Example 1: Consider the following square wave functions on interval \( (0, 2\ell ) : \)

The Fourier series for the characteristic function is

Example 2: On the interval [-ℓ, ℓ], consider three saw-tooth functions

Example 3: We start with the Heaviside function \[ H(t) = \begin{cases} 1, \qquad& \mbox{for } \ t > 0 , \\ \frac{1}{2} , \qquad& \mbox{if } \ t = 0 , \\ 0, \qquad& \mbox{for } t < 0 . \end{cases} \] We restrict this function on interval [−ℓ, ℓ] and then extend it periodically with period 2ℓ, where ℓ is any positive number. The periodic Heaviside function is piecewise continuous ℭ¹. Its derivative is almost everywhere zero except points nℓ (n ∈ ℤ), where its is undefined. If you are familiar with generating functions, you should recognize that the derivative of the Heaviside function (in terms of distributions) is the Dirac delta function.

The Fourier series is pointwise convergent everywhere with the sum function H(t): \[ H(t) = \frac{1}{2} + \frac{2}{\pi} \ \sum_{k\ge 1}\, \frac{1}{2k-1}\ \sin \left( (2k-1)\,\frac{\pi}{\ell}\,t \right) , \] because its coefficients are \begin{align*} a_0 &= \frac{1}{\ell} \,\int_0^{\ell}\, 1\,{\text d}t = 1 , \\ a_n &= \frac{1}{\ell} \,\int_0^{\ell}\, \cos \left( n\,\frac{\pi}{\ell}\,t \right) \,{\text d}t = 0, \\ b_n &= \frac{1}{\ell} \,\int_0^{\ell}\, \sin \left( n\,\frac{\pi}{\ell}\,t \right) \,{\text d}t = \frac{1 - \cos (n\pi )}{n\pi} , \qquad n=1,2,\ldots . \end{align*}

Example B: Now we consider a shifted Heaviside function, H(𝑡 − 𝑎), where 𝑎 ∈ (−ℓ, ℓ). Using Mathematica, we find the corresponding Fourier coefficients: \begin{align*} a_0 &= \frac{1}{\ell} \,\int_a^{\ell}\, 1\,{\text d}t = \frac{\ell - a}{\ell} , \\ a_n &= \frac{1}{\ell} \,\int_a^{\ell}\, \cos \left( n\,\frac{\pi}{\ell}\,t \right) \,{\text d}t = - \frac{1}{n\pi}\, \sin \left( \frac{an\pi}{\ell} \right), \\ b_n &= \frac{1}{\ell} \,\int_a^{\ell}\, \sin \left( n\,\frac{\pi}{\ell}\,t \right) \,{\text d}t = \frac{1}{n\pi} \left( \cos \left( \frac{an\pi}{\ell} \right) - (-1)^n \right) , \qquad n=1,2,\ldots . \end{align*}

Example C: Our next function is f(𝑡) = H(𝑡)sin(𝑡). It is natural to consider this function on interval {−π, π] and the expand it periodically with period 2π. Then 𝑓(𝑡) will be continuous on whole line ℝ. Moreover, it is piecewise smooth ℭ¹, and its derivative has finite discontinuities at 𝑡 = nπ (n ∈ ℤ) being corner points of the given function 𝑓(𝑡).

Since 𝑓(𝑡) is piecerwise smooth, it has convergent everywhere Fourier series, and its coefficients are evaluated by the standard Euler--Fourier formulas: \begin{align*} a_0 &= \frac{1}{\pi} \,\int_0^{\pi}\, \sin x\,{\text d}t = \frac{2}{\pi} , \\ a_n &= \frac{1}{\pi} \,\int_0^{\pi}\ \sin t\, \cos \left( n\,t \right) \,{\text d}t = \frac{1 + (-1)^n}{\left( 1 - n^2 \right) \pi} , \qquad n\ne 1, \\ b_n &= \frac{1}{\pi} \,\int_0^{\pi}\, \sin t\,\sin \left( n\,t \right) \,{\text d}t = \frac{1}{2}\,\delta_{n,1} , \qquad n=1,2,\ldots . \end{align*}

If we set 𝑡 = 0, we get \[ 0 = H(0)\,\sin (0) = \frac{1}{\pi} + \frac{2}{\pi}\ \sum_{k\ge 1}\,\frac{1}{1 - 4 k^2} . \] This leads to the identity: \[ \sum_{k\ge 1}\,\frac{1}{4 k^2 -1} = \frac{1}{2} . \] 🔶 ❖ ♠ ♣ ♥ ♦ ❖ 🔶 ﹡ ⁎ ✱ ✲ ✳ ✺ ✻ ✼ ✽ ❋

Example D: We consider difference of two Heaviside functions that defines a window function: \[ f_4 (t) = H(t-\pi /2) -H(t+\pi /2) , \qquad t \in [-\pi , pi] , \] with natural periodic extension that makes 𝑓₄ 2π-periodic function. This function is piecerwise continuous with jumps at points 𝑡 = π/2 + 𝑚π, 𝑚 = 0, ±1, ±2, …. Therefore, we expect Fourier coefficients of function 𝑓₄(𝑡) to decay linearly as O(1/n). Note that its derivative is zero almost everywhere. First, using Mathematica, we evaluate the Fourier coefficients: \begin{align*} a_0 &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi /2}\, \,{\text d}t = 0 , \\ a_n &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi}\ \cos \left( n\,t \right) \,{\text d}t = \frac{2}{n\pi} \cdot \sin \left( \frac{n\pi}{2} \right)n , \qquad n\ne 1, \\ b_n &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi}\,\,\sin \left( n\,t \right) \,{\text d}t = 0 , \qquad n=1,2,\ldots . \end{align*}

Example E: Our next function is \[ f_5 (t) = \left[ H(t-\pi /2) -H(t+\pi /2)\right] \sin (t) , \qquad t \in [-\pi , pi] , \] extended periodically with period 2π. According to Euler--Fourier formulas, its Fourier coefficients are \begin{align*} a_0 &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi/2}\, \sin x\,{\text d}t = 0 , \\ a_n &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi /2}\ \sin t\, \cos \left( n\,t \right) \,{\text d}t = 0 , \qquad n\ne 1, \\ b_n &= \frac{1}{\pi} \,\int_{-\pi /2}^{\pi /2}\ \sin t\,\sin \left( n\,t \right) \,{\text d}t = \frac{2}{\pi} \cdot \frac{n}{n^2 -1}\ \cos \left( \frac{n\pi}{2} \right) , \qquad n\ne 1 . \end{align*}

Example 4:    ■

1. Show that \( \displaystyle \quad \sum_{k\ge 1}\,\frac{(-1)^{k-1}}{(2k-1)^3} \, = \,\frac{\pi^3}{32} \ . \)

1. M. Abramowitz and I. A. Stegun (eds.), Handbook of Mathematical Functions.
2. Lewin, L., Polylogarithms and Associated Functions, Elsevier Science Ltd.