calculus - 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.

Answer

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

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