# integration of a periodic function by substitution [duplicate]

Given a periodic function $f$ with period $L$ I want to prove the following:

$$\int_a^{a+L}f(x) \,dx=\int_0^Lf(x)\,dx$$

I've found some nice proofs here but I have tried by my own doing the following:

\begin{align*}\int_a^{a+L}f(x)dx&=\int_a^Lf(x)dx+\int_L^{a+L}f(x)dx\\ &=\int_a^Lf(x)dx+\int_0^af(y)dy\\ &=\int_0^Lf(x)dx \end{align*}

by using the substitution $y=x-L$.

So I was wondering if this is enough? Because all the proofs there say that this only holds for $a\in[0,L)$ but not for $a$ in general.

So my question is why is this not enough and only holds for $a\in[0,L)$ but not $a$ in general?

## marked as duplicate by 6005, user99914, Jennifer, Daniel W. Farlow, ShaileshJul 18 '16 at 0:08

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## 1 Answer

You're right, this holds for general $a$, and your proof is correct. Note that for $a\notin[0,L]$ one of your integrals has reversed limits, but that's OK. Perhaps the other proofs restricted to $a\in[0,L)$ to avoid that.