Laplace transform of $\dfrac{\sin2t} t$

Question

So I'm taking a look at my notes and the professor wrote this:

${\scr L}(\frac {\sin2t}{t}) = \arctan \frac 2s$

But I can't see this anywhere in the tables. So, where does this come from?

Thanks in advance!

Answer 1

It is in the tables. Just use these two rules:

${\scr L}(\sin{at})=\frac{a}{s^2+a^2}$ and ${\scr L}(\frac{f(t)}{t})=\int_{s}^\infty F(\sigma)d\sigma$ where $F(s)={\scr L}(f(t))$

Therefore;

$$F(s)=\frac{2}{s^2+4}$$

$${\scr L}(\frac{f(t)}{t})=\int_s^\infty \frac{2}{\sigma^2+4}d\sigma = \frac{\pi}{2}-\arctan{\frac{s}{2}} = \arctan{\frac{2}{s}}$$

See, it is very simple.

Answer 2

I can only deduce it from the series: $$ f(t) = \frac{\sin t}{t} = \sum_{k=0}^\infty (-1)^k \frac{t^{2k}}{(2k+1)!} $$ then $$ \mathcal{L}\left\{f(t)\right\} = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \mathcal{L}\left\{t^{2k}\right\} = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} \frac{(2k)!}{s^{2k+1}} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \left(\frac{1}{s}\right)^{2k+1} = \arctan\frac{1}{s} $$ Using $$ \mathcal{L}\left\{ f(2t) \right\} = \frac{1}{2} F\left(\frac{s}{2}\right) $$ we get $$ \frac{\sin 2t}{2t} = \frac{1}{2}\arctan \frac{2}{s} $$

Answer 3

Compute $$\phi(x)=\int_0^\infty \mathrm e^{-st}\cos(xt)\mathrm dt =\Re\mathit e\left[\int_0^\infty\mathrm e^{-st-\mathrm ixt}\mathrm dt\right] =\Re\mathit e\frac1{s+\mathrm ix}=\frac{s}{s^2+x^2}.$$ Then notice that the Laplace transform you are looking for is the $x$-antiderivative of $\phi$ taken at $x=2$. To see it, consider $\Phi(x)=\int_0^\infty \mathrm e^{-st}\frac{\sin(xt)}t\mathrm dt$. Derivating under the integral gives $\Phi'(x)=\phi(x)$.

Of course $\Phi(0)=0$. The antiderivative of $\phi$ is therefore $$\Phi(x)=\int_0^x\phi(y)\mathrm dy=\int_0^x\frac{s}{s^2+y^2}\mathrm dy=\arctan\frac xs$$ and this gives the result.