TITLE: Dominated convergence theorem on $F(x) = \int_{0}^{\infty} e^{-xt}\frac{\sin(t)}{t}dt$
QUESTION [2 upvotes]: 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.

REPLY [1 votes]: 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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