# Help in integration in MATLAB?

4 vues (au cours des 30 derniers jours)
Ali Almakhmari le 26 Oct 2023
Commenté : John D'Errico le 26 Oct 2023
syms w
L = (5.*(i.*w).^5 - 170.*(i.*w).^3 + 1125.*(i.*w))./(5.*(i.*w).^7 + 7.5.*(i.*w).^6 - 170.*(i.*w).^5 - 255.*(i.*w).^4 + 1125.*(i.*w).^3 + 1688.*(i.*w).^2);
S = 1./(1 + L);
int(log(abs(S)),[0 inf])
I need to integrate the log of the absolute value of S (w being the variable) from 0 to infinity. But MATLAB cannot do it, anyways to approximate this integral in MATLAB or any other approaches?
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Réponse acceptée

Star Strider le 26 Oct 2023
If you want a numeric result, use the vpaintegral function —
syms w
L = (5.*(i.*w).^5 - 170.*(i.*w).^3 + 1125.*(i.*w))./(5.*(i.*w).^7 + 7.5.*(i.*w).^6 - 170.*(i.*w).^5 - 255.*(i.*w).^4 + 1125.*(i.*w).^3 + 1688.*(i.*w).^2);
S = 1./(1 + L);
LaS = simplify(log(abs(S)), 500);
LaS =
IntLaS = int(LaS,[0 Inf])
IntLaS =
IntS = vpaintegral(log(abs(S)),[0 inf])
IntS =
0.0000411287
It probably does not have an analytic (symbolic) solution.
.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
John D'Errico le 26 Oct 2023
Probably? That seems a mild understatement. :)

Connectez-vous pour commenter.

### Catégories

En savoir plus sur Calculus dans Help Center et File Exchange

R2022b

### Community Treasure Hunt

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

Start Hunting!

Translated by