How to recognize an improper integral? 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?
 A: 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.
A: 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.
A: 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.  
