Dirichlet Kernel
Definition
Dirichlet Kernel $D_{n}$ is defined as follows.
$$ \begin{equation} D_{n}(t) := \dfrac{1}{2}+\sum \limits_{k=1}^{n} \cos kt \end{equation} $$
Explanation
The Dirichlet kernel is related to delta functions, exponential functions, etc., and appears in Fourier analysis. Here are some related theorems and their proofs.
Theorem 1
The Dirichlet Kernel satisfies the following equation.
$$ D_{n}(t)=\dfrac{\sin\left(n+\frac{1}{2}\right) t}{2\sin \frac{1}{2}t} $$
Proof
If we express the cosine function as a complex exponential form, we get the following.
$$ \begin{align*} D_{n}(t) =&\ \dfrac{1}{2}+\dfrac{1}{2}\sum \limits_{k=1}^n( e^{ikt}+e^{-ikt} ) \\ =&\ \dfrac{1}{2} \left[ 1+\sum \limits_{k=1}^{n} (e^{ikt}+e^{-ikt} ) \right] \\ =&\ \dfrac{1}{2} \sum \limits_{k=-n}^{n} e^{ikt} \end{align*} $$
In this case
Geometric Series Summation Formula
$$ \sum_{k=1}^{n} a_{k}= \dfrac{a (r^{n} -1) }{ r-1 } $$
Using this, with the first term being $a_{1}=e^{-int}$ and the common ratio $r=e^{it}$, we can arrange it as follows.
$$ \begin{align*} D_{n}(t) =&\ \dfrac{1}{2} \sum \limits_{k=-n}^{n} e^{ikt} = \dfrac{1}{2} \sum \limits_{k=1}^{2n+1} e^{i(k-n-1)t} \\ =&\ \dfrac{1}{2} \dfrac{ ( e^{-int} ) \left( e^{i(2n+1)t -1} \right) }{e^{it}-1} \\ =&\ \dfrac{1}{2}e^{-int}\dfrac{e^{i(n+\frac{1}{2}) t }-e^{-i(n+\frac{1}{2})t} }{e^{i\frac{1}{2}t}-e^{-i\frac{1}{2}t }} \dfrac{e^{i(n+\frac{1}{2})t}} {e^{i\frac{1}{2}t}} \\ =&\ \dfrac{1}{2}\dfrac{e^{i(n+\frac{1}{2}) t }-e^{-i(n+\frac{1}{2})t} }{e^{i\frac{1}{2}t}-e^{-i\frac{1}{2}t }} \dfrac{e^{i(n+\frac{1}{2})t}} {e^{i(n+\frac{1}{2})t}} \\ =&\ \dfrac{1}{2}\dfrac{\sin (n+\frac{1}{2})t} {\sin \frac{1}{2} t} \end{align*} $$
In the last equality, we used that $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$.
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Theorem 2
Let’s say the following equation is the $2L$-periodic function $f(t)$’s Fourier series partial sum.
$$ \begin{equation} S_{N} ^{f} (t)=\dfrac{1}{2}a_{0}+\sum \limits_{n=1}^{N} \left( a_{n}\cos\dfrac{n\pi t}{L}+b_{n}\sin\frac{n\pi t}{L} \right) \end{equation} $$
Then, the partial sum $S_{N}^{f}(t)$ can be represented as an integral including the Dirichlet kernel as follows.
$$ S_{N}^{f} (t)=\dfrac{1}{L}\int_{-L}^{L}f(x)D_{n}\left(\dfrac{\pi (x-t)}{L}\right)dx $$
Proof
After calculating the Fourier coefficients $a_{0}$, $a_{n}$, $b_{n}$, we get the following.
$$ \begin{align*} a_{0} &= \dfrac{1}{L}\int_{-L}^{L} f(x) dx \\ a_{n} &= \dfrac{1}{L}\int_{-L}^{L}f(x)\cos\dfrac{n\pi x}{L} dx \\ b_{n} &= \dfrac{1}{L}\int_{-L}^{L}f(x)\sin\dfrac{n\pi x}{L} dx \end{align*} $$
Then we obtain the following equation.
$$ \begin{align*} & a_{n} \cos\dfrac{n\pi t}{L} + b_{n} \sin\dfrac{n\pi t}{L} \\ =&\ \left( \dfrac{1}{L}\int_{-L}^{L}f(x)\cos\dfrac{n\pi x}{L} dx \right) \cos\dfrac{n\pi t}{L} + \left( \dfrac{1}{L}\int_{-L}^{L}f(x)\sin\dfrac{n\pi x}{L} dx \right)\sin\dfrac{n\pi t}{L} \\ =&\ \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi x}{L} \cos\dfrac{n\pi t}{L} + \sin\dfrac{n\pi x}{L} \sin\dfrac{n\pi t}{L} \right] dx \end{align*} $$
Then, by the sum and difference identities for trigonometric functions, we get the following equation.
$$ a_{n}\cos\dfrac{n\pi t}{L} + b_{n}\cos\dfrac{n\pi t}{L} = \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi (x-t)}{L} \right] dx $$
Substituting this into $(2)$ yields the following.
$$ \begin{align*} S_{N} ^{f} (t) =&\ \dfrac{1}{2}a_{0}+\sum \limits_{n=1}^{N} \left( a_{n}\cos\dfrac{n\pi t}{L}+b_{n}\sin\frac{n\pi t}{L} \right) \\ =&\ \dfrac{1}{2}\dfrac{1}{L}\int_{-L}^{L} f(x)dx+\sum\limits_{n=1}^{N}\left( \dfrac{1}{L}\int_{-L}^{L}f(x) \left[ \cos\dfrac{n\pi (x-t)}{L} \right] dx \right) \\ =&\ \dfrac{1}{L} \int_{-L}^{L} f(x) \left[ \dfrac{1}{2} + \sum\limits_{n=1}^{N} \cos \dfrac{n\pi (x-t)}{L}\right]dx \\ =&\ \dfrac{1}{L}\int_{-L}^{L}f(x)D_{n}\left(\dfrac{\pi (x-t)}{L}\right)dx \end{align*} $$
In the last equality, the definition of the Dirichlet kernel was used.
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Theorem 3
For any integer $n \in \mathbb{Z}$, the following equation holds.
$$ \begin{equation} \dfrac{1}{L}\int_{-L}^{L}D_{n}\left( \dfrac{\pi (x-t)}{L} \right)dx=1 \end{equation} $$
Proof
Let’s substitute $\dfrac{\pi (x-t)}{L}=y$. Then, the left side of $(3)$ is as follows.
$$ \begin{align*} &\dfrac{1}{L}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} D_{n}(y) \dfrac{L}{\pi}dy \\ =&\ \dfrac{1}{\pi}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \left( \dfrac{1}{2} + \sum \limits_{n=1}^{N} \cos ny \right) dy & \text{ by } (1) \\ =&\ \dfrac{1}{2\pi}\int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} dy+ \sum \limits_{n=1}^{N} \int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \cos ny dy \\ =&\ \dfrac{1}{2\pi} 2\pi + \sum \limits_{n=1}^{N} \int_{-\pi -\frac{\pi}{L}t}^{\pi-\frac{\pi}{L}t} \cos ny dy \\ =&\ 1 \end{align*} $$
At this point, the reason why the second term integrates to become $0$ is because $\cos 0y=1$ and $\cos ny (n\ne 0)$ are orthogonal to each other.
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