Check whether the integrand is continuous when evaluating improper integrals In order to evaluate improper integrals, I need to know whether the integrand is continuous between the limits of the integral. For the lower and upper limits, I believe you find out if it's continuous at the points if the limit of the integrand as x tends to either the upper or lower limit exists, but how would I find out if the integrand is continuous over the whole range and not just at the upper and lower limits?

E.g for this integrand how would I find out if it's continuous over the whole range of the limits given?

$$\int_0^\infty e^{-ax}\,\frac{\sin x}{x}\,dx$$

Any help would be much appreciated.
Some functions are, by definition, continuous, and others are not. For example: 


polynomials are continuous on their domain
rational functions have a discontinuity when the denominator is equal to $0$
exponential functions are continuous on their domain
split functions often have a discontinuity, when the function jumps or has a removable discontinuity
trigonometric functions such as $\sin$ and $\cos$ are continuous but $
\csc$, $\sec$, $\tan$ and $\cot$ are not continuous. In the latter case, the functions are still integrable because the integral becomes improper yet can still be evaluated.


In your example, $\lim_{x \to 0}\frac{\sin x}{x}$ exists, so the function is not discontinuous at $0$.
Some functions are, by definition, continuous, and others are not. For example: 


polynomials are continuous on their domain
rational functions have a discontinuity when the denominator is equal to $0$
exponential functions are continuous on their domain
split functions often have a discontinuity, when the function jumps or has a removable discontinuity
trigonometric functions such as $\sin$ and $\cos$ are continuous but $
\csc$, $\sec$, $\tan$ and $\cot$ are not continuous. In the latter case, the functions are still integrable because the integral becomes improper yet can still be evaluated.


In your example, $\lim_{x \to 0}\frac{\sin x}{x}$ exists, so the function is not discontinuous at $0$.