We begin by using the well-known geometric series formula to derive the following formula for the sum of powers of a number #a\ne1# with exponents ranging from #k=-n# to #k=n#:

\[\begin{array}{c}\displaystyle \sum_{k=-n}^{n} a^{k} = \frac{a^{-n-\frac12}-a^{n+\frac12}}{a^{-\frac12}-a^{\frac12}} \end{array}\]

Indeed,

\[\begin{array}{rcl}\displaystyle \sum_{k=-n}^{n} a^{k} &=&\displaystyle a^{-n} \cdot \frac{1-a^{2n+1}}{1-a} \\&&\phantom{xxx}\color{blue}{\text{geometric series formula}}\\&=&\displaystyle \frac{a^{-n-\frac12}}{a^{-\frac12}} \cdot \frac{1-a^{2n+1}}{1-a}\\&&\phantom{xxx}\color{blue}{\text{both denominator and numerator multiplied by }a^{-\frac12}}\\ &=&\displaystyle \frac{a^{-n-\frac12}-a^{n+\frac12}}{a^{-\frac12}-a^{\frac12}}\\ &&\phantom{xxx}\color{blue}{\text{simplified}}\end{array}\]

By de Moivre's formula, we have #(\cos(x) + \ii\sin(x))^n =\cos(nx) + \ii\sin(nx)=\ee^{n\ii x}#. Since #\sin# is odd and #\cos# is even, we find, for #x# such that #\ee^{x\ii}\ne1#,

\[\begin{array}{rcl} D_n(x) &=&\displaystyle \frac{1}{\pi} \left( \frac{1}{2} + \sum_{k=1}^{n} \cos(kx) \right)\\&&\phantom{xxx}\color{blue}{\text{definition of }D_n}\\ &=&\displaystyle \frac{1}{2\pi} \sum_{k=-n}^{n} (\cos(kx) +\ii\sin(kx))\\&&\phantom{xxx}\color{blue}{\cos\text{ is even and }\sin\text{ is odd}}\\&=&\displaystyle \frac{1}{2\pi} \sum_{k=-n}^{n} \ee^{kx\ii} \\&&\phantom{xxx}\color{blue}{\text{ De Moivre's formula}}\\&=&\displaystyle\frac{1}{2\pi}\frac{\ee^{-(n+\frac12)\ii x}-\ee^{(n+\frac12)\ii x}}{\ee^{-\ii \frac{x}{2}}-\ee^{\ii \frac{x}{2}}} \\&&\phantom{xxx}\color{blue}{\text{ above version of geometric series formula with }a = \ee^{\ii x}}\\&=&\displaystyle\frac{1}{2\pi}\frac{-2\ii\sin((n+\frac12)x)}{-2\ii\sin(\frac{x}{2})} \\&&\phantom{xxx}\color{blue}{\ee^{x\ii}-\ee^{-x\ii} = 2\ii\sin(x)}\\ &=&\displaystyle\frac{1}{2\pi}\frac{\sin((n+\frac12)x)}{\sin(\frac{x}{2})} \\&&\phantom{xxx}\color{blue}{\text{ simplified}}\\ \end{array}\]

Since #\ee^{x\ii}\ne1# if and only if #x# is not an integer multiple of #2\pi#, this establishes the formula for #D_n(x)#.

We continue by calculating the first integral involving #D_n#:

\[\begin{array}{rcl}\displaystyle \int_{-\pi}^0 D_n(x)\,\dd x & =& \displaystyle\frac{1}{2\pi} \int_{-\pi}^0 \dd x +\frac{1}{\pi}\sum_{k=1}^{n} \int_{-\pi}^0\cos(kx)\dd x\\&&\phantom{xxx}\color{blue}{\text{definition of }D_n}\\ & =& \displaystyle\frac{1}{2\pi} \left[x\right]_{x=-\pi}^{x=0} +\sum_{k=1}^{n} \frac{1}{k\pi}\left[\sin(kx)\right]_{x=-\pi}^{x=0}\\&&\phantom{xxx}\color{blue}{\text{antiderivatives determined}}\\& =& \displaystyle\frac{1}{2\pi} \pi +\sum_{k=1}^{n} \frac{1}{k\pi}\cdot 0\\&&\phantom{xxx}\color{blue}{\text{antiderivatives evaluated at boundaries}}\\& =& \displaystyle\frac{1}{2}\\&&\phantom{xxx}\color{blue}{\text{simplified}}\\ \end{array}\]

Since #D_n# is an even function, the second definite integral has the same value. This ends the proof of the theorem.

