For a periodic function we have: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$ , but I don't understand how we obtain $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt=\int _b^{na}\:f\left(t\right)dt$ in our equality?


marked as duplicate by 6005, user99914, Daniel W. Farlow, Shailesh, Leucippus Jul 18 '16 at 0:18

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    $\begingroup$ Is $a$ the length of one period? $\endgroup$ – Matthias Apr 16 '15 at 7:25
  • $\begingroup$ yes, a is the length of one period $\endgroup$ – Lucas Apr 16 '15 at 7:42
  • $\begingroup$ You add $a$ to both the lower and the upper limit. You start with $b$ and $na$ and end up with $b+a$ and $na+a=(n+1)a$. $\endgroup$ – mickep Apr 16 '15 at 7:54
  • $\begingroup$ and how we split that? can you show me ? how you split $\int _b^{na}\:f\left(t\right)dt$ to obtain $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt$ ? $\endgroup$ – Lucas Apr 16 '15 at 7:57

What does it mean for a function to have a period $a$ ? Informally, it means that the function values gets repeated after an increment of $a$ in the $x-$value. (domain value). Notationally,

$$a\textrm{ is period of }f\iff f(x)=f(x+a)~\forall~x,x+a\in\textrm{Dom(f)}$$

That gives us $na=na+a=n(a+1)$ and $b=b+a$ since $na$ and $b$ are domain values ($x-$values) for $f$ and $a$ is the period.

So, the definite integral remains the same.

Here's a simple diagram:

Image http://img.prntscr.com/img?url=http://i.imgur.com/iF1Msyw.png

  • $\begingroup$ $\int _b^{b+a}\:f\left(t\right)\:dt=\int _0^a\:f\left(t\right)dt$ is general case for a periodic functions? $\endgroup$ – Lucas Apr 16 '15 at 8:14
  • $\begingroup$ @Lucas, I have added a diagram for better understanding. $\endgroup$ – Prasun Biswas Apr 16 '15 at 8:17
  • $\begingroup$ Take the curve as $f$ $\endgroup$ – Prasun Biswas Apr 16 '15 at 8:18
  • $\begingroup$ @Lucas, didn't it help you? $\endgroup$ – Prasun Biswas Apr 16 '15 at 8:29
  • $\begingroup$ I can't see diagram...don't show me $\endgroup$ – Lucas Apr 16 '15 at 8:31

If $a$ is the length of one period, then $f(t)=f(t-a)$ and so on. Then

$\int_{b+a}^{na+a}f(t)dt=\int_{b+a}^{na+a}f(t-a)dt=\text{(take $u=t-a$)}=\int_{b}^{na}f(u)du$

  • $\begingroup$ f(t) is not equal with f(t+a) ? or not matter? $\endgroup$ – Lucas Apr 16 '15 at 8:09
  • $\begingroup$ @Lucas Well, $f(t)=f(t+a)=f(t+2a)=f(t+3a)=...=f(t-a)=f(t-2a)=...$ It's all the same. $\endgroup$ – mathifold.org Apr 16 '15 at 8:11
  • $\begingroup$ I don't know that... my teacher don't tell me about this, thank you ! $\endgroup$ – Lucas Apr 16 '15 at 8:13
  • $\begingroup$ $\int _b^{b+a}\:f\left(t\right)\:dt=\int _0^a\:f\left(t\right)dt$ is general case for a periodic functions? $\endgroup$ – Lucas Apr 16 '15 at 8:15
  • $\begingroup$ @Lucas This would be true if the period is $b$. Translations not related to the period do not preserve the function and neither the integral. $\endgroup$ – mathifold.org Apr 16 '15 at 8:21

Let $F$ be a primitive of $f$. Then, by the fundamental theorem of calculus, one has

$ \frac{d}{db} \int_{b}^{b+a}f(t)dt= \frac{d}{db} [F(b+a)-F(b)]=F^ \prime (b+a)-F^ \prime (b)=f(b+a)-f(b)=0.$

since $f$ is $a$-periodic. Then, the integral $\int_{b}^{b+a}f(t)dt$ is independent of $b$, so one can take $b=0$.

  • $\begingroup$ Thank you,but I want to understand something else, how we demonstrate that $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt$ is equal with $\int _b^{na}\:f\left(t\right)dt$ ? $\endgroup$ – Lucas Apr 16 '15 at 22:26
  • $\begingroup$ This is a simpler question: it suffices to perform the change of variable $u=t-a$ in your right hand integral, you will got the left one after using the fact that $f$ is $a$-periodic, namely $f(t-a)=f(t)$. $\endgroup$ – Idris Apr 16 '15 at 22:36
  • $\begingroup$ I know that was a simple question, but is my first time after a long time when I comeback to "play math", I forgot some things.. :D $\endgroup$ – Lucas Apr 16 '15 at 22:42

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