integration of (1/x-x)^alpha
4 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Camille
am 11 Jun. 2014
Kommentiert: Camille
am 11 Jun. 2014
Hi,
here is my code
function y = f(r, alpha)
F = @(x) (1./x-x).^alpha;
y = integral(F,0,r);
end
Then if i type
f(1/2,0.2)
i get
ans =
0.707008459146274
but the problems start when alpha gets closer to 1, for example:
f(1/2, 0.8)
Warning: Infinite or Not-a-Number value
encountered.
> In funfun\private\integralCalc>iterateScalarValued at 349
In funfun\private\integralCalc>vadapt at 132
In funfun\private\integralCalc at 75
In integral at 88
In f at 3
ans =
Inf
How can I compute this integral? Thank you for reading
0 Kommentare
Akzeptierte Antwort
Roger Stafford
am 11 Jun. 2014
Yes, you are right, Camille. If alpha < 1, your function is integrable, but the infinity in its integrand at x = 0 is apparently proving to be too difficult for matlab's 'integral' function to evaluate it properly.
However, if you make the following simple change of variable, you should encounter no such difficulty. To simplify notation use 'a' instead of 'alpha'. Define variable z as:
z = x^(1-a)
Then we have
dz = (1-a)/x^a*dx
and
(1/x-x)^a*dx = (1-x^2)^a/x^a*dx = (1-z^(2/(1-a)))^a/(1-a)*dz
Hence you should evaluate the integral
F = @(z) (1-z^(2/(1-a)))^a/(1-a);
y = integral(F,0,r^(1-a));
Matlab should have no trouble with this integral provided a < 1, since it has no singularities in its integrand.
0 Kommentare
Weitere Antworten (1)
Sean de Wolski
am 11 Jun. 2014
1./1-1 yields a 0 and then 0 raised to any negative power is inf. What do you expect at 1?
Siehe auch
Kategorien
Mehr zu Numerical Integration and Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!