Is the integral $\int^{π/2}_0 \frac{\sin x}{x} \,\mathrm{d}x$ improper and why?

Question

I am unable to understand the following question.
Is the integral $$\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x$$ improper and why?
Also what is an impromper integral?

Answer 1

Without any other information the integral $\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x$ is to be considered improper since $\frac{\sin x}{x}$ is not defined for $x=0$ and we need to define it by limit

$$\int^\frac{π}{2}_0 \frac{\sin x}{x} \,\mathrm{d}x=\lim_{a\to 0^+} \int^\frac{π}{2}_a \frac{\sin x}{x} \,\mathrm{d}x$$

and it converges since $x\to 0$ $\frac{\sin x}{x}\to1$.

But it is not improper if we define the integrand $\frac{\sin x}{x}=a\in \mathbb{R}$ for $x=0$ and in particular $a=1$ which leads to a continuos integrand function; in this case indeed we are not forced to define the integral by limit.

Answer 2

Since $\frac{sin(x)}{x} \to 1$ as $x \to 0$, we can define $f: \mathbb R \to \mathbb R$ by

$f(x):=\frac{sin(x)}{x}$ if $x \ne 0$ and $f(0):=1$. Then $f$ is continuous and $ \int_0^{\pi/2}\frac{sin(x)}{x} dx = \int_0^{\pi/2}f(x) dx$.

Consequence: $ \int_0^{\pi/2}\frac{sin(x)}{x} dx $ is not improper.