An integrable and periodic function f(x) [duplicate]

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, LeucippusJul 18 '16 at 0:18

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

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

• f(t) is not equal with f(t+a) ? or not matter? – Lucas Apr 16 '15 at 8:09
• @Lucas Well, $f(t)=f(t+a)=f(t+2a)=f(t+3a)=...=f(t-a)=f(t-2a)=...$ It's all the same. – mathifold.org Apr 16 '15 at 8:11
• I don't know that... my teacher don't tell me about this, thank you ! – Lucas Apr 16 '15 at 8:13
• $\int _b^{b+a}\:f\left(t\right)\:dt=\int _0^a\:f\left(t\right)dt$ is general case for a periodic functions? – Lucas Apr 16 '15 at 8:15
• @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. – 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 $b=0$.

• 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 $\int _b^{na}\:f\left(t\right)dt$ ? – Lucas Apr 16 '15 at 22:26
• 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 $f(t-a)=f(t)$. – Idris Apr 16 '15 at 22:36
• 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 – Lucas Apr 16 '15 at 22:42