## UnkleRhaukus 3 years ago Is there a simpler method to show from the definition of the Laplace transform. $\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}$

1. LolWolf

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

2. LolWolf

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

3. LolWolf
4. hartnn

so u can't use properties of LT ?

5. UnkleRhaukus

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

6. UnkleRhaukus

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

7. mahmit2012

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

8. LolWolf

From definition...?

9. UnkleRhaukus

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

10. LolWolf

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

11. mahmit2012

Yes, that is correct.

12. LolWolf

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

13. UnkleRhaukus

$\mathcal L\left\{t sin(nt) \right\}$ $\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}$

14. UnkleRhaukus

that is much quicker

15. hartnn

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

16. LolWolf

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

17. hartnn

u used a property that follows from definition

18. UnkleRhaukus

i did prove $f(t)=sin(nt)$ $F(p)= \frac{n}{p^2+n^2}$ in a earlier question

19. hartnn

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

20. LolWolf

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.

21. UnkleRhaukus

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

22. LolWolf

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.