# Improper Integrals, Infinite Limits

1. Mar 10, 2013

### Magnawolf

1. The problem statement, all variables and given/known data

∫e-Sxsin(ax) dx, S and A are constants, upper limit is ∞ lower is 0

2. Relevant equations

∫ u dv = uv - ∫ vdu

3. 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)?

Last edited: Mar 10, 2013
2. Mar 10, 2013

### SammyS

Staff Emeritus
sin(ax) is bounded.

-1 ≤ sin(ax) ≤ 1

3. Mar 11, 2013

### 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.