Compute improper integral $\lim_{A\to \infty}\int_{[0,A]} \frac{\sin(x)}{x}\ \mathrm dx$

Question

I'm trying to compute the improper integral $\displaystyle\lim_{A\to \infty}\int_{[0,A]} \dfrac{\sin(x)}{x}\ \mathrm dx.$

I assumed $\displaystyle\int_0^{\infty}e^{-tx}\ \mathrm dt=\frac{1}{x}\; \forall x>0$ (per a hint) and used Fubini and integration by parts to get that the answer is $\dfrac{\pi}{2}$. The issue I'm having is that I don't know how to show that $\displaystyle\int_0^{\infty}e^{-tx}\ \mathrm dt =\frac{1}{x} \text{ for } x>0$.

Answer 1

\begin{align} I&=\int_0^{\infty} e^{-tx}\,dt \qquad u=tx\\ \\ &=\frac{1}{x}\int_0^{\infty} e^{-u}\,du\\ \\ &=\frac{1}{x} \left(-e^{-u} \Big |_0^{\infty}\right)\\ &=\frac{1}{x} \end{align}