integration of a piecewise function

3 views (last 30 days)
Richard
Richard on 11 Sep 2017
Answered: Sarah Mohamed on 14 Sep 2017
Can anyone help me understand why
integral(@(mu) (1.0+0.0*mu).*(mu>0)+(1-exp(1.0./mu)).*(mu<0),0,1)
gives a "NaN", while
integral(@(mu) (1.0+0.0*mu).*(mu>0)+(1-(1.0./mu)).*(mu<0),0,1)
gives the correct answer "1.0"?
It's bugging me for so long.
Background, I started with a two variate function
psi_MMS =@(x,mu) (1.0+0.0*x+0.0*mu).*(mu>0) + (1-exp((Tau-x)./(mu+1e-16))).*(mu<0);
And I wanted to integrate it over mu to get a function of x:
phi0_MMS =@(x) integral(@(mu) psi_MMS(x,mu), -1,1);
Thanks!

Answers (1)

Sarah Mohamed
Sarah Mohamed on 14 Sep 2017
It looks like NaN values are generated in the first function when you try to multiply -INF and logical 0’s.
For ease of demonstration, let me go ahead and create a handle for the first function as follows:
f1 = @(mu) (1.0+0.0*mu).*(mu>0)+(1-exp(1.0./mu)).*(mu<0);
Let me also go ahead and create a range of values for “mu”, 0.1, 0.01, 0.001, etc, that are close to 0:
mu = 10.^(-1*[1:10]);
If I plug these values into f1, I get the following:
>> f1(mu)
ans =
1 1 NaN NaN NaN NaN NaN NaN NaN NaN
Let me try to localize the problem slightly more:
>> (1-exp(1.0./mu))
ans =
1.0e+43 *
-0.0000 -2.6881 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf
At mu = 0.001 and onwards, we find that "exp(1/mu)" evaluates to INF.
Let me now add in the piecewise part of the expression. “mu<0” should give a vector of logical 0’s:
>> (1-exp(1.0./mu)).*(mu<0)
ans =
0 0 NaN NaN NaN NaN NaN NaN NaN NaN
In summary, multiplying INF or -INF by a logical 0 (or just zero, for that matter) evaluates to NaN. As a quick check, we can confirm this quickly via:
>> inf*logical(0)
ans =
NaN
And as a final sanity check, let's try the original integration, but setting the lower limit to 0.01:
>> integral(f1, 0.01, 1)
ans =
0.9900

Categories

Find more on Multidimensional Arrays in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!