Laplace transform of $\sin (t) / t$

Question

I need to find the Laplace transform of $$\frac{\sin(t)}{t}$$ without using the following rule $\mathcal{L}\left(\frac{f(t)}{t}\right)=\int_s^{\infty}F(u)du$.

We aren't allowed to use this rule unless we can prove it, and I'm assuming our lecturer does not want us to use it.

However, we were given the following hint "For this question, you may need to evaluate $ \int^\infty_0 \frac{\sin(t)}{t}dt$. Try substituting $\frac{1}{t} = \int^\infty_0 e^{-st} ds$."

I'm not sure what exactly this is supposed to mean and would appreciate it if someone could show me what this would look like. Does it mean evaluating $ \int^\infty_0 e^{-st} \int^\infty_0 e^{-st}\sin(t) dsdt$? I'm really confused.

Answer 1

I think you should write $$ \int_0^\infty e^{-ut} \frac{\sin{t}}{t} = \int_0^\infty \int_0^\infty e^{-(u+s)t} \frac1{2i}( e^{it}-e^{-it}) \;ds \; dt$$ and change the order of integration. Details depends on the expected level of math rigor.

Answer 2

Assuming $s>0$, the functions $\frac{\sin t}{t}e^{-st}$ and $\sin(t)e^{-st}$ belong to $L^1(\mathbb{R}^+)$ and $$ g(s)=\int_{0}^{+\infty}\frac{\sin t}{t}e^{-st}\,dt\tag{1}$$ fulfills (by the dominated convergence theorem) $$ g'(s) = -\int_{0}^{+\infty}\sin(t)e^{-st}\,dt = -\frac{1}{1+s^2}\tag{2} $$ hence $$ g(s) = C-\arctan(s) \tag{3} $$ for some constant $C$. Since, by (1) and the fact that $\frac{\sin t}{t}$ is a bounded smooth function, it is obvious that $\lim_{s\to +\infty}g(s)=0$, such a constant is $C=\frac{\pi}{2}$ and $$ \mathcal{L}\left(\frac{\sin t}{t}\right)(s) = \color{red}{\arctan\left(\frac{1}{s}\right)}.\tag{4} $$