In what follows we can fix #L = \pi# without loss of generality, as the properties we exploit only depend on the integration interval being symmetric.

Consider the trigonometric series

\[\displaystyle s_n(x) = \frac{a_0}{2}+\sum_{k=1}^n\left(a_k\cos(kx)+b_k\sin(kx)\right)\]

where #a_0#, #a_1,\ldots,a_n#, #b_1,\ldots,b_n# are given by the Euler formulas for #f#.

We will show that \(\lim_{n\to\infty}s_n \) exists and is a Fourier series of #f#. Using the Euler formulas, we can write

\[ s_n(x) = \frac{1}{\pi} \int_{-\pi}^{\pi}f(y)\, \left(\frac{1}{2} + \sum_{k=1}^{n}\left( \cos(kx)\cos(ky) + \sin(kx)\sin(ky)\right)\right) \dd y \]

Recall from Dirichlet kernel that the function #D_n# is defined by

\[\begin{array}{c}\displaystyle D_n(x) = \frac{1}{\pi} \left( \frac{1}{2} + \sum_{k=1}^{n} \cos(kx) \right) \end{array}\]

Applying the addition formula for the cosine of a difference, we find

\[\begin{array}{c}D_n(x-y) = \displaystyle \frac{1}{\pi} \left( \frac{1}{2} + \sum_{k=1}^{n} \cos(kx)\cos(ky) + \sin(kx)\sin(ky) \right) \end{array}\]

We use this identity to rewrite #s_n(x)#.

\[\begin{array}{rcl}\displaystyle s_n(x) &=&\displaystyle\int_{-\pi}^{\pi} f(y)D_n(x-y) \dd y \\&&\phantom{xxx}\color{blue}{\text{use of the identity for }D_n(x-y) \text{ gives the above expression for }s_n(x)} \\ & =&\displaystyle \int_{-\pi+x}^{\pi+x} f(x+z)D_n(z) \dd z \\&&\phantom{xxx}\color{blue}{\text{change of variable }y = x+z\text{ since }D_n\text{ is even}}\\& =&\displaystyle \int_{-\pi}^{\pi}f(x+z) D_n(z) \dd z \\&&\phantom{xxx}\color{blue}{f(x+z) D_n(z)\text{ is periodic in }{z}\text{ with period }2\pi}\end{array}\]

By using properties of the Dirichlet kernel we continue to rewrite #s_n(x)# as follows.

\[\begin{array}{rcl}\displaystyle s_n(x) &=& \displaystyle\int_{-\pi}^{0} f(x+z)D_n(z)\dd z + \int_{0}^{\pi} f(x+z)D_n(z)\dd z \\ &&\phantom{xxx}\color{blue}{\text{property of definite integrals}}\\ &=& \displaystyle\int_{-\pi}^{0} \left(f(x+z)-f_-(x)\right)D_n(x) \dd z +\frac{1}{2}f_-(x) \\ &&\,+ \displaystyle\int_{0}^{\pi} \left(f(x+z)-f_+(x)\right)D_n(x)\dd z +\frac{1}{2}f_+(x) \\&&\phantom{xxx}\color{blue}{\int_{-\pi}^0D_n(x)\,\dd x=\frac{1}{2}= \int_{0}^{\pi}D_n(x)}\\ &=& \displaystyle\frac{1}{2\pi}\int_{-\pi}^{0} \left(f(x+z)-f_-(x)\right) \frac{\sin((n+\frac12)z)}{\sin(\frac{z}{2})} \dd z +\frac{1}{2}f_-(x) \\ &&\,+ \displaystyle\frac{1}{2\pi}\int_{0}^{\pi} \left(f(x+z)-f_+(x)\right)\frac{\sin((n+\frac12)z)}{\sin(\frac{z}{2})} \dd z +\frac{1}{2}f_+(x) \\&&\phantom{xxx}\color{blue}{D_n(z) =\frac{1}{2\pi} \frac{\sin(((n+\frac12)z)}{\sin(\frac{z}{2})} \text{ for } z \neq 2k\pi }\\ &=& \displaystyle\frac{1}{2\pi}\int_{0}^{\pi}\left(g_{L}(z)+g_{R}(z)\right)\sin((n+\frac12)z) \dd z + \frac{f_-(x) +f_+(x)}{2}\end{array}\]

where

\[\begin{array}{c}\displaystyle g_{L}(y) = \frac{f(x-y)-f_-(x)}{\sin\left(\frac{y}{2}\right)}\phantom{xxx}\text{ and }\phantom{xxx}\displaystyle g_{R}(y) = \frac{f(x+y)-f_+(x)}{\sin\left(\frac{y}{2}\right)}\end{array}\]

