infinite loop and the term within loop again contains infinity

I want to solve a loop given below

prev_val=0; 
  for m=0:inf
    curr_val=(beta/(2*m+1)*pi)*sin((2*m+1)*pi/beta)*exp((2*m+1)^2)*alpha;
    prev_val=prev_val+curr_val;
  endfor
gt=const*prev_val;

I am facing two problems with this. In this, the upper limit of the loop, as per the equation to be solved, is infinity. In addition, the value of beta is again infinity. How can I solve this problem?

[Rik]

Do you have access to the Symbolic Computing toolbox?

[math seeker]

no.

[Walter Roberson]

When you say that beta is infinity, do you mean that you are executing one time wanting the limit as beta approaches infinity? Or do you mean that you need the sum or integral of all of these gt values for some range of beta that includes infinity?

[math seeker]

I mean beta is a constant and its value is infinity.

[Walter Roberson]

In that case, as I posted, the sum over m is sign(alpha) times infinity

Answer 1

[Rik]

I found an answer to your question on stackoverflow: if you don't have the symbolic toolbox, quadgk can still solve your integral.

f = @(x) x.*exp(-x);
a = quadgk(f, 0, inf) 

a =
    1.0000

How you can deal with beta being infinity is something I don't have any ideas for. Maybe you can try with just a big number, maybe enlarging it in a while loop to simulate the limit function.

f=@(m) (beta/(2*m+1)*pi)*sin((2*m+1)*pi/beta)*exp((2*m+1)^2)*alpha;

[Ameer Hamza]

You can also use integral(). The function name integral() appears to be more intuitive as compared to quadgk() as himself mentioned by Mathwork cofounder.

Answer 2

[Walter Roberson]

For the case of limit as beta approaches infinity, then under the assumption that m is a non-negative integer, the individual terms are

Pi^2*exp(4*m^2+4*m+1)*alpha

The sum of this over m = 0 to infinity is sign(alpha) * infinity