Periodic functions and properties of integration

Question

This is from a book in Fourier series.

I don't understand how the integral can be split as in equality 2 and 3. Which properties of integration justify that?

$F(x)$ and $f(x)$ are two functions with the periodicity $2l$. The integral $F(x)=\int_{-l}^x f(t) \, \mathrm dt$ is equal to zero at $x=-l$ and at $x=l$. \begin{align*} F(x+2l)= \int_{-l}^{x+2l} f(t) \,\mathrm dt &= \underbrace{ \int_{-l}^x f(t) \, \mathrm dt+ \int_x^{x+2l}f(t)\, \mathrm dt }_\text{How?}\\ &= \underbrace{ \int_{-l}^x f(t)\, \mathrm dt + \int_0^{2l} f(t) \, dt}_\text{How?} \\ &=\int_{-l}^x f(t) \, \mathrm dt= F(x). \end{align*}

Answer 1

• The first How ? is because of Chasles relation:

$$\int_a^b=\int_a^c+\int_c^b,$$

• the second How ? is because the integral of a periodic function does not depend on what part of the period your are integrating. Just do the changement of variable $t=u+2\ell$ to prove it, and use the fact that $f(u+2\ell)=f(u)$ since $f$ is $2\ell$-periodic.

Answer 2

Let us consider the function f (x)=cos x. At x=-phi and+phi f (x) is zero. Here,l=phi and let us consider x=phi/6. When considered the area of f (x) from -phi to 13phi/6. In the region 0 to phi the curve area is positive. In the region phi to 2phi the curve area is negative. Now cosider the area from -phi to 13phi/6.we can divide this area as (-phi to 0), (0 to phi), (phi to 2phi), (2phi to 13phi/6).since the areas (0 to phi), (phi to 2phi) are equal and opposite in direction they cancel each other.Now we are left with the areas (-phi to 0) and (2phi to 13phi/6).this is what you got