How to recognize an improper integral?

Question

I think $\displaystyle \int ^{4} _0 \frac{\sin x}{x} dx$ is not an improper integral. Is this true? If not true, then how can one recongnise improper integrals in general?

Answer 1

It's not an improper integral because $\dfrac{\sin x} {x}$ has a removable discontinuity at $0$. If $f$ has an infinite essential discontinuity at some point in an interval $I$, then you have an improper integral $\displaystyle\int_If(x) dx$. You also get an improper integral if the interval $I$ where you are integrating is unbounded.

Answer 2

To put it simply, an improper integral is one where there is some problem with the integrand on the interval, or where the interval is unbounded. Thus, since

$$ \frac{\sin(x)}{x} $$ is undefined at $x=0$, your integral is indeed improper. What you are really evaluating is the limit $$ \lim_{a\rightarrow 0+} \int_a^4 \frac{\sin(x)}{x} dx $$

That the undefined point is on the boundary does not matter, if you had been asked to evaluate an integral like $$ \int_{-1}^4 \frac{\sin(x)}{x} dx $$ you'd really be evaluating $$ \lim_{a\rightarrow 0-} \int_{-1}^a \frac{\sin(x)}{x} dx+ \lim_{b\rightarrow 0+} \int_b^4 \frac{\sin(x)}{x} dx $$ (note that the limits are independent; here it does not really matter as the improper integrals evaluate to finite numbers, but in other cases, it does).

Another example of an improper integral would be $$ \int_{1}^\infty \frac{\sin(x)}{x} dx $$ which is shorthand for $$ \lim_{b\rightarrow \infty} \int_1^b \frac{\sin(x)}{x} dx $$

Some improper integrals involve integrands which are unbounded; these should also be evaluated as above, with limits. Sometimes they yield something finite, sometimes they don't.

Answer 3

Partly I think it is because of the fact that $(\sin 4 )/ 4$ is a real number as well as $ \displaystyle \lim_{x \rightarrow 0 ^{-} } (\sin x )/ x =1$.

I also think the general method is not that of finding discontinuities but that of finding points of singularities.