# question on integration bounds

by joshmccraney
Tags: bounds, integration
 P: 115 hey everyone is integrating from $-L  P: 1,218 If you're speaking about a Riemman Integral, then last I checked there was zero contribution from the endpoints.  P: 696 Which definition of integration are you using? PF Patron HW Helper Sci Advisor P: 2,562 ## question on integration bounds Assuming the function is integrable on ##[-L,L]##, it will have the same integral on ##(-L,L)##. For the Lebesgue integral, the reason is simple: the two endpoints have measure zero, so they do not contribute to the value of the integral. The Riemann integral is only defined on closed intervals, so we have to define what is meant by integrating over ##(-L,L)##. The usual way to do this is to use improper integrals: \lim_{\epsilon \rightarrow 0^+} \int_{-L + \epsilon}^0 f(x) dx + \lim_{\epsilon \rightarrow 0^+} \int_0^{L - \epsilon} f(x) dx Note that we have \int_{-L}^{L} f(x) dx = \int_{-L}^{-L + \epsilon} f(x) dx + \int_{-L + \epsilon}^0 f(x) dx + \int_0^{L - \epsilon} f(x) dx + \int_{L - \epsilon}^{L} f(x) dx so to prove what we want, we simply need to show that \lim_{\epsilon \rightarrow 0^+} \int_{-L}^{-L + \epsilon} f(x) dx = \lim_{\epsilon \rightarrow 0^+} \int_{L - \epsilon}^{L} f(x) dx = 0 But this is quite straightforward; since ##f## is integrable over ##[-L,L]##, by definition it is bounded on that interval, say ##|f(x)| \leq M## for all ##x \in [-L,L]##. Then \left| \int_{-L}^{-L + \epsilon} f(x) dx \right| \leq \int_{-L}^{-L + \epsilon} |f(x)| dx \leq \int_{-L}^{-L + \epsilon} M dx = M\epsilon which converges to 0 as ##\epsilon \rightarrow 0^+##. We can do the same thing for the other small interval.  P: 204 pwsnafu, I'm curious what integral definitions produce different answers in this case? P: 696  Quote by bossman27 pwsnafu, I'm curious what integral definitions produce different answers in this case? If ##\mu## is a measure on ℝ, then the integral wrt ##\mu## gives ##\int_{[a,b]} f \, d\mu = \int_{(a,b)} f \, d\mu + f(a)\, \mu(\{a\}) + f(b) \, \mu(\{b\})##. So for any atomless measure (including the Lebesgue measure) the integrals are equal. But if ##\mu(\{a\}) \neq 0## then the integrals are different. P: 115  Quote by jbunniii Assuming the function is integrable on ##[-L,L]##, it will have the same integral on ##(-L,L)##. and for [itex]f(x)$ to be integrable on ##[-L,L]##, is it sufficient $f(x)$ need only exist on that interval provided only a finite amount of discontinuities exist? i ask because i am integrating over a piecewise-smooth function.

for the record i am talking about a reimann integral. apologies for the ambiguity.

P: 696
 Quote by joshmccraney and for $f(x)$ to be integrable on ##[-L,L]##, is it sufficient $f(x)$ need only exist on that interval provided only a finite amount of discontinuities exist? i ask because i am integrating over a piecewise-smooth function. for the record i am talking about a reimann integral. apologies for the ambiguity. thanks i appreciate your response(s)!
A function is Riemann integrable if the set of discontinuities has measure zero. In particular if there are a finite number of jump discontinuities, you have no problem.
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