Using laplace transform to calculate $\int^{\infty}_0 \sin(t)\,dt$

Question

I've learned that:

$$\int^{\infty}_0 \frac{f(t)}{t}\,dt=\, \int^{\infty}_0 \, \mathcal{L}(f (t))\, ds$$

Where $\mathcal{L}(f (t))$ is the laplace transform of the funcion $f(t)$

So I started playing with this and I did the following: $$\int^{\infty}_0 \sin(t)\,dt = \int^{\infty}_0 \frac{t \sin(t)}{t}\,dt=\int^{\infty}_0 \, \mathcal{L}(t\sin(t))\, ds$$

Turns out that: $$\mathcal{L}(t\sin(t)) = \frac{2s}{(s^2 + 1)^2}$$

and

$$\int^{\infty}_0 \,\frac{2s}{(s^2 + 1)^2}\, ds = 1$$

thus:

$$\int^{\infty}_0 \sin(t)\,dt = 1$$

because $\int^{\infty}_0 \sin(t)\,dt = \cos(0) - \lim_{t \rightarrow \infty} \cos(t)= 1 - \lim_{t \rightarrow \infty} \cos(t)$, this would imply that:

$$ \lim_{t \rightarrow \infty} \cos(t) = 0$$

This can't be correct because $\cos(t)$ does not approach any number as $t \rightarrow \infty$. Where is the mistake?

Answer 1

Firstly, $\dfrac{f(t)}{t}$ must be integrable to apply this rule. Also, the limit of cosine is not well defined, since for any $x>>0$, you can still attain any point of $[-1,1]$ for some $y>x$ so this limit doesn’t exist.

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