# Sign problem in laplace transform

1. Sep 17, 2009

### Mechdude

1. The problem statement, all variables and given/known data
$$\int_{\pi/2}^{\infty}cos(t)e^{-st}dt$$

2. Relevant equations

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

3. 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
$$f(x)=\left\{\begin{array}{cc}0,&\mbox{ if }0\leq x\leq \pi/2 \\ cos(t), & \mbox{ if } x>\pi/2 \end{array}\right.$$

Last edited: Sep 17, 2009
2. Sep 17, 2009

### 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?

3. Sep 17, 2009

### Mechdude

well typo on my part , let me edit it,

4. Sep 17, 2009

### 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

5. Sep 17, 2009

### Dick

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.

6. Sep 17, 2009

### Mechdude

So the answers for them would be different?

7. Sep 17, 2009

### 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)

8. Sep 17, 2009

### Dick

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?

9. Sep 17, 2009

### Mechdude

Thats sensible thanks,