Using the integral function for large upper limit

9 次查看（过去 30 天）

will steel 2022-1-8

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1625140-using-the-integral-function-for-large-upper-limit

评论： Torsten 2022-1-10

how do I know if integral(fun,x0,inf) is being accurate or not? Ive noticed that replacing inf by a very large number (given i know the fun decays sharply with x) gives a different and smaller answer? Why is this the case and what kind of magic is matlab doing when I give it an infinite limit?

Also, I notice a rescaling of my variables gives a different answer still. In SI units the lower bound x0 is of order 10^15. I have rescaled my units such that x0 is of order 1. Each gives a very different answer (yes i have accounted for converting back the units)...which can i trust?

8 个评论
显示 6更早的评论隐藏 6更早的评论

David Goodmanson 2022-1-9

在 MATLAB Online 中打开

Hi will,

this all can be reduced to two specific situations with the integral function. Your integral falls off like 1 / ( x*exp(x) ), but x varies slowly enough that the effects show up with just the use of exp(-x).

[1] inf replaced by a large number

fun = @(x) exp(-x);
integral(fun,1,inf)
    ans = 0.3679        % 1/e, correct   
integral(fun,1,1e7)
    ans = 0.3679        % correct
integral(fun,1,1e8)
    ans = 1.8175e-22

I have wondered in the past why that is.

[2] Scaling the variable of integration. Should still give 1/e. Note that at the lower limit of integration, the argument of exp is always -1, so it's not like exp is immediately forced into overflow or underflow..

fun = @(x,A) A*exp(-A*x);
A = 1;    integral(@(x) fun(x,A),1/A,inf)
    ans = 0.3679        % correct
other values of A:
A = 1e-13
    ans = 0.3679        % correct
A = 1e-14
    Warning: Minimum step size reached near x = 1e+14, etc.
    ans = 1.0972e-07
A = 1e8
    ans = 0.3679        % correct
A = 1e9
    ans = 1.6570e-77