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I am trying to evaluate $\int_{0}^{\infty} t \sin (t) dt$ by using Laplace transformation but I am getting inconsistent results.

Here is my solution:

$I=\int_{0}^{\infty} t \sin (t) dt$

$J(s):=\int_{0}^{\infty} t \sin (t)e^{-st} dt \implies J(0) =I$

Since $J(s)$ is the Laplace transform of $t\sin(t)$, therefore it is equal to $\dfrac{2s}{(s^2+1)^2} \implies J(s)=\dfrac{2s}{(s^2+1)^2}\implies J(0) =0$

On the other hand the integral of $t\sin(t)$ is $\sin(t)-t\cos(t)$ whose value keeps oscilating, so the integral should be undefined rather $0$.

What went wrong with the laplace transformation and how can I avoid it in future?

PS: Since I am not familiar with this forum, I am not sure what tags to add. I would be grateful if somebody could do it for me.

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    $\begingroup$ The integral $\int_0^{\infty} t\sin t \,dt$ doesn't exist (as a Lebesgue integral or an improper Riemann integral). Thus the existence of $\lim_{s \to 0} J(s)$ doesn't help. $\endgroup$ – Daniel Fischer Sep 29 '16 at 13:40
  • $\begingroup$ @DanielFischer How can I get this information by looking at the Laplace transform of the function? $\endgroup$ – Guest Sep 29 '16 at 13:42
  • $\begingroup$ You can't. At least not easily. You need to look at whether the integral $\int_0^{\infty} f(t)\,dt$ exists first. If it exists (as a Lebesgue integral or an improper Riemann integral), then you can [maybe, only if you can determine the LT of course] use the Laplace transformation to evaluate it. $\endgroup$ – Daniel Fischer Sep 29 '16 at 13:48
  • $\begingroup$ I was thinking that the domain of $J(s)$ (the values of $s$ for which the transformation is valid) may not include $0$,perhaps this is why I got the inconsistent result. $\endgroup$ – Guest Sep 29 '16 at 13:59
  • $\begingroup$ Yes, the integral defining the Laplace transformation exists for $\operatorname{Re} s > 0$. In this case, $J$ can be meromorphically continued to all of $\mathbb{C}$ (with poles at $\pm i$ and nowhere else), but the continuation doesn't say the corresponding integral exists. You can use the continuation to associate a value to the divergent integrals $\int_0^{\infty} e^{-st} t \sin t \,dt$, akin to some summation methods for divergent series. $\endgroup$ – Daniel Fischer Sep 29 '16 at 14:06

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