Question

Is there a simpler method to show from the definition of the Laplace transform.
\[\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}\]

Answer 1

I haven't come across a nicer one, myself, but it'd be nice to see.

Answer 2

(If what I'm seeing is right, of course, considering it's cut off)

Answer 3

http://mathurl.com/aq279pu.png

Answer 4

so u can't use properties of LT ?

Answer 5

\[\mathcal L\left\{t\sin(nt)\right\}=\int\limits_0^\infty t\sin(nt)e^{-pt}\text dt\]
\[=\int_0^\infty t\frac{e^{int}-e^{-int}}{2i}e^{-pt}\text dt\]
\[=\frac 1{2i}\int_0^\infty t({e^{int}-e^{-int}})e^{-pt}\text dt\]
\[=\frac 1{2i}\left(\int_0^\infty t\cdot e^{-(p-in)t}\text dt-\int_0^\infty t\cdot e^{-(p+in)t}\text dt\right)\]
\[=\frac 1{2i}\left[\left(\left.\frac{t\cdot e^{-(p-in)}}{-(p-in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p-in)t}}{-(p-in)}\text dt\right)\right.\]\[\qquad\left.-\left(\left.\frac{t\cdot e^{-(p+in)}}{-(p+in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p+in)t}}{-(p+in)}\text dt\right)\right]\]
\[=\frac 1{2i}\left[\left(0+\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt\right)-\left(0+\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right)\right]\]
\[=\frac 1{2i}\left[\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt-\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right]\]
\[=\frac 1{2i}\left[\left.\frac{e^{-(p-in)t}}{-(p-in)^2}-\frac{e^{-(p+in)t}}{-(p+in)^2}\right|_0^\infty\right]\]
\[=\frac 1{2i}\left[\frac{1}{(p-in)^2}-\frac{1}{(p+in)^2}\right]\]
\[=\frac 1{2i}\left[\frac{(p+in)^2-(p+in)^2}{(p-in)^2(p-in)^2}\right]\]
\[=\frac 1{2i}\left[\frac{(p^2-2ipn-n^2)-(p^2+2ipn-n^2)}{\left((p-in)(p+in)\right)^2}\right]\]
\[=\frac 1{2i}\left[\frac{-4ipn}{\left(p^2+n^2\right)^2}\right]\]
\[=\frac{2pn}{\left(p^2+n^2\right)^2}\]

Answer 6

the question was
"From the definition, evaluate the following Laplace transforms."

Answer 7

0
L(sinnt)=n/p2+n2
L(tsinnt)=-d/dp(n/p2+n2)=2np/(p2_n2)2

Answer 8

From definition...?

Answer 9

\[\mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\]
@mahmit2012 does that follow from the definition?

Answer 10

I guess you could write it as a Lemma... don't know how that will fly, with your prof. though.

Answer 11

Yes, that is correct.

Answer 12

I'm still thinking it requires some work, it doesn't just follow from definition.

Answer 13

\[\mathcal L\left\{t sin(nt) \right\}\]
\[\qquad\qquad\text{let } f(t)=sin(nt)\]\[\qquad\qquad F(t)= \frac{n}{p^2+n^2}\]\[\qquad\qquad \mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\]\[\mathcal L\left\{tf(p)\right\}=-\frac{\text d}{\text dp}\left(\frac{n}{p^2+n^2}\right)\]\[=-\frac{\text d}{\text dp}\left({n}({p^2+n^2})^{-1}\right)\]\[=-2pn(p^2+n^2)^{-2}\]\[=\frac{2pn}{(p^2+n^2)^2}\]

Answer 14

that is much quicker

Answer 15

but u didn't use definition.....

Answer 16

I agree, but I thought you said it had to be from definition...?

Answer 17

u used a property that follows from definition

Answer 18

i did prove

\[f(t)=sin(nt)\]
\[F(p)= \frac{n}{p^2+n^2}\]
in a earlier question

Answer 19

with the given question, i would go with your earlier response, though its longer

Answer 20

What I was trying to say, as with @hartnn is, just go ahead and prove this:
\[\mathcal {L}\{ t f(t)\}=-\frac{d}{dp}F(p)\] as a lemma, and you know that you're working from definition.

Answer 21

i havent proved \[\mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\] yet

Answer 22

Yes, and you should as it's not a corollary of the definition, it's actually a lemma. Once you've got that, just reference it, I haven't had professors nag me about doing such like this, yet, but you never know.