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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.

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    $\begingroup$ $$\lim_{s\to0}\int_0^\infty f(s,x)dx\neq \int_0^\infty\lim_{s\to0}f(s,x)dx$$ $\endgroup$ Jul 25, 2021 at 8:57
  • $\begingroup$ I agree with @NinadMunshi ... $$\int_0^{\infty} e^{-sx}\sin x dx=\frac1{1+s^2}$$ is true for $s>0$, but $\int_0^\infty \sin x dx = 1$ is false. $\endgroup$
    – GEdgar
    Jul 25, 2021 at 11:57
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    $\begingroup$ Next time that you post a question, please don't use titles that consists of mathematical expressions or equations only. These are discouraged for technical reasons - see the second item from Guidelines for good use of $\rm\LaTeX$ in question titles. $\endgroup$
    – soupless
    Jul 25, 2021 at 15:33

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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 $\infty$.

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  • $\begingroup$ You beat me to it (the first version of my answer was wrong). 🙂 $\endgroup$ Jul 25, 2021 at 9:00
  • $\begingroup$ Yeah, I double checked whether OP divided by $s$ when I saw your answer. $\endgroup$ Jul 25, 2021 at 9:00
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    $\begingroup$ You mean $-\cos x$. $\endgroup$
    – TonyK
    Jul 25, 2021 at 9:38
  • $\begingroup$ @TonyK Thanks for catching the error. $\endgroup$ Jul 25, 2021 at 9:52

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