Sign problem in laplace transform

[Mechdude]

Homework Statement

\int_{\pi/2}^{\infty}cos(t)e^{-st}dt

Homework Equations

\int vdu=uv-\int udv
v=cos(t) , dv=-sin(t)dt, du=e^{-st} , u=\frac{-e^{-st}} {s}

The Attempt at a Solution

\int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac{-e^{-st}}{s}cos(t) -\int (-)\frac{-e^{-st}}{s}(-)sin(t)dt
the right side becomes:
\frac{-e^{-st}}{s}cos(t) - \frac{1}{s} \int e^{-st}sin(t)dt
for the second integral on the right:v=sin(t) , dv=cos(t)dt and du=e^{-st} , u=\frac{-e^{-st}} {s}
\int e^{-st}sin(t)dt =- \frac{1}{s}e^{-st}sin(t)dt - \int \frac{-e^{-st}}{s}cos(t)
now \int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac{-e^{-st}}{s}cos(t) - \frac {1}{s} \left[ - \frac{1}{s}e^{-st}sin(t) + \int \frac{e^{-st}}{s}cos(t)dt \right]
thus
\frac{-e^{-st}}{s}cos(t) + \frac {1}{s^2}e^{-st}sin(t) - \frac {1}{s^2} \int e^{-st}cos(t)dt
simplifying
\int_{\pi/2}^{\infty}cos(t)e^{-st}dt \frac {s^2 +1} {s^2}= \frac {1}{s^2}e^{-st}sin(t) - \frac{e^{-st}}{s}cos(t)
going on :
\int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac {s^2}{s^2 +1} \left[ \frac {1}{s^2}e^{-st}sin(t) - \frac{e^{-st}}{s}cos(t) \right]
the right side is:
= \frac {1}{s^2 +1}e^{-st}sin(t) + \frac{e^{-st}}{s^2+1}scos(t)
now when i evaluate the limits:
=(0-0) - ( \frac{e^{-s\pi/2}}{s^2+1} - 0 )
=- \frac{e^{-s\pi/2}}{s^2+1}
im sure this is wrong because of the sign (its supposed to be positive now where did i drop the sign? can u help? this was the function whose transform i was looking for
<br /> f(x)=\left\{\begin{array}{cc}0,&amp;\mbox{ if }0\leq x\leq \pi/2 \\<br /> cos(t), &amp; \mbox{ if } x&gt;\pi/2 \end{array}\right.<br />

 

[Dick]

I get the same thing you got. Why would you think the answer should be positive? cos(t) is negative for pi/2<t<3pi/2. If you were trying to transform that step function f(x), where did the cos(t) come from?

 

[Mechdude]

, where did the cos(t) come from?

well typo on my part , let me edit it,

 

[Mechdude]

Here's the culprit :
L \left\{ U_{a}(t) f(t-a) \right\} = e^{-as} F(s)
which implies my result might be wrong

 

[Dick]

Here's the culprit :
L \left\{ U_{a}(t) f(t-a) \right\} = e^{-as} F(s)
which implies my result might be wrong

That formula would be helpful for integrating cos(t-pi/2)*exp(-ts) from pi/2 to infinity. But that's not the same as your problem.

 

[Mechdude]

So the answers for them would be different?

 

[Mechdude]

So the answers for them would be different? As an aside how would you determine L(t^2u(t-1)) with the hint to write t^2u(t-1) in the form f(t-a)u(t-a)

 

[Dick]

So the answers for them would be different? As an aside how would you determine L(t^2u(t-1)) with the hint to write t^2u(t-1) in the form f(t-a)u(t-a)

Sure the answers to cos(t)e^(-st) and cos(t-pi/2)*e^(-st) integrated from pi/2 to infinity are different. If you want t^2 to be f(t-1) then I guess f(t) would have to be (t+1)^2, right?

 

[Mechdude]

Thats sensible thanks,