3
$\begingroup$

Possible Duplicate:
An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.

I have to show if $f\in \mathcal L[-\pi,\pi]$ and if $$f(u+2\pi)=f(u), (-\infty then, for any real $x$, $$\int_{-\pi+x}^{\pi+x}f(u) du=\int_{-\pi}^{\pi}f(u) du.$$ I am approaching the answer of this question in the following manner, $$ \int_{-\pi}^{\pi}f(u) du=\int_{-\pi}^{-\pi+x}f(u)du+\int_{-\pi+x}^{\pi}f(u)du= I_{1}+I_{2}.\tag{1}$$ Now, $$I_{1}=\int_{-\pi}^{-\pi+x}f(u)du$$ Taking $u=t-2\pi$, we get, $$I_{1}=\int_{\pi}^{\pi+x}f(t-2\pi) dt=\int_{\pi}^{\pi+x}f(t)dt. (\because f\ \text{is periodic})$$ Thus, $$I_{1}=\int_{\pi}^{\pi+x}f(t) dt.$$ From $(1)$, we get, $$ \int_{-\pi}^{\pi}f(u)du=\int_{\pi}^{\pi+x}f(u)du+\int_{-\pi+x}^{\pi}f(u)du=\int_{-\pi+x}^{\pi+x}f(u)du.$$ Am i approach right?

  • 1
    This is basically the same question as: [An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.](http://math.stackexchange.com/questions/94233/an-integrable-and-periodic-function-fx-satisfies-int-0tfxdx-int)2012-06-03
  • 0
    I don't know that this question already asked! Sorry for that! But you can tell me am i justify the answer of this question?2012-06-03

1 Answers 1