Is it true that $\int_a^{a+T}f(x)\,dx \,=\, \int_0^Tf(x)\,dx $ for function $f$ with period $T$?

Question

I was wondering if the following property holds true for the definite integrals of periodic functions:

Consider a function $f(x)$ with period $T$, ie;

$$f (x + T) = f (x) $$

If $a$ is a real number in the domain of the function, then

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

Is this true? It seems intuitive, especially for functions like $\sin(x)$, $\cos(x)$ or $\{x\}$, but I was unable to prove it.

Any kind of hint/help is appreciated.

Answer 1

The difference of the two sides is $$\int_a^{a+T}f(x)\,dx-\int_0^T f(x)\,dx =\int_T^{a+T}f(x)\,dx-\int_0^af(x)\,dx$$ which vanishes by periodicity.

Answer 2

Yes, this is true (provided of course that the integrals exist).

Write $a=mT+b$ with $m\in\Bbb Z$ and $0\le b<T$. Then $$\begin{align} \int_a^{a+T}f(x)\,\mathrm dx&=\int_a^{(m+1)T}f(x)\,\mathrm dx+\int_{(m+1)T}^{a+T}f(x)\,\mathrm dx\\ &=\int_a^{(m+1)T}f(x-mT)\,\mathrm dx+\int_{(m+1)T}^{a+T}f(x-(m+1)T)\,\mathrm dx\\ &=\int_b^{T}f(x)\,\mathrm dx+\int_{0}^{b}f(x)\,\mathrm dx\\ &=\int_{0}^{b}f(x)\,\mathrm dx+\int_b^{T}f(x)\,\mathrm dx\\ &=\int_0^Tf(x)\,\mathrm dx.\end{align}$$