This is a question from an old paper exam, where the ultimate goal is to prove that $$\int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}$$ There are a lot of answers to that question, and I believe I've seen most of then, but none following the argument given in this problem (despite some similarities). The question is divided in three parts:
Question 1: Show that, for $x>0$, $$\frac{1}{x} = \int_0^{\infty}e^{-xt}dt$$
This I believe is can be proved just by integrating. Since $\int e^{-xt}dt= -\frac{1}{x}e^{-xt}+K$, $$\int_0^{\infty} e^{-xt}dt = \lim_{t \to \infty}\left(-\frac{1}{x}\right)e^{-xt} - \left(\frac{1}{x}\right) = 0 - \left(-\frac{1}{x}\right)=\frac{1}{x}$$ Now:
Question 2: Show that, if $A>0$, the function $[0,A]\times [0,\infty)\rightarrow \mathbb{R}$ defined by $(x,t)\mapsto \sin x\, e^{-xt}$ is integrable in $[0,A]\times[0,\infty)$.
I am not sure if what I've done here is correct, but this is my attempt: $$\int_0^A \int_0^{\infty}\sin x\, e^{-xt}dt\ dx = \int_0^{\infty}\sin x \left(\int_0^A e^{-xt} dt \right) dx = \int_0^{\infty}\frac{\sin x}{x}dx$$ where interchanging the integral is (I think) justified by Tonelli's theorem, and now the function integrable since $\frac{\sin x}{x}$ is continuous in $[0, \infty)$.
The last question asks the following:
Question 3: Use Question 1, Question 2 and Fubini's Theorem to show that \begin{align} \int_0^A \frac{\sin x}{x}dx = \frac{\pi}{2} + \int_0^{\infty}g_A(t)dt, \, A>0 \tag{1} \end{align} where $$\left \vert \int_0^{\infty}g_A(t)dt\right\vert \leq \frac{K}{A},$$ and $K$ is some constant, and deduce that $$\lim_{A\to\infty}\int_0^A \frac{\sin x}{x}dx = \frac{\pi}{2}$$
I see how the result follows from $(1)$, but I don't have any ideas on how to get to that equation. Some help regarding that would be helpful.
$\int_0^A \frac {\sin x}{x} dx = \int_0^A \int_0^\infty e^{-xt}\sin x\ dt\ dx = \int_0^\infty\int_0^A e^{-xt}\sin x\ dx\ dt\\ \int_0^\infty -\frac {e^{-xt}(\cos x + t\sin x)}{1+t^2}|_0^A dt\\ \int_0^\infty \frac {1}{1+t^2} - \frac {e^{-At}(\cos A + t \sin A)}{1+t^2}\ dt\\ \frac {\pi}{2} + \int_0^\infty g(A,t)\ dt$
Now we need to show that $\int_0^\infty g(A,t) \ dt$ is bounded as described above.
$|\frac {\cos A + t\sin A}{1+t^2}| < |\cos A + \sin A| < \sqrt 2$
alternatively
$|\frac {\cos A + t\sin A}{1+t^2}| < \frac {1 + t}{1+t^2} \le \frac 12 + \frac 1{\sqrt 2}$
$|\int_0^\infty g(A,t)\ dt| < \int_0^{\infty} \sqrt 2 e^{-At}\ dt =\frac {\sqrt 2}{A}$