Integral and convergence of the sinc function: $\lim_{t \to \infty} \int_\frac{1}{t}^t \frac{\sin(ax)}{x}dx $ I am asked to show that  $$\lim_{t \to \infty} \int_\frac{1}{t}^t \frac{\sin(ax)}{x}dx $$ converges for all real numbers $a$ and that the value of the converged integral is the same for all $a>0$.
For the first part, I want to apply a previously proven result that if $f$ is continuous and $g'$ is locally integrable on [a,b) with $\lim_{x \to b^- }g(x) = 0$ and $F(t)=\int_a^tf$ is bounded on $[a,b)$, and $g'$ has a constant sign then $\int_a^bfg$ converges. Can I use this theorem here? If we take $g(x)=\frac{1}{x}$ and $f(x)=\sin(ax)$ then clearly $F(t)$ is bounded and $g'$ has a constant sign and $g(x)$ tends to 0. But the condition of local integrability is throwing me off.
If I cannot use this theorem. How may I alternatively prove that the above integral converges? additionally, the proof that the value of the integral is the same for all positive $a$ is eluding me. Any help is greatly appreciated.
 A: We assume $a>0$, $t>0$. 
Hint. One may first perform an integration by parts,
$$
\int_{1/t}^t \frac{\sin(ax)}{x}dx =\left[\frac1{x}\cdot\frac{1- \cos(ax)}{a} \right]_{1/t}^t+\int_{1/t}^t \frac{1- \cos(ax)}{ax^2}dx
$$ giving
$$
\lim_{t \to \infty}\int_{1/t}^t \frac{\sin(ax)}{x}dx =\lim_{t \to \infty}\int_{1/t}^t \frac{1- \cos(ax)}{ax^2}dx\tag1
$$ the latter integrand may be extended as a continuous function over $[0,b]$ ($b\ge1$) satisfying
$$
\left| \frac{1- \cos(ax)}{ax^2}\right| \le  \frac{2}{ax^2}, \qquad x\ge1,
$$ the latter dominating function being integrable over $[1,\infty)$, one deduces the convergence of the initial integral.
From $(1)$, using a standard result one has

$$
\lim_{t \to \infty}\int_{1/t}^t \frac{\sin(ax)}{x}dx =\int_0^\infty \frac{1- \cos(ax)}{ax^2}dx=\frac{\pi}{2} ,\quad a>0.
$$

A: Another way is to note that, for $a>0$, we have $$I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(ax\right)}{x}dx=\frac{1}{2i}\int_{0}^{\infty}\frac{e^{iax}-e^{-iax}}{x}dx
 $$ and now we can use the Frullani theorem for complex parameters and get $$I\left(a\right)=\frac{\log\left(\frac{ia}{-ia}\right)}{2i}=\frac{\log\left(-1\right)}{2i}=\color{red}{\frac{\pi}{2}}.$$
