# Proof that this integral is independent of $a$

I indeed don't how to treat to this problem. If $f$ is a continuous periodic function with period $T$ then $I=\int_a^{a+T}f(x)dx$ is independent of $a$. Thanks!

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What is $T$? The period? –  Daniel Littlewood Dec 1 '12 at 14:46
Hint: Show that $I = \int_0^k f(x)\,\mathrm dx$ regardless of the value of $a$. Splitting the integral into two integrals (the split being determined by the value of $a$) or a change of variables (the specific change also being determined by the value of $a$) might be needed. –  Dilip Sarwate Dec 1 '12 at 14:47
@DilipSarwate: D o you think we can take $I$ as an function of $a$? –  B. S. Dec 1 '12 at 14:54
@BabakSorouh Yes, of course, as Grapth's answer shows, that is a very easy way of solving the problem. The way suggested in my hint is developed in more detail in the answer by Didier. –  Dilip Sarwate Dec 1 '12 at 19:10

Let $nT$ denote the unique multiple of $T$ in $[a,a+T)$. Then $I=J+K$ where $$J=\int_a^{nT}f(x)\mathrm dx,\qquad K=\int_{nT}^{a+T}f(x)\mathrm dx.$$ The change of variables $x=t-T+nT$ in $J$ yields $$J=\int_{a-nT+T}^Tf(t)\mathrm dt.$$ The change of variables $x=s+nT$ in $K$ yields $$K=\int_0^{a-nT+T}f(s)\mathrm ds.$$ Thus $I=\displaystyle\int_0^Tf(u)\mathrm du$ is independent of $a$.
So, clearly, $I$ is a function of $a$. Let us find $I'(a)$. Since $f(x)$ is continuous on the closed interval $[a, a + k]$, the First Fundamental Theorem of Calculus applies to give $$I'(a) = f(a + T) - f(a).$$ Since $f$ is periodic of period $T$, $f(a + T) = f(a)$, so $I'(a) = 0$. Therefore, $I(a)$ is the constant function. In other words, as $a$ changes, $I(a)$ does not, so $I$ does not depend on $a$.