1
$\begingroup$

I was wondering if anyone can help me with this, if f(x) is a periodic function with period T then it satisfies $$\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)\;dx$$ for all $a \in \Bbb R$. It is clear that this must be true, but if you differentiate both sides with respect to $T$ do you not get $$f(T)=f(T+a)$$ and so because $$f(T+a)=f(a)$$ this implies that $$f(T)=f(a)$$ for all $a \in \Bbb R$, but does this not imply $f$ is constant? I am struggling to understand what is going wrong.

$\endgroup$
  • 2
    $\begingroup$ The relation holds only if T is a period of f, not for arbitrary T. Therefore you cannot differentiate with respect to T. $\endgroup$ – Martin R May 5 '15 at 9:54
  • $\begingroup$ Related $\endgroup$ – 6005 Jul 17 '16 at 16:38
1
$\begingroup$

The equality $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$ holds for all $a \in \Bbb R$, but only for values $T$ which are a period of $f$.

The derivative is only defined for functions defined on an interval. So what you actually have proved is that if the set of periods contains an open interval then $f$ is constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.