Question

What is the laplace transform of tcosh(at)? (Using integration by parts)

Answer 1

\[\Large\bf \mathscr{L}\left[t\color{#DD4747 }{\cosh at}\right]\quad=\quad \mathscr{L}\left[\color{#DD4747 }{\frac{1}{2}}t\color{#DD4747 }{(e^{at}+e^{-at})}\right]\]So then throwing that into our definition gives us a couple of integrals to deal with, yes?\[\Large\bf=\quad \frac{1}{2}\bf \int\limits_0^{\infty}t e^{at}e^{-st}dt\quad +\quad \frac{1}{2}\bf \int\limits_0^{\infty}t e^{-at}e^{-st}dt\]

Answer 2

We can combine the exponentials in each integral to make them easier to work with:\[\Large \bf \frac{1}{2}\int\limits_0^{\infty}t e^{-(s-a)t}\;dt\]

Answer 3

ok then ... ?

Answer 4

So you want the t to go away, so that will be the `u` in our parts.\[\Large\bf u=t\qquad\qquad\qquad dv=e^{-(s-a)t}dt\]\[\Large\bf du=?\qquad\qquad\qquad v=?\]

Answer 5

then what to do with the new integral?

Answer 6

For the integral we're left with, we can pull the denominator out front since it's constant,
\[\Large\bf \frac{1}{s-a}\int\limits_0^{\infty}e^{-(s-a)t}\;dt\]And we can simply integrate here, no parts needed for this.

Answer 7

That was my point.
Got it, awesome !
Thanks alot!

Answer 8

cool c: