Application of a Laplace transform property to solve a definite (improper) integral

Question

I want to solve the following integral using Laplace transform properties:

$$\int_0^\infty \frac{\sin(t)}{t} dt $$

The result is $\frac\pi2\ $. There is a Laplace transform property concerning integrals:

$$ \mathcal{L}\left(\int_0^x f(t) dt\right)=\frac{\mathcal{L}(f)(s)}{s} $$

I saw a procedure from a colleague which has a step that goes like this:

$$ \mathcal{L}\left(\int_0^\infty \frac{\sin(t)}{t} dt\right)=\int_0^\infty \frac{1}{1+s^2} ds $$

The above expression seems like if the transform could be introduced inside the integral and applied to the integrand, with integration limits unaffected by this operation. I would like to know whether this step is correct and, if it isn't, what is the correct application of the aforementioned property to solve the integral? Also, does the final value theorem hold in this case?

Thanks in advance.

Answer 1

Let $I=\int_0^{\infty}e^{-st}\frac{sint}{t}dt , (s=0)$

or $I=\mathcal{L}[\frac{sint}{t}]=\int_s^\infty \frac{1}{s^2+1}ds=\frac{\pi}{2}-tan^{-1}s=\frac{\pi}{2} $ at $s=0$.

The property is $\mathcal{L}[f(t)/t]=\int_s^\infty F(s)ds$ where $\mathcal{L}[f(t)]=F(s)$.

Answer 2

Consider the following integral: $$I(s)=\int_0^\infty \frac{\sin(t)}{t} e^{-st}dt$$ $$I(s)=\mathcal{L}\left (\frac{\sin(t)}{t}\right )$$ $$I(s)=\int_s^\infty \frac{du}{u^2+1} $$ $$I(s)=\bigg|\arctan u \bigg|_s^\infty $$ $$I(s)=\dfrac {\pi}2-\arctan s $$ You can evalute your intergal now: $$I=\int_0^\infty \frac{\sin(t)}{t} dt=I(0)=\dfrac {\pi}2$$