# Dominated convergence theorem on $F(x) = \int_{0}^{\infty} e^{-xt}\frac{\sin(t)}{t}dt$

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.

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 $$x\geq 1$$.
• @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 $0$. – Song Jan 23 at 16:52
• @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. – xbh Jan 23 at 17:05