## question on integration bounds

hey everyone

is integrating from $-L<x<L$ the same as integrating over $-L \leq x \leq L$? im looking for a rigorous response, so if you have time could you explain why, rather than simply yes or no?

thanks!
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 If you're speaking about a Riemman Integral, then last I checked there was zero contribution from the endpoints.
 Which definition of integration are you using?

Blog Entries: 1
Recognitions:
Gold Member
Homework Help

## 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.
 Recognitions: Gold Member pwsnafu, I'm curious what integral definitions produce different answers in this case?

 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.

 Quote by jbunniii Assuming the function is integrable on ##[-L,L]##, it will have the same integral on ##(-L,L)##.
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.

 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.

Blog Entries: 1
Recognitions:
Gold Member
Homework Help