TITLE: Quick Integration Clarification
QUESTION [5 upvotes]: $ \int_ {-\pi}^{\pi}\ f(x) \ dx  $ is equivalent to integrating $f(x)$ over $-\pi < x < \pi$ (strictly less than) and also $-\pi \le x \le \pi$ (less than or equal). So I just need to reassure myself that the bounds of integration don't change when considering either less than or less than or equal. A quick confirmation will do.
Cheers.
UPDATE:Just applying this to a question to do with fourier series: $\
f(x) = \begin{cases}x+π & -π<x\le 0 \\ π & 0 < x < π\\  \end{cases} $ with 2π periodicity.
Can I proceed to compute:
$a_0= {1\over 2L}[\displaystyle \int \limits_{-π}^{0} \ x+π \ dx + \int \limits_{0}^{π} π\ dx]$

REPLY [3 votes]: With respect to your edit, Yes, you can compute this particular integral (and evaluate "at" the bounds $I(x) \big|_b^a$, but keep in mind Jon Claus's answer: that what you are really doing is evaluating the limit of $I(x)$ as $x \to a^-, x\to b^+$, and that's crucial to remember in the case of improper integrals.