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Consider the function: $$F(x) = \int_{0}^{\infty} e^{-xt}\frac{\sin(t)}{t}dt$$ Here $x \in \mathbb{R}^{+}$. I want to use the dominated convergence theorem to show that: $$\lim_{x \to \infty} F(x) = \lim_{x \to \infty} \int_{0}^{\infty} e^{-xt}\frac{\sin(t)}{t}dt = \int_{0}^{\infty} \lim_{x \to \infty} e^{-xt}\frac{\sin(t)}{t}dt = 0$$

So to get the limit in the integral, there are $2$ conditions. First, the sequence of integrable functions $f_x(t) = e^{-xt}\frac{\sin(t)}{t} $ needs to converge to $f(t) = 0$ for $x \to \infty$, which is satisfied. Secondly, you need to estimate the function $f_x(t)$ from above so that: $$|f_x(t)| \leq g(t)$$ Here $g: \mathbb{R} \to [0, \infty]$ is a integrable function. In this case we get: $$|f_x(t)| = |e^{-xt}\frac{\sin(t)}{t}| \leq |e^{-xt}| \leq g(t) $$ The problem is, is that I can't find a function $g(t)$ which fulfills this condition and is independent of $x$. If such a $g(t)$ exists, then we can use the dominated convergence theorem.

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Hint: Since we are taking limit as $x\to \infty$, we can assume that $x\ge 1$. Observe that $$ e^{-xt}\leq e^{-t} $$ for all $x\geq 1$.

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  • $\begingroup$ I see this is true, but I thought that the condition for the estimation needed to be for every x>0 $\endgroup$ – Belgium_Physics Jan 23 at 16:50
  • $\begingroup$ @Belgium_Physics And in fact, since $\sup\limits_{x>0}e^{-xt} = 1$ for all $t>0$, it must be that $1\le g(t)$. So, $g$ that you are looking for cannot exist as a $L^1$-function. This means we have to discard some values of $x$ in a neighborhood of $0$. $\endgroup$ – Song Jan 23 at 16:52
  • $\begingroup$ @Belgium_Physics Then you use DCT on the set, say, $[\pi/2, +\infty)$ and you simply compute $\lim_{x\to +\infty} \vert \int_0^{\pi/2} \mathrm e^{-xt} \sin t \cdot t^{-1} \mathrm dt\vert $ by basic estimates. $\endgroup$ – xbh Jan 23 at 17:05

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