Let be an integralable function on , and let be periodic with period . Prove

for all .

* Proof. * For any with we know from a previous exercise that there exists a unique such that

Then, we start by splitting the integral into two pieces. (The goal here is to rearrange the integral so that it starts at and ends at , and then use the fact that is periodic to conclude that this integral is the same as the one from 0 to .)

This completes the proof

Do we have to assume that exists for all ?

Alt sol:

= – +

= IFF =

= =

Hi Jamie, Nice solution. I did it similarly to RoRi but this is a very elegant approach.

Hi jamiehlusko,

I think your solution is wrong because you don’t know if a0 so you are not allowed to use the translation property of integrals.

I checked the book now and you are right.

How do you know the integral exists on [0, a]?

I guess it is because it exists for [0, p] and since it is periodic with period p it exists for all real values. The function just repeats itself and thus the area following p is congruent to the area following 0.

It seems that the proof is assuming what Vivek Suraiya wrote. For a complete proof it would be nice to see a proof of that too.

Namely, theorem 1.17 assumes existance of 2 intervals out the 3 (and its proof in the book too). Therefore, it seems that the proof here cannot simply rely on it, because it relaxes that and assumes what Vivek Suraiya wrote.