No. You should NOT make it inf. Even 1e24 is incredibly large, wild overkill.
As to why you are using integral to compute the area under what loos like a normal PDF is completely beyond me.
d1 = 8;
d2 = 1.4142;
apPDF = @(x1) (1./sqrt(2*pi)) .* exp(-power((d2.*x1./sqrt(2)) - d1,2) ./ d2^2);
You need to understand what integral does when it sees a function with limits that wide. It evaluates the function at a variety of points in the interval. Lets try a few, just for kicks.
apPDF(0)
ans =
5.04917109360107e-15
>> apPDF(1e24)
ans =
0
>> apPDF(1e24 / 2)
ans =
0
>> apPDF(1e24 / 100000)
ans =
0
>> apPDF(1e24 / 10000000000000)
ans =
0
Do you see anything significant? Do you see a function that seems to be everywhere zero on the interval [0,1e24]? And even when not identically zero, it deviats from zero on the order of the convergence tolerance. Should it somehow, magically know that in effectively a tiny corner of that HUGE interval, it is non-zero?
apPDF(5)
ans =
0.00443082846737379
So now if we do this:
integral(apPDF,0,100)
ans =
1
What a surprise! It integrates to 1.
