Editing Dirichlet integral (section)

=== Dirichlet kernel ===
Consider the well-known formula for the [[Dirichlet kernel]]:<ref>{{cite report |url=https://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/ChenGuo.pdf |title=A Treatment of the Dirichlet Integral Via the Methods of Real Analysis |author=Chen, Guo |date=26 June 2009}}</ref><math display="block">
D_n(x) = 1 + 2\sum_{k=1}^n \cos(2kx) = \frac{\sin[(2n+1)x]}{\sin(x)}.
</math>

It immediately follows that:<math display="block">
\int_0^{\frac{\pi}{2}} D_n(x)\, dx = \frac{\pi}{2}.
</math>

Define
<math display="block">f(x) =
\begin{cases}
\frac{1}{x} - \frac{1}{\sin(x)} & x \neq 0 \\[6pt]
0 & x = 0
\end{cases}
</math>

Clearly, <math>f</math> is continuous when <math> x \in (0,\pi/2] ;</math> to see its continuity at 0 apply [[L'Hopital's Rule]]:
<math display="block">
\lim_{x\to 0} \frac{\sin(x) - x}{x\sin(x)} = 
\lim_{x\to 0} \frac{\cos(x) - 1}{\sin(x) + x\cos(x)} = 
\lim_{x\to 0} \frac{-\sin(x)}{2\cos(x) - x\sin(x)} = 0.
</math>

Hence, <math>f</math> fulfills the requirements of the [[Riemann-Lebesgue Lemma]]. This means:
<math display="block">
\lim_{\lambda \to \infty} \int_0^{\pi/2} f(x)\sin(\lambda x)dx = 0 
\quad\Longrightarrow\quad
\lim_{\lambda \to \infty} \int_0^{\pi/2} \frac{\sin(\lambda x)}{x}dx = 
    \lim_{\lambda \to \infty} \int_0^{\pi/2} \frac{\sin(\lambda x)}{\sin(x)}dx.
</math>

(The form of the Riemann-Lebesgue Lemma used here is proven in the article cited.)

We would like to compute:
<math display="block">
\begin{align}
\int_0^\infty \frac{\sin(t)}{t}dt
= & \lim_{\lambda \to \infty} \int_0^{\lambda\frac{\pi}{2}} \frac{\sin(t)}{t}dt \\[6pt]
= & \lim_{\lambda \to \infty} \int_0^{\frac{\pi}{2}} \frac{\sin(\lambda x)}{x}dx \\[6pt]
= & \lim_{\lambda \to \infty} \int_0^{\frac{\pi}{2}} \frac{\sin(\lambda x)}{\sin(x)}dx \\[6pt]
= & \lim_{n\to \infty} \int_0^{\frac{\pi}{2}} \frac{\sin((2n+1)x)}{\sin(x)}dx \\[6pt]
= & \lim_{n\to \infty} \int_0^{\frac{\pi}{2}} D_n(x) dx = \frac{\pi}{2}
\end{align} </math>

However, we must justify switching the real limit in <math>\lambda</math> to the integral limit in <math>n,</math> which will follow from showing that the limit does exist.

Using [[integration by parts]], we have:
<math display="block">
\int_a^b \frac{\sin(x)}{x}dx = 
\int_a^b \frac{d(1-\cos(x))}{x}dx = 
\left. \frac{1-\cos(x)}{x}\right|_a^b + \int_a^b \frac{1-\cos(x)}{x^2}dx
</math>

Now, as <math>a \to 0</math> and <math> b \to \infty</math> the term on the left converges with no problem. See the [[List of limits#Trigonometric functions|list of limits of trigonometric functions]].  We now show that <math> \int_{-\infty}^{\infty} \frac{1-\cos(x)}{x^2}dx </math> is absolutely integrable, which implies that the limit exists.<ref>{{cite report |url=http://ramanujan.math.trinity.edu/rdaileda/teach/m4342f10/improper_integrals.pdf |title=Improper Integrals |author=R.C. Daileda}}</ref>

First, we seek to bound the integral near the origin. Using the Taylor-series expansion of the cosine about zero,
<math display="block">
1 - \cos(x) = 1 - \sum_{k\geq 0}\frac{{(-1)^{(k+1)}}x^{2k}}{2k!} = \sum_{k\geq 1}\frac{{(-1)^{(k+1)}}x^{2k}}{2k!}.
</math>

Therefore,
<math display="block">
\left|\frac{1 - \cos(x)}{x^2}\right| 
  =  \left|-\sum_{k\geq 0}\frac{x^{2k}}{2(k+1)!}\right|
\leq \sum_{k\geq 0} \frac{|x|^k}{k!}
  =  e^{|x|}.
</math>

Splitting the integral into pieces, we have
<math display="block">
     \int_{-\infty}^{\infty}\left|\frac{1-\cos(x)}{x^2}\right|dx
\leq \int_{-\infty}^{-\varepsilon} \frac{2}{x^2}dx + 
     \int_{-\varepsilon}^{\varepsilon} e^{|x|}dx + 
     \int_{\varepsilon}^{\infty} \frac{2}{x^2}dx
\leq K,
</math>

for some constant <math>K > 0.</math>  This shows that the integral is absolutely integrable, which implies the original integral exists, and switching from <math>\lambda</math> to <math>n</math> was in fact justified, and the proof is complete.