Array of integrals within loop

Hello everyone!

For the last few days, I am stuck with a problem with parametrized integration in GNU Octave. I would like to to evaluate an integral of some f(x) function that contains an exponential constant parameter alpha:

eq1 = @(x) 1 ./ (x.^ 2 + 2 .* x * exp(alpha) .+ 1);
eq2 = integral(f,0,inf)

The problem comes when I want to check and plot how the result will look over the defined range of the parameter alpha, so, I would like to get an array of eq2(alpha).

I have defined the range, k. Let's assume alpha will change from -1.0 to 1.0 by 0.1:

k = -1.0:0.1:1.0;

and then alpha will be a matrix with k-values:

for n = length(k)
    alpha(k) = n;
end

I would expect the outcome in form of matrix containing calculated integrals with diffrent value of alpha(k). Unfortunately, I am not able to put function

eq1(alpha(k)) = @(x) 1 ./ (x.^ 2 .+ x .* exp(alpha(n)) .+ 1)

within the for loop and integrate it as

eq2(k) = int(eq1(k))

I tried to get a symbolical equation and it works, but this form is not satisfying me enough. I also tried to create a function with loop, but also without success. Do you have in mind the solution that could help me construct and evaluate such an array of integrals? Thank you in advance.

[Jan]

Is this a typo?

for n = length(k)
    %   1:length(k)  is required
    alpha(k) = n;
end

But you cannot use the floating point values of the vector k as an index.

There is no .+ operator.

You did not define f in eq2 = integral(f,0,inf) .

Please post some running code. "Does not satisfy me enough" and "but also without success" are not clear enough. We cannot guess, what the problem is.

[Kamil Mickiewicz]

You are right, I am sorry, I forgot to define the range of for loop. I apologize for not putting another line of code.

I was defining a number of iterations as length(alpha) instead of using numel and just put the integral within the loop, so it was looking like this:

k = -1.0:0.1:0.0;


for n = 1:length(k)
    alpha(n) = n;
    
    eq1(n) = @(x) 1 ./ (x.^ 2 + 2 .* x * exp(alpha(n)) + 1); % in this case eq1 should not be dependent on (n).
    eq2(n) = integral(eq1,0,inf);
   
end

eq2

After your comment I understood why it had no chance to work correctly.

Answer 1

[Jan]

alpha = -1.0:0.1:1.0;
for k = 1:numel(alpha)
    f      = @(x) 1 ./ (x.^ 2 + 2 .* x * exp(alpha(k)) + 1);
    eq2(k) = integral(f, 0, inf);
end

[Kamil Mickiewicz]

Dear Jan, thank you so much! I did not consider using the numel function, this is what I was looking for!