Is Integrating from -L<x<L the Same as -L ≤ x ≤ L?

[member 428835]

hey everyone

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

thanks!

 

[STEMucator]

If you're speaking about a Riemman Integral, then last I checked there was zero contribution from the endpoints.

 

[pwsnafu]

Which definition of integration are you using?

 

[jbunniii]

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.

 

[bossman27]

pwsnafu, I'm curious what integral definitions produce different answers in this case?

 

[pwsnafu]

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.

 

[member 428835]

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.

thanks i appreciate your response(s)!

 

[pwsnafu]

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.

 

[jbunniii]

A function is Riemann integrable if the set of discontinuities has measure zero.

Just for the sake of completeness I'll mention that the function must be bounded, and that the condition is both necessary and sufficient: a bounded function on a finite length closed interval [a,b] is Riemann integrable if and only if the set of discontinuities has measure zero.

 

[jbunniii]

By the way, another way to see this is to observe that if $f$ is integrable over $[-L,L]$, then integrating $f$ over $(-L,L)$ is equivalent to setting $f(L) = f(-L) = 0$, and integrating over $[-L, L]$. So the problem reduces to showing that if you change the value of a function at a finite number of points, the integral does not change. It's easy to see that this in turn is equivalent to showing that if you integrate a function which is zero everywhere except at a finite number of points (it suffices to consider just one point, by linearity), the result is zero. This is an easy Riemann sum argument.

 

[member 428835]

By the way, another way to see this is to observe that if $f$ is integrable over $[-L,L]$, then integrating $f$ over $(-L,L)$ is equivalent to setting $f(L) = f(-L) = 0$, and integrating over $[-L, L]$.

thanks. i suddenly feel like an idiot for not realizing this.