# Differentiation and Integration of Fourier Series

## Differentiation of Fourier Series

Let $$f\left( x \right)$$ be a $$2\pi$$-periodic piecewise continuous function defined on the closed interval $$\left[ { - \pi ,\pi } \right].$$ As we know, the Fourier series expansion of such a function exists and is given by

$f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} .$

If the derivative $$f'\left( x \right)$$ of this function is also piecewise continuous and the function $$f\left( x \right)$$ satisfies the periodicity conditions

$f\left( { - \pi } \right) = f\left( \pi \right),\;\; f'\left( { - \pi } \right) = f'\left( \pi \right),$

then the Fourier series expansion of the derivative $$f'\left( x \right)$$ is expressed by the formula

$f'\left( x \right) = \sum\limits_{n = 1}^\infty {\left( {n{b_n}\cos nx - n{a_n}\sin nx} \right)} .$

## Integration of Fourier Series

Let $$g\left( x \right)$$ be a $$2\pi$$-periodic piecewise continuous function on the interval $$\left[ { - \pi ,\pi } \right].$$ Then this function can be integrated term by term on this interval. The Fourier series for $$g\left( x \right)$$ is given by

$g\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)} .$

Consider the function

$G\left( x \right) = \int\limits_0^x {g\left( t \right)dt} \sim \frac{{{A_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos nx + {B_n}\sin nx} \right)},$

where $${A_n} = - {\frac{{{b_n}}}{n}},$$ $${B_n} = {\frac{{{a_n}}}{n}}.$$

By setting $$x = 0,$$ we see that

$G\left( 0 \right) = 0 = \frac{{{A_0}}}{2} + \sum\limits_{n = 1}^\infty {{A_n}} = \frac{{{A_0}}}{2} - \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} \;\;\text{or}\;\; \frac{{{A_0}}}{2} = \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} .$

Hence, the Fourier series expansion of the function $$G\left( x \right)$$ is defined by

$G\left( x \right) = \int\limits_0^x {g\left( t \right)dt} = \int\limits_0^x {\frac{{{a_0}}}{2}dx} + \sum\limits_{n = 1}^\infty {\int\limits_0^x {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)dx} } = \frac{{{a_0}x}}{2} + \sum\limits_{n = 1}^\infty {\frac{{{a_n}\sin nx + {b_n}\left( {1 - \cos nx} \right)}}{n}},$

where the series on the right-hand side is obtained by the formal term-by-term integration of the Fourier series for $$g\left( x \right).$$

Because of the presence of the term depending on $$x$$ on the right-hand side, this is not clearly a Fourier series expansion of the integral of $$g\left( x \right).$$ The result can be rearranged to be a Fourier series expansion of the function

$\Phi \left( x \right) = \int\limits_0^x {g\left( t \right)dt} - \frac{{{a_0}x}}{2}.$

The Fourier series of the function $$\Phi\left( x \right)$$ is given by the expression

$\Phi \left( x \right) = \int\limits_0^x {g\left( t \right)dt} - \frac{{{a_0}x}}{2} = \frac{{{A_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{A_n}\cos nx + {B_n}\sin nx} \right)} ,$

where the Fourier coefficients are defined by the relationships:

$\frac{{{A_0}}}{2} = \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{n}} ,\;\; {A_n} = - \frac{{{b_n}}}{n},\;\; {B_n} = \frac{{{a_n}}}{n}.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the Fourier series of the sign function

$f\left( x \right) = \text{sign}\,x = \begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases},$

knowing that the Fourier series expansion of the function $$F\left( x \right) = \left| x \right|$$ on the interval $$\left[ { - \pi ,\pi } \right]$$ is given by

$F\left( x \right) = \left| x \right| = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}}.$

### Example 2

Find the Fourier series expansion of the function $$f\left( x \right) = {x^2}$$ knowing that

$x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx}$

for $$- \pi \le x \le \pi.$$

### Example 1.

Find the Fourier series of the sign function

$f\left( x \right) = \text{sign}\,x = \begin{cases} -1, & -\pi \le x \le 0 \\ 1, & 0 \lt x \le \pi \end{cases},$

knowing that the Fourier series expansion of the function $$F\left( x \right) = \left| x \right|$$ on the interval $$\left[ { - \pi ,\pi } \right]$$ is given by

$F\left( x \right) = \left| x \right| = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}}.$

Solution.

As $$f\left( x \right) = F'\left( x \right)$$ for all $$x \ne 0,$$ we obtain

$f\left( x \right) = \frac{d}{{dx}}\left[ {\frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\cos \left( {2n + 1} \right)x}}{{{{\left( {2n + 1} \right)}^2}}}} } \right]$

or

$f\left( x \right) = \frac{4}{\pi }\sum\limits_{n = 0}^\infty {\frac{{\sin\left( {2n + 1} \right)x}}{{2n + 1}}} .$

The graphs of the given function and its Fourier approximation are shown in Figure $$1.$$

### Example 2.

Find the Fourier series expansion of the function $$f\left( x \right) = {x^2}$$ knowing that

$x = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nx}$

for $$- \pi \le x \le \pi.$$

Solution.

As the function $$f\left( x \right)$$ is piecewise continuous on the interval $$\left[ { - \pi ,\pi } \right],$$ we may integrate its Fourier series term by term to obtain

$\int\limits_{ - \pi }^x {tdt} = 2\sum\limits_{n = 1}^\infty {\int\limits_{ - \pi }^x {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}\sin nt\,dt} } .$

Consequently,

$\frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{2} = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\left[ {\left. {\left( { - \frac{{\cos nt}}{{{n^2}}}} \right)} \right|_{ - \pi }^x } \right]} ,\;\; \Rightarrow \frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{2} = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\left[ {\cos nx - \cos \left( { - \pi n} \right)} \right]} ,\;\; \Rightarrow \frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{2} = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} - 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\left( { - 1} \right)}^n}}}{{{n^2}}}} ,\;\; \Rightarrow \frac{{{x^2}}}{2} - \frac{{{\pi ^2}}}{2} = 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} - 2\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} .$

We know from Example $$1$$ on the page Bessel's Inequality and Parseval's Theorem that $$\zeta \left( 2 \right) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = {\frac{{{\pi ^2}}}{6}}.$$ Then we get

${x^2} - {\pi ^2} = 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} - \frac{{2{\pi ^2}}}{3}$

or

${x^2} = \frac{{{\pi ^2}}}{3} + 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{n^2}}}\cos nx} .$