## UnkleRhaukus 2 years ago Can someone please check this; $\begin{equation*} f(t)=\cos(t)\big(h(t-\pi)-h(t)\big) \end{equation*}$ \newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t \end{align*}

1. UnkleRhaukus

\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\ &=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\ &=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\ &=\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_\pi^\infty-\left.\frac{e^{-pt}\big(-p\cos t+\sin t\big)}{(-p)^2+1^2}\right|_0^\infty\\ &=\frac{-e^{-p\pi}\big(-p\cos \pi+\sin \pi\big)}{p^2+1}+\frac{\big(-p\cos 0+\sin 0\big)}{p^2+1}\\ &=\frac{-pe^{-p\pi}}{p^2+1}-\frac{p}{p^2+1}\\ \\ &=\frac{-p\big(1+e^{-p\pi }\big)}{p^2+1}\\ \end{align*}

2. whpalmer4

I haven't had my coffee yet, but I don't quite see how you managed to make h(t) vanish

3. phi

I think h is the step function

4. whpalmer4

ah, okay, sure, I didn't realize that we could assume anything about h(t)

5. experimentX

isn't that Laplace transform? .. let p = s

6. UnkleRhaukus

The unit heaviside step function $h(t-a) =\begin{cases}0&t<a\\ 1&t≥a\end{cases}$

7. experimentX

|dw:1358356261290:dw|

8. UnkleRhaukus

h=H=u=θ

9. experimentX

just integrate negative of cos(x) e^(-px) from 0 to pi

10. experimentX

yes!! multiple ways of representing unit step function.

11. UnkleRhaukus

12. experimentX

think (h(t−π)−h(t)) as a switch (toward negative) that get's on at t=0 and off at t=pi .

13. experimentX

yes that's correct!!

14. experimentX
15. experimentX

there's a short formula.for int e^(ax) cos(bx) dx ...also note that you can shift integral for step function.

16. experimentX
17. experimentX

|dw:1358358498070:dw|

18. experimentX

|dw:1358358737716:dw| this should be simpler.

19. UnkleRhaukus

Ah yeah I can see that now, one integral is simpler than two

20. UnkleRhaukus

\newcommand\dd[1]{\,\mathrm d#1} % infinitesimal \newcommand\intl[4]{\int\limits_{#1}^{#2}{#3}{\dd #4}} % integral _{a}^{b}{f(x)}\dd x \begin{align*} \intl0\infty{f(t)e^{-pt}}t&=\intl0\infty{\cos(t)\big(h(t-\pi)-h(t)\big)e^{-pt}}t\\ &=\intl0\infty{\cos(t)h(t-\pi)e^{-pt}}t-\intl0\infty{\cos(t)h(t)e^{-pt}}t\\ &=\intl\pi\infty{\cos(t)e^{-pt}}t-\intl0\infty{\cos(t)e^{-pt}}t\\ &=\intl\pi0{\cos(t)e^{-pt}}t\\ &=\left.\frac{e^{-pt}\big(\sin (t)+p\cos (t)\big)}{(-p)^2+1^2}\right|_\pi^0\\ &=\frac{\sin 0+p\cos 0-e^{-p\pi}\big(\sin \pi+p\cos \pi\big)}{p^2+1^2}\\ &=\frac{p+pe^{-p\pi}}{p^2+1^2}\\ &=\frac{p\left(1+e^{-p\pi}\right)}{p^2+1^2}\\&\color{red}\checkmark \end{align*}

21. UnkleRhaukus

thanks experimentX

22. experimentX

yw!! you work hard man!!