It is said that $f(x+T) = f(x)$ for all $x$ in domain of $fx$ then we have to show $\int_{a}^{a+T} f(x)dx$ is independent in $a$.
My method says that let $F(a) = \int_{a}^{a+T} f(x)dx$ then $$F'(a) = f(a+T)-f(a) + \int_{a}^{a+T} \frac{\partial}{\partial a} f(x ) dx = 0$$
so it means $F(a)$ is independent in $a$. This proof is not very good?? Give good proof please. thank.
Miss said dont try derivative as it can be non differentiable, use property of integration. I also try $x-a = u$ to get $\int_{0}^{T} f(u+a) du$ but what to do now. Thank.
Let $$F(a) =\int_{a} ^{a+T} f(x) \, dx\tag{1}$$ and via Fundamental Theorem of Calculus we have $$F'(a) =f(a+T) - f(a) =0$$ therefore $F(a) $ is independent of $a$. You don't need to give any more explanation.
The above can be proved without differentiation also. We have $$F(a) =\int_{0}^{T}f(x)\,dx+\int_{T} ^{a+T} f(x) \, dx-\int_{0}^{a}f(x)\,dx$$ and in the second integral put $t=x-T$ to get $$F(a) =\int_{0}^{T}f(x)\,dx+\int_{0}^{a}f(t+T)\,dt-\int_{0}^{a}f(x)\,dx$$ Since $f(t+T) =f(t) $ second and third integral are equal and cancel each other so $$F(a) =\int_{0}^{T}f(x)\,dx$$ which is independent of $a$.