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Let $f:R\rightarrow R$ with a period of $T\gt0$, and $f(x)=f(x+T)$ for every x. Assume that $f$ is integrable on $[0,T].$

1.) Prove that $f$ is integrable on $[a,a+T]$

2.) Prove that: $$ \int_{a}^{a+T}f(x)dx=\int_{0}^{T}f(x)dx$$

Attempts:

1.) According to the "invariance" property of integrals, if $f$ is integrable on [0,T] then we can define $g(x)=f(x-a)$ such that $g:[0+a,a+T] \rightarrow R$ which is also integrable on [0+a,a+T]

$$ \int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x+a)dx$$

For question #1, is my thought process correct? And could I somehow use some sort pf substitution to prove #2?

Again, thanks for any help and hints are preferred over solutions.

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Your first argument is not correct; it just shows that some other function is integrable.

Hint for #2: Substitution, $u = x-a$ or $x = u + a$.

Hint for #1: For $0 \le a \le T$, split the integral into $$ \int_a^T + \int_T^{T+a} $$ In the second, replace $f(x)$ with $f(x-T)$ throughout, and then observe that this second integral becomes $$ \int_0^a f(x) ~dx $$ and now sum with the first to get the answer. A similar trick works for $T < a < 2T$, and so on, except that you need to do two simplifications instead of just one.

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  • $\begingroup$ Using your hint for #1, I solved #2. Splitting the integral up then switching $f(x)=f(x-T)$ for the second one gives us two integrals: $\int_0^a f(x) dx+\int_a^T f(x) ~dx$ which gives up $\int_0^T f(x) ~dx$. Maybe I'm not understanding a step? $\endgroup$ – RonaldB Jun 11 '16 at 11:40
  • $\begingroup$ could you please give another hint for a solution for #2? Been stuck on it for quite some time now. $\endgroup$ – RonaldB Jun 11 '16 at 15:57
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    $\begingroup$ Substitution: $u = x - a, du = dx$. The limits of integration for $x$ are $x = a$ to $x = a + T$; for $x = a$, we get $u = 0$; for $x = a + T$, we get $u = T$. (Sorry about writing $u = x + a$ in my earlier hint...I meant "$x = u + a$", which gives $u = x - a$.) $\endgroup$ – John Hughes Jun 11 '16 at 17:36
  • $\begingroup$ Thanks! I figured I had tried every possibility and was just overlooking something simple. That new substitution solved it quickly. $\endgroup$ – RonaldB Jun 11 '16 at 17:41

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