Improper Integrals, Infinite Limits

[Magnawolf]

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] |^{C}_{0}

Okay, now how on Earth do I take the lim c→∞ if sin(ax) is periodic? Since limx→∞ e^-x=0 would it just be (S²)/S(S²+a²) [(0+0) - (0+1)] which becomes
-(S²)/S(S²+a²)?

 

[SammyS]

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] |^{C}_{0}

Okay, now how on Earth do I take the lim c→∞ if sin(ax) is periodic?

sin(ax) is bounded.

-1 ≤ sin(ax) ≤ 1

 

[Magnawolf]

Yeah I know but how do you take the limit of something that's bounded by two numbers? I'm assuming you can't. What I got was that since limx→∞ e^-x=0 would it just be (S²)/S(S²+a²) [(0+0) - (0+1)]? which becomes -(S²)/S(S²+a²). I just want to verify that this is the answer.