L_bar=25;
sigma_d =[0.25, 0.5, 0.75, 1.0];
mu = log(L_bar) - 0.5*(sigma_d).^2;
L_tilda = exp(mu);
% eta_r = P.*exp(-r./L);
% I want to integrate eta from zero to infinity where P has changing
% variables to give 4 different plots.
r = 0:100;
% I was just testing below to see if a loop would run before attempting to
% plot anything
for ii=1:numel(sigma_d)
P= 1./(L.*sqrt(2*pi)*sigma_d(ii)).*exp(-log(L/L_tilda(ii)).^2/(2*sigma_d(ii)^2));
fun = @(x) P.*exp(-x./L);
eta_r=integral(fun,0,Inf,'ArrayValued',true)
end
Warning: Inf or NaN value encountered.
eta_r = 1×101
NaN 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0006 0.0030 0.0109 0.0307 0.0718 0.1437 0.2535 0.4022 0.5842 0.7870 0.9943 1.1886 1.3546 1.4810 1.5614 1.5946 1.5834 1.5336 1.4531 1.3500 1.2326
Warning: Inf or NaN value encountered.
eta_r = 1×101
NaN 0.0000 0.0000 0.0003 0.0023 0.0097 0.0269 0.0572 0.1019 0.1598 0.2281 0.3028 0.3800 0.4560 0.5276 0.5924 0.6491 0.6965 0.7345 0.7630 0.7827 0.7940 0.7979 0.7951 0.7867 0.7733 0.7560 0.7354 0.7122 0.6871
Warning: Inf or NaN value encountered.
eta_r = 1×101
NaN 0.0002 0.0060 0.0263 0.0626 0.1109 0.1656 0.2219 0.2764 0.3268 0.3717 0.4106 0.4433 0.4701 0.4914 0.5076 0.5192 0.5268 0.5309 0.5319 0.5303 0.5265 0.5209 0.5137 0.5053 0.4958 0.4855 0.4746 0.4632 0.4514
Warning: Inf or NaN value encountered.
eta_r = 1×101
NaN 0.0099 0.0513 0.1074 0.1642 0.2156 0.2596 0.2959 0.3252 0.3482 0.3658 0.3789 0.3882 0.3942 0.3977 0.3989 0.3984 0.3963 0.3931 0.3889 0.3839 0.3783 0.3722 0.3658 0.3590 0.3521 0.3450 0.3378 0.3305 0.3233
Since you are dividing by 0 (first element of L), you get a NaN and thus the warning from the integral() solver.
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Dyuman Joshi
Dyuman Joshi 2023-2-9
"I need to be integrating with respect to L"
You should have mentioned that before. By the definition of the function handle, I took it as you are integrating w.r.t r
Now as you want to integrate w.r.t L, is there a need to define L=0:100?
Or L is a variable in the definition of P as well?
David Harra
David Harra 2023-2-9
Aplogies
So my definition of P is
P= 1./(L.*sqrt(2*pi)*sigma_d).*exp(-log(L/L_tilda).^2/(2*sigma_d^2))
So L is definied within P and L defined as a length. I think I could maybe make this a fixed constant of L=100 and keep r=0:100
The integral I am trying to calculate is
I am changing values of L_tilda and sigma 4 times to get 4 different plots.
Hopefully this is more clear :)

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