Integration

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Bill Luckow
Bill Luckow 2012-1-11
Does matlab have a function of Integration? For example, I have a Gaussian density function, if I use the integration function on the density function over the time of (-infinity, +infinity), then we should obtain one. Thank you very much

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回答（2 个）

Walter Roberson
Walter Roberson 2012-1-11
MATLAB has several numeric integration routines, such as quad(), quadv(), quadgk(), triplequad()
The Symbolic Toolbox offers symbolic integration.
Mike Hosea
Mike Hosea 2012-1-11
QUADGK can handle infinite limits.
>> quadgk(@(x)exp(-x.^2./2)./sqrt(2*pi),-inf,inf)
ans =
1.000000000000038
The default tolerances used there were, BTW, AbsTol = 1e-10 and RelTol = 1e-6. Since the numerical answer is about 1, and RelTol > AbsTol, only the RelTol was important. We can crank down both tolerances
>> quadgk(@(x)exp(-x.^2./2)./sqrt(2*pi),-inf,inf,'AbsTol',5e-14,'RelTol',5e-14)
ans =
1
Getting 1 exactly in that case was just luck. Evaluating CDFs is one example where pure absolute error control might make sense to you:
>> quadgk(@(x)exp(-x.^2./2)./sqrt(2*pi),-inf,inf,'AbsTol',1e-10,'RelTol',0)
ans =
1.000000000000000
If you have a very small variance, the integrator might not see the non-zero part of the curve.
>> sigma = 1e-10; f = @(x)exp(-x.^2./(2*sigma))./sqrt(2*pi*sigma)
f =
@(x)exp(-x.^2./(2*sigma))./sqrt(2*pi*sigma)
>> quadgk(f,-inf,inf,'AbsTol',1e-10,'RelTol',0)
ans =
0
This is because the function was zero everywhere the integrator "looked". If you know you have a problem like that, you can cope with it in one of two ways. You can split the interval where the action is:
>> quadgk(f,-inf,0,'AbsTol',1e-10,'RelTol',0)+quadgk(f,0,inf,'AbsTol',1e-10,'RelTol',0)
ans =
1
or you can toss in some waypoints to force the integrator to use a finer mesh where you believe the action is:
>> quadgk(f,-inf,inf,'AbsTol',1e-10,'RelTol',0,'Waypoints',linspace(-1e-3,1e-3,101))
ans =
1.000000000000000
although I have to say that using Waypoints for this purpose may require some trial and error. Obviously for such a function there is hardly any need to integrate from -inf to inf because the integrand evaluates to exactly zero for abs(x) => 4e-4 or so.
>> quadgk(f,-4e-4,4e-4,'AbsTol',1e-10,'RelTol',0)
ans =
1

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