I need some help solving numerical integration.
The integral is:
fun = @(x) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* 1 .* x)).^2 ./ sqrt(1 - (0.0003 .* 1 .* x).^2))); Note: asind = sin^-1(....) --> (I am not sure if this is correct)
C = @(T) integral(fun, 0, 517.39 ./ T);
The limit of x is from: x = 0 to x = (500/T)
T ranges from (1 - 1000) (which is T = logspace(0, 3))
Can anyone help me solve this integral. I want to plot a loglog plot of (C as a function of T). I am getting some errors which I am not able to overcome.
Thanks in advance.
Raj.

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Star Strider
Star Strider 2020년 10월 13일

1 개 추천

The integral function will produce one value for the specific limits of integration. To create a vector from its outputs, a loop is necessary.
Try this:
fun = @(x) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* 1 .* x)).^2 ./ sqrt(1 - (0.0003 .* 1 .* x).^2)));
T = logspace(0, 3);
C = @(T) integral(fun, 0, 517.39 ./ T);
for k = 1:numel(T)
Cint(k) = C(T(k));
end
figure
plot(T, Cint)
grid
.

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Raj Patel
Raj Patel 2020년 10월 13일
Oh I am sorry, the function is:
fun = @(x) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* T .* x)).^2 ./ sqrt(1 - (0.0003 .* T .* x).^2)));
Can you see the changes?
Thanks
Raj
My pleasure!
No worries!
The ‘fun’ function now needs to be coded as:
fun = @(x,T) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* T .* x)).^2 ./ sqrt(1 - (0.0003 .* T .* x).^2)));
and ‘C’ now needs to be made compatible with it:
C = @(T) integral(@(x)fun(x,T), 0, 517.39 ./ T);
Then, with those changes, the complete code is now:
fun = @(x,T) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* T .* x)).^2 ./ sqrt(1 - (0.0003 .* T .* x).^2)));
T = logspace(0, 3);
C = @(T) integral(@(x)fun(x,T), 0, 517.39 ./ T);
for k = 1:numel(T)
Cint(k) = C(T(k));
end
figure
plot(T, Cint) % Linear ‘x’
grid
figure
semilogx(T, Cint) % Logarithmic ‘x’
grid
I forgot about the possibility of a logarithmic scale for the x-axis previously. I include it here as an option. .
Raj Patel
Raj Patel 2020년 10월 13일
Thank you so much. Apprecite your effort.
Raj.
Star Strider
Star Strider 2020년 10월 13일
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Raj Patel
Raj Patel 2020년 10월 13일
I wanted to ask that if I have a prefactor of (5 * (1/T)) before the integral, where should I put it? I tried putting it like this, but it is not working.
fun = @(x,T) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* T .* x)).^2 ./ sqrt(1 - (0.0003 .* T .* x).^2)));
T = logspace(0, 3);
C = 2.35 .* 10.^-23 .* (1 ./ T) .* @(T) integral(@(x)fun(x,T), 0, 517.39 ./ T);
Thanks,
Raj.
As always, my pleasure!
The code to do that needs to be something like this:
fun = @(x,T) (x .* (exp(x) ./ (exp(x) - 1).^2) .* ((asind(0.0003 .* T .* x)).^2 ./ sqrt(1 - (0.0003 .* T .* x).^2)));
T = logspace(0, 3);
C = @(T) 2.35 .* 1E-23 .* (1 ./ T) .* integral(@(x)fun(x,T), 0, 517.39 ./ T);
for k = 1:numel(T)
Cint(k) = C(T(k));
end
figure
semilogx(T, Cint) % Logarithmic ‘x’
grid
I changed the ‘C’ function to be compatible with the premultiplication requirement. Note that now ‘C’ is a second anonymous function. An explicit for loop is still the most efficient option here.
Raj Patel
Raj Patel 2020년 10월 13일
Thanks Star.
Raj
Star Strider
Star Strider 2020년 10월 13일
As always, my pleasure!
Hi Star,
I wanted some help with a Matlab coding. I am trying to integrate a function and this function has a variable T which ranges from 1:1000. How do I get numerical integral for all these values of T and then plot it as a function of T? Could you help me? The Matlab code is:
kb = 1.38 .* 10.^-23;
h = 1.05 .* 10.^-34;
wd = 6.94 .* 10.^13;
vs = 6084;
B1 = 2.7 .* 10.^-19;
B2 = 170;
A1 = 2 .* 10.^-45;
D = 0.004;
c = (h.^2 .* D) ./ (2 .* pi * pi * vs * kb .* T .* T);
T = 1:1000;
fun = @(x) ((x.^4 .* exp((h.*x) ./ (kb .* T))) ./ (((exp((h.*x) ./ (kb .* T))-1).^2) .* ((D.*B1 .* x.^2 .* T .* exp(-B2/T)) + (D.*A1 .* x.^4) + vs)));
K = c .* integral(fun,0,wd)
plot(T, K)
Thanks in advance,
Raj Patel.
As always, my pleasure!
I thought we just solvedd that exact problem? I only posted the few lines that changed.
The full code is:
kb = 1.38 .* 10.^-23;
h = 1.05 .* 10.^-34;
wd = 6.94 .* 10.^13;
vs = 6084;
B1 = 2.7 .* 10.^-19;
B2 = 170;
A1 = 2 .* 10.^-45;
D = 0.004;
T = 300;
c = @(T) (h.^2 .* D) ./ (2 .* pi * pi * vs * kb .* T .* T);
fun = @(x,T) ((x.^4 .* exp((h.*x) ./ (kb .* T))) ./ (((exp((h.*x) ./ (kb .* T))-1).^2) .* ((D.*B1 .* x.^2 .* T .* exp(-B2/T)) + (D.*A1 .* x.^4) + vs)));
T = 1:1000;
for k = 1:numel(T)
K(k) = c(T(k)) .* integral(@(x)fun(x,T(k)),0,wd);
end
figure
plot(T, K)
grid
xlabel('T')
ylabel('K')
It produces the same plot, so I will not re-post it.
This also works (without the loop, using 'ArrayValued'):
c = @(T) (h.^2 .* D) ./ (2 .* pi * pi * vs * kb .* T .* T);
fun = @(x,T) ((x.^4 .* exp((h.*x) ./ (kb .* T))) ./ (((exp((h.*x) ./ (kb .* T))-1).^2) .* ((D.*B1 .* x.^2 .* T .* exp(-B2./T)) + (D.*A1 .* x.^4) + vs)));
T = 1:1000;
K = c(T) .* integral(@(x)fun(x,T),0,wd,'ArrayValued',1);
figure
plot(T, K)
grid
xlabel('T')
ylabel('K')
producing the same plot.
The ‘fun’ code changed slightly to make it fully vectorised.
.

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