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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.

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  • $\begingroup$ There are multiple ways to show continuity. Could you give us some specific examples? $\endgroup$ Commented Apr 26, 2015 at 21:02
  • $\begingroup$ Do you need to see how to evaluate the integral? $\endgroup$ Commented Apr 26, 2015 at 21:26

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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 $0$.

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  • $\begingroup$ "a discontinuity when the denominator" $\: \mapsto \:$ "a point not in their domain when the denominator" $\hspace{.72 in}$ $\endgroup$ Commented Apr 26, 2015 at 21:29
  • $\begingroup$ Are you saying that the point does exist, on another domain? $\endgroup$ Commented Apr 26, 2015 at 21:44
  • $\begingroup$ No, I'm saying that [continuous at] and [discontinuous at] are just defined for points in the domain. $\hspace{.5 in}$ $\endgroup$ Commented Apr 26, 2015 at 21:49
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    $\begingroup$ Ahorn, I don't agree with your statement that some function are continuous ‘by definition’. You certainly have to prove that polynomials, and others, are continuous; that doesn't immediately follow from the definition of the function. Furthermore, all trigonometric functions, including $tan$, $csc$, $sec$ and $cot$, are continuous, because the points where they ‘become infinite’ are not included in the domain; and as Ricky Demer correctly noted: the property of being continuous at a point is only defined for points inside the domain. $\endgroup$ Commented Apr 26, 2015 at 21:58

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