# laplace

• Feb 25th 2013, 01:05 AM
jomex42
laplace
how you get the laplace of 1/tsin(wt)
pls show me the solution
• Feb 25th 2013, 03:37 AM
TheEmptySet
Re: laplace
Quote:

Originally Posted by jomex42
how you get the laplace of 1/tsin(wt)
pls show me the solution

I will leave the details of why to you.

Given $f(w,t)=\frac{1}{t}\sin(wt) \implies \frac{\partial f}{\partial w}=\cos(wt)$

$\mathcal{L}\{ \frac{\partial f}{\partial x}\}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}$

$\frac{s}{s^2+w^2}=\frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}$

Now integrate both sides with respect to w to solve for the Laplace transform you want

$\int \frac{s}{s^2+w^2}dw=\int \frac{\partial }{\partial w}\mathcal{L}\{ f(w,t)\}dw$

$\tan^{-1}\left( \frac{w}{s}\right)=\mathcal{L}\{ f(w,t)\}$
• Mar 8th 2013, 07:19 AM
jomex42
Re: laplace
could you show me the solution of laplace of 1/tsinwt using the formula
intergral of limits 0 to infinity e^st(ft)dt
then you will get the arctanw/s
plss the exact solution my teacher want it
• Mar 9th 2013, 12:16 AM
jomex42
Re: laplace
Attachment 27443

can you solve this and you will get the result of
arctan(α/s)
• Mar 9th 2013, 01:18 PM
TheEmptySet
Re: laplace
Quote:

Originally Posted by jomex42
Attachment 27443

can you solve this and you will get the result of
arctan(α/s)

Another method is to note that the integral converges uniformly so limits and integrals can be interchanged.

Note that

$\sin(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!} \quad \tan^{-1}(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)}$

This gives that

$\int_{0}^{\infty}\frac{e^{-st}}{t}\left( \sum_{n=0}^{\infty}\frac{(-1)^n(\alpha t)^{2n+1}}{(2n+1)!}\right)dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\int_{0}^{\infty} e^{-st} t^{2n}dt = \sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!} \mathcal{L}\{t^{2n}\}$

Now integrate or use the definition of the Laplace transform to get

$\sum_{n=0}^{\infty}\frac{(-1)^n\alpha^{2n+1}}{(2n+1)!}\frac{(2n)!}{s^{2n+1}}= \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)}\left( \frac{\alpha}{s}\right)^{2n+1}=\tan^{-1}\left( \frac{\alpha}{s} \right)$
• Mar 10th 2013, 05:34 AM
jomex42
Re: laplace
What do you called that big letter e in your equation
what method did you use?

Because im not familiar in that equation i have not encounter that kind of function could you explain it
• Mar 10th 2013, 05:45 AM
jomex42
Re: laplace
what method did you use so i can explain it to my teacher how it been solved
• Mar 10th 2013, 05:01 PM
jomex42
Re: laplace
how the t in your first function gone in the next function

and the summation get out of the integral sign

What did you do in the first function to comeup with the second function
Attachment 27473