Prove that $\int_{0}^{\pi/2} \frac{\sin x}{x} dx$ converges

Question

Prove that $\int_{0}^{\pi/2} \frac{\sin x}{x} dx$ converges.

Shall I compare it with 1/x?

Answer 1

If you define $\frac {\sin\, x} x$ to be $1$ when $x=0$ you get a continuous function on $[0,\frac {\pi} 2]$. Hence it is Riemann integrable.

Answer 2

As noticed in the comments since as $x\to 0^+$ we have

$$\frac{\sin x}{x}\to 1$$

the function is bounded and the integral is finite.

As an alternative, not necessary in that case, we could also proceed by limit comparison test with the convergent $\int_0^1 \frac1{\sqrt x}$, indeed

$$\frac{\frac{\sin x}{x}}{\frac1{\sqrt x}}=\sqrt x \cdot \frac{\sin x}{x}\to 0$$

Answer 3

For any acute $x$, $\frac{\sin x}{x}\in [0,\,1]$. We therefore know in particular that $\int_0^{\pi/2}\frac{\sin x dx}{x}\in [0,\,\frac{\pi}{2}]$.