Suppose i have f is an odd function.
How to show that $\displaystyle \int_{[-\pi,\pi]} f(x) $ =0 ?
An odd function that is not the null function is positive in a part and negative in a different part of the definition interval. Let's define $\displaystyle f^{+} (x)= f(x)$ where $\displaystyle f(x)>0$ and $\displaystyle f^{+} (x)= 0$ elsewhere and $\displaystyle f^{-} (x)= -f(x)$ where $\displaystyle f(x)<0$ and $\displaystyle f^{-} (x)= 0$ elsewhere. The Lebesgue integral of f(x) is defined as...
$\displaystyle \int_{\mu} f(x) dx = \int_{\mu} f^{+}(x) dx - \int_{\mu} f^{-}(x) dx $ (1)
Now the integral (1) exists only if both the integrals in (1) exist and that is not true [for example...] for $\displaystyle f(x)= \frac{1}{x}$ in $\displaystyle [- \pi, +\pi]$...
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$
Well, $\displaystyle \frac{1}{x}$ isn't Lebesgue integrable on $\displaystyle [-\pi,\pi]$, so it's a moot point, no? The basic idea is that change of variables still works for Lebsgue integrals. What if you let $\displaystyle x=-z$?
The original question was to demonstrate that if f(x) is and odd function, then ...
$\displaystyle \int_{-\pi}^{+\pi} f(x) dx =0$ (1)
... where the integral is 'Lebesgue Integral'. My replay has been [symply...] that an odd function isn't necessarly Lebesgue integrable and the example $\displaystyle f(x)= \frac{1}{x}$ has been supplied. Of course if we add the condition that f(x) must be 'regular', then the demonstration of (1) is very comfortable...
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$
Clearly we are to assume that the function is integrable, otherwise the integral doesn't make sense! That's like saying "Prove that the limit of a sequence in $\displaystyle [0,\infty)$ is in $\displaystyle [0,\infty)$ is false since $\displaystyle \lim n$ is infinite--it's clear by wording that we are to assume the sequence converges!
... if f(x) is odd then $\displaystyle f(x)=-f(-x)$ so that is...
$\displaystyle \int_{-\pi}^{0} f(x)\ dx = - \int_{0}^{\pi} f(x)\ dx \implies \int_{-\pi}^{0} f(x)\ dx + \int_{0}^{\pi} f(x)\ dx = 0$
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$