calculus - Evaluating $lim_{btoinfty} int_0^b frac{sin x}{x}, dx= frac{pi}{2}$

Using the identity $$\lim_{a\to\infty} \int_0^a e^{-xt}\, dt = \frac{1}{x}, x\gt 0,$$ can I get a hint to show that $$\lim_{b\to\infty} \int_0^b \frac{\sin x}{x} \,dx= \frac{\pi}{2}.$$

Answer

Hint:
$$\begin{align} \lim_{b\to \infty}\int_{0}^{b}\frac{\sin x}{x}dx &= \lim_{a,b\to \infty}\int_{0}^{b}\int_{0}^{a}e^{-xt}dt\sin x dx\\& = \lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}e^{-xt}\frac{e^{ix}-e^{-ix}}{2i} dx \\&=\lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}\frac{e^{-(t-i)x}-e^{-(i+t)x}}{2i} dx\end{align}$$.