Solving an integral with variable as upper limit and plotting the solution

How can I solve an integral when the upper limit is a variable, and then plot the solution? My integral is as follows: Integral of e^(-(y^2))dy and my upper limit is x, while the lower limit is negative infinity. X and y are variables. I have tried using the trapz function, but then you must have a constant as upper limit. I have also tried using the int-funtion, but that is only solving for an indefinite integral. Does anyone know what I should do?

Answer 1

[Torsten]

Your function is equal to

sqrt(pi)/2*(1+erf(x))

Best wishes

Torsten.

[john zhang]

what if the function is very complex and has no general antiderivative? Thank you

Answer 2

[Alessandro Masullo]

You can use cumsum or cumtrapz.

Answer 3

[David]

Ok, thank you both for your answers. But I do not understand how I should write when I want to use a variable as the upper limit. How is that possible?

Answer 4

[David]

I have now tried to use the while loop for some time and it looks like this:

upper_limit=-1;

while upper_limit<1

    upper_limit=upper_limit+0.1;
    theta=[-10^5:upper_limit];
    y=(pi^(-(0.5)))*exp(-(theta.^2));
    z=trapz(theta,y);
end

The problem now is that I only get the final answer to be z. I am not sure if this is correct, but is it possible to get all the answers into a vector som that it can be used in the plotting?

David

[Torsten]

Why don't you use the simple formula I provided ?

Look at the definition of the error function erf(x) in MATLAB ; all your attempts for integration are already implemented in this function.

Best wishes

Torsten.

[Alessandro Masullo]

    upper_limit=upper_limit+0.1;
    theta=[-10^5:upper_limit];
    y=(pi^(-(0.5)))*exp(-(theta.^2));
    z=cumtrapz(theta,y);

[David]

Thank you very much for new anwers. I have now tried to use the erf function for some time, but I have a problem. My error function had limits from negative infinity to x. X=[-0.5:0.5]. But the erf function has a lower boundary equal to 0. How is it possible to find the value when the lower boundary is negative infinity?

David

[John D'Errico]

You could just use basic calculus.

erf gives you the integral from 0 to x. If you want the integral from -inf to x, then add 1. This is because the integral from -inf to 0 is 1.

erf(-inf)
ans =
    -1

(Beware of the order of those limits, which gives me here -1.)

So the integral from -inf to x is

1 + erf(x)

Which is valid for any real x as your upper limit.

Answer 5

[David]

Thank you very much all of you! Your help is very much appreciated. Now I understand how it can be computed in matlab, and how to find the integral from negative infinity to zero.

David