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

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David
David 2015-1-20
评论： john zhang 2020-8-12
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?

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采纳的回答

Torsten
Torsten 2015-1-20
Your function is equal to
sqrt(pi)/2*(1+erf(x))
Best wishes
Torsten.

更多回答（4 个）

Alessandro Masullo
Alessandro Masullo 2015-1-20
You can use cumsum or cumtrapz.
David
David 2015-1-20
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?
David
David 2015-1-20
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
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David
David 2015-1-21
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
John D'Errico 2015-1-21
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.

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David
David 2015-1-21
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

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