Can someone find the flaw in an argument that $\int_0^{\infty}\sin x dx=1$?

Question

Consinder the following indefinite integrals,
$$I=\int e^{-sx}\sin x dx$$ $$\displaystyle J=\int e^{-sx}\cos x dx$$ Using integration by parts on the first and second integrals we get, $$I=-e^{-sx}\cos x-sJ$$ $$J=e^{-sx}\sin x+sI$$ $$\begin{align*}\implies I&=-e^{-sx}\cos x-s(e^{-sx}\sin x+sI) \\\\\implies (1+s^2)I&=-e^{-sx}(\cos x+s\sin x) \\\\\implies I&=\frac{-e^{-sx}(\cos x+s\sin x)}{1+s^2}\end{align*}$$ Therefore, $$\begin{align*}\int_0^{\infty} e^{-sx}\sin x dx&=\bigg[\frac{-e^{-sx}(\cos x+s\sin x)}{1+s^2}\bigg]_0^{\infty}\\&=0-\frac{-1}{1+s^2}\\&=\frac1{1+s^2}\end{align*}$$ Substituting $s=0$ we get that, $$\int_0^{\infty} \sin x dx=1$$ Obviously this is not true because $\displaystyle\int_0^{n} \sin x dx$ oscillates between $0$ and $2$ when we increase the value of $n$. Can someone tell me the flaw in the above argument. Thanks in advance.

Answer 1

I believe the part before evaluating $\frac{-e^{-sx}(\cos x+s\sin x)}{1+s^2}$ at $\infty$ is correct. The thing is, when $s=0$ this is $-\cos x$ which does not have a limit at $\infty$.