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

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

Mathematics
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I haven't come across a nicer one, myself, but it'd be nice to see.
(If what I'm seeing is right, of course, considering it's cut off)
http://mathurl.com/aq279pu.png

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so u can't use properties of LT ?
\[\mathcal L\left\{t\sin(nt)\right\}=\int\limits_0^\infty t\sin(nt)e^{-pt}\text dt\] \[=\int_0^\infty t\frac{e^{int}-e^{-int}}{2i}e^{-pt}\text dt\] \[=\frac 1{2i}\int_0^\infty t({e^{int}-e^{-int}})e^{-pt}\text dt\] \[=\frac 1{2i}\left(\int_0^\infty t\cdot e^{-(p-in)t}\text dt-\int_0^\infty t\cdot e^{-(p+in)t}\text dt\right)\] \[=\frac 1{2i}\left[\left(\left.\frac{t\cdot e^{-(p-in)}}{-(p-in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p-in)t}}{-(p-in)}\text dt\right)\right.\]\[\qquad\left.-\left(\left.\frac{t\cdot e^{-(p+in)}}{-(p+in)}\right|_0^\infty-\int_0^\infty\frac{e^{-(p+in)t}}{-(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\left(0+\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt\right)-\left(0+\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\int_0^\infty\frac{e^{-(p-in)t}}{(p-in)}\text dt-\int_0^\infty\frac{e^{-(p+in)t}}{(p+in)}\text dt\right]\] \[=\frac 1{2i}\left[\left.\frac{e^{-(p-in)t}}{-(p-in)^2}-\frac{e^{-(p+in)t}}{-(p+in)^2}\right|_0^\infty\right]\] \[=\frac 1{2i}\left[\frac{1}{(p-in)^2}-\frac{1}{(p+in)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p+in)^2-(p+in)^2}{(p-in)^2(p-in)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p^2-2ipn-n^2)-(p^2+2ipn-n^2)}{\left((p-in)(p+in)\right)^2}\right]\] \[=\frac 1{2i}\left[\frac{-4ipn}{\left(p^2+n^2\right)^2}\right]\] \[=\frac{2pn}{\left(p^2+n^2\right)^2}\]
the question was "From the definition, evaluate the following Laplace transforms."
0 L(sinnt)=n/p2+n2 L(tsinnt)=-d/dp(n/p2+n2)=2np/(p2_n2)2
From definition...?
\[\mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\] @mahmit2012 does that follow from the definition?
I guess you could write it as a Lemma... don't know how that will fly, with your prof. though.
Yes, that is correct.
I'm still thinking it requires some work, it doesn't just follow from definition.
\[\mathcal L\left\{t sin(nt) \right\}\] \[\qquad\qquad\text{let } f(t)=sin(nt)\]\[\qquad\qquad F(t)= \frac{n}{p^2+n^2}\]\[\qquad\qquad \mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\]\[\mathcal L\left\{tf(p)\right\}=-\frac{\text d}{\text dp}\left(\frac{n}{p^2+n^2}\right)\]\[=-\frac{\text d}{\text dp}\left({n}({p^2+n^2})^{-1}\right)\]\[=-2pn(p^2+n^2)^{-2}\]\[=\frac{2pn}{(p^2+n^2)^2}\]
that is much quicker
but u didn't use definition.....
I agree, but I thought you said it had to be from definition...?
u used a property that follows from definition
i did prove \[f(t)=sin(nt)\] \[F(p)= \frac{n}{p^2+n^2}\] in a earlier question
with the given question, i would go with your earlier response, though its longer
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.
i havent proved \[\mathcal {L}\{ t f(t)\}=-\frac{\text d}{\text dp}F(p)\] yet
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.

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