sin(ax) is bounded.Magnawolf said:Homework Statement
∫e-Sxsin(ax) dx, S and A are constants, upper limit is ∞ lower is 0
Homework Equations
∫ u dv = uv - ∫ vdu
The Attempt at a Solution
After integrating by parts twice I got:
(S2)/S(S2+a2) lim c→∞ [-sin(ax)e-Sx + acos(ax)e-Sx] |[itex]^{C}_{0}[/itex]
Okay, now how on Earth do I take the lim c→∞ if sin(ax) is periodic?
An improper integral is an integral that cannot be evaluated using the standard methods of integration. It occurs when one or both of the limits of integration are infinite or when the function being integrated is not defined at certain points within the interval of integration.
To determine if an improper integral converges or diverges, you must evaluate the integral using a limit. If the limit exists and is a finite number, then the integral converges. If the limit does not exist or is infinite, then the integral diverges.
Yes, an improper integral can have a finite value even if one or both of the limits of integration are infinite. This can occur if the function being integrated approaches zero faster than the limits approach infinity.
The two types of improper integrals are Type 1 and Type 2. Type 1 improper integrals have one or both limits of integration as infinity, while Type 2 improper integrals have a discontinuity or infinite value within the interval of integration.
To evaluate an improper integral, you must first determine if it converges or diverges. If it converges, you can use a limit to evaluate the integral. If it diverges, you can use a comparison test, limit comparison test, or the Cauchy condensation test to determine the behavior of the integral.