- #1
Mechdude
- 117
- 1
Homework Statement
[tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt [/tex]
Homework Equations
[tex] \int vdu=uv-\int udv [/tex][itex] v=cos(t) , dv=-sin(t)dt [/itex], [itex] du=e^{-st} , u=\frac{-e^{-st}} {s}[/itex]
The Attempt at a Solution
[tex] \int_{\pi/2}^{\infty}cos(t)e^{-st}dt = \frac{-e^{-st}}{s}cos(t) -\int (-)\frac{-e^{-st}}{s}(-)sin(t)dt [/tex]
the right side becomes:
[tex] \frac{-e^{-st}}{s}cos(t) - \frac{1}{s} \int e^{-st}sin(t)dt [/tex]
for the second integral on the right:[itex] v=sin(t) , dv=cos(t)dt [/itex] and [itex] du=e^{-st} , u=\frac{-e^{-st}} {s} [/itex]
[tex] \int e^{-st}sin(t)dt =- \frac{1}{s}e^{-st}sin(t)dt - \int \frac{-e^{-st}}{s}cos(t) [/tex]
now [tex] \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] [/tex]
thus
[tex] \frac{-e^{-st}}{s}cos(t) + \frac {1}{s^2}e^{-st}sin(t) - \frac {1}{s^2} \int e^{-st}cos(t)dt [/tex]
simplifying
[tex] \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) [/tex]
going on :
[tex] \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] [/tex]
the right side is:
[tex] = \frac {1}{s^2 +1}e^{-st}sin(t) + \frac{e^{-st}}{s^2+1}scos(t) [/tex]
now when i evaluate the limits:
[tex] =(0-0) - ( \frac{e^{-s\pi/2}}{s^2+1} - 0 ) [/tex]
[tex] =- \frac{e^{-s\pi/2}}{s^2+1} [/tex]
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
[tex]
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.
[/tex]
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