We will finish the proof by establishing that the first term of the expression found for #s_n(x)# is equal to zero. Both functions #g_L# and #g_R# are piecewise continuous on #\ivcc{0}{\pi}#. For instance, #g_{R}# is the quotient of the piecewise continuous function #f(x+y)-f_+(x)# of #y# by the continuous function #\sin\left(\frac{y}{2}\right)# that has a zero only at #0#. But this function is well-behaved at #0# too. Before proving this, we first need to analyse the limit of #\frac{f(x+y)-f_+(x)}{y}# for #y\downarrow0#.

By the assumption that #f# is piecewise smooth, the right-hand limit #f'_+(x)=(f')_+(x)# of #f'(x+y)# for #y\downarrow0# exists. We assert that

\[\lim_{y\downarrow0}\frac{f(x+y)-f_+(x)}{y} = f'_+(x)\]

In order to see this, let #\varepsilon\gt0#. By the definition of right-hand limit for #f'# at #x#, there is #\delta\gt0# such that #f# is continuous on #\ivoo{0}{\delta}# and such that, for all #y\in\ivoo{0}{\delta}#, we have #\left|f'(x+y)-f'_+(x)\right|\lt \varepsilon#. Now let #y\in\ivoo{0}{\delta}#. By the Mean value theorem, there is a number #\eta\in\ivoo{0}{1}# such that

\[\frac{f(x+y)-f_+(x)}{y}= f'(x+\eta \cdot y)\]As a consequence,

\[\begin{array}{rcl}\displaystyle \left|\dfrac{f(x+y)-f_+(x)}{y}-f'_+(x)\right|&=& \displaystyle \left|f'(x+\eta \cdot y)-f'_+(x)\right | \lt \varepsilon\end{array}\]which proves the assertion. With this assertion, we calculate

\[\begin{array}{rcl}\displaystyle \lim_{y \downarrow 0} g_{R}(y) &=& \displaystyle\lim_{y \downarrow 0} \frac{f(x+y)-f_+(x)}{\sin(\frac{y}{2})} \\ &&\phantom{xxx}\blue{\text{function rule for }g_{R}} \\ &=& \displaystyle\lim_{y \downarrow 0}\left(\frac{f(x+y)-f_+(x)}{y}\cdot\frac{y}{\sin(\frac{y}{2})}\right)\\ &=& \displaystyle\lim_{y \downarrow 0}\frac{f(x+y)-f_+(x)}{y} \cdot\lim_{y \downarrow 0}\frac{y}{\sin(\frac{y}{2})}\\ &=& f'_+(x)\cdot 2 \\ &&\phantom{xxx}\blue{\text{by the above assertion and }\lim_{y \downarrow 0}\frac{\sin(y)}{y} =1} \\ &=&2 f'_+(x) \end{array}\]

Similarly, we can prove #\lim_{y \downarrow 0} g_{L}(y)=2 f'_-(x)#. In particular, the function #g_{L}+g_{R}# is Riemann integrable on #\ivcc{0}{\pi}#, so the Riemann-Lebesgue lemma gives

\[\begin{array}{l}\displaystyle \lim_{n \rightarrow \infty} \int_{0}^{\pi}\left(g_{L}(y)+g_{R}(y)\right)\cdot \sin((n+\frac12)y)\, \dd y = 0 \end{array}\]

As a consequence,

\[\displaystyle \lim_{n \rightarrow \infty} s_n(x)=\frac{f_-(x)+f_+(x)}{2} \]

If #f# is continuous at #x#, then #\frac{1}{2} \left(f_-(x)+f_+(x)\right)# coincides with #f(x)# and the Fourier series converges to #f(x)#.