Dirichlet integral

 In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet, one of which is the improper integral of the sinc function over the positive real line:

${\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\,dx={\frac {\pi }{2}}.}$

Peter Gustav Lejeune Dirichlet

This integral is not absolutely convergent, meaning ${\displaystyle {\Biggl |}{\frac {\sin x}{x}}{\Biggl |}}$ is not Lebesgue-integrable, and so the Dirichlet integral is undefined in the sense of Lebesgue integration. It is, however, defined in the sense of the improper Riemann integral or the generalized Riemann or Henstock–Kurzweil integral.^[1][2] The value of the integral (in the Riemann or Henstock sense) can be derived using various ways, including the Laplace transform, double integration, differentiating under the integral sign, contour integration, and the Dirichlet kernel.

EvaluationEdit

Laplace transformEdit

Let ${\displaystyle f(t)}$ be a function defined whenever ${\displaystyle t\geq 0}$. Then its Laplace transform is given by

${\displaystyle {\mathcal {L}}\{f(t)\}=F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt,}$

if the integral exists.^[3]

A property of the Laplace transform useful for evaluating improper integrals is

${\displaystyle {\mathcal {L}}{\Biggl [}{\frac {f(t)}{t}}{\Biggl ]}=\int _{s}^{\infty }F(u)\,du,}$

provided ${\displaystyle \lim _{t\rightarrow 0}{\frac {f(t)}{t}}}$ exists.

One may use this property to evaluate the Dirichlet integral as follows:

${\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin t}{t}}\,dt&=\lim _{s\rightarrow 0}\int _{0}^{\infty }e^{-st}{\frac {\sin t}{t}}\,dt=\lim _{s\rightarrow 0}{\mathcal {L}}{\Biggl [}{\frac {\sin t}{t}}{\Biggl ]}\\[6pt]&=\lim _{s\rightarrow 0}\int _{s}^{\infty }{\frac {du}{u^{2}+1}}=\lim _{s\rightarrow 0}\arctan u{\Biggl |}_{s}^{\infty }\\[6pt]&=\lim _{s\rightarrow 0}{\Biggl [}{\frac {\pi }{2}}-\arctan(s){\Biggl ]}={\frac {\pi }{2}},\end{aligned}}}$

because ${\displaystyle {\mathcal {L}}\{\sin t\}={\frac {1}{s^{2}+1}}}$ is the Laplace transform of the function ${\displaystyle \sin t}$. (See the section 'Differentiating under the integral sign' for a derivation.)

Double integrationEdit

Evaluating the Dirichlet integral using the Laplace transform is equivalent to attempting to evaluate the same double definite integral in two different ways, by reversal of the order of integration, namely:

${\displaystyle \left(I_{1}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,dt\,ds\right)=\left(I_{2}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,ds\,dt\right),}$

${\displaystyle \left(I_{1}=\int _{0}^{\infty }{\frac {1}{s^{2}+1}}\,ds={\frac {\pi }{2}}\right)=\left(I_{2}=\int _{0}^{\infty }{\frac {\sin t}{t}}\,dt\right),{\text{ provided }}s>0.}$

Differentiation under the integral sign (Feynman's trick)Edit

First rewrite the integral as a function of the additional variable ${\displaystyle a}$. Let

${\displaystyle f(a)=\int _{0}^{\infty }e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega .}$

In order to evaluate the Dirichlet integral, we need to determine${\displaystyle f(0)}$.

Differentiate with respect to ${\displaystyle a}$ and apply the Leibniz rule for differentiating under the integral sign to obtain

${\displaystyle {\begin{aligned}{\frac {df}{da}}&={\frac {d}{da}}\int _{0}^{\infty }e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega =\int _{0}^{\infty }{\frac {\partial }{\partial a}}e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega \\[6pt]&=-\int _{0}^{\infty }e^{-a\omega }\sin \omega \,d\omega .\end{aligned}}}$

Now, using Euler's formula ${\displaystyle e^{i\omega }=\cos \omega +i\sin \omega ,}$ one can express a sinusoid in terms of complex exponential functions. We thus have

${\displaystyle \sin(\omega )={\frac {1}{2i}}\left(e^{i\omega }-e^{-i\omega }\right).}$

Therefore,

${\displaystyle {\begin{aligned}{\frac {df}{da}}&=-\int _{0}^{\infty }e^{-a\omega }\sin \omega \,d\omega =-\int _{0}^{\infty }e^{-a\omega }{\frac {e^{i\omega }-e^{-i\omega }}{2i}}d\omega \\[6pt]&=-{\frac {1}{2i}}\int _{0}^{\infty }\left[e^{-\omega (a-i)}-e^{-\omega (a+i)}\right]d\omega \\[6pt]&=-{\frac {1}{2i}}\left[{\frac {-1}{a-i}}e^{-\omega (a-i)}-{\frac {-1}{a+i}}e^{-\omega (a+i)}\right]{\Biggl |}_{0}^{\infty }\\[6pt]&=-{\frac {1}{2i}}\left[0-\left({\frac {-1}{a-i}}+{\frac {1}{a+i}}\right)\right]=-{\frac {1}{2i}}\left({\frac {1}{a-i}}-{\frac {1}{a+i}}\right)\\[6pt]&=-{\frac {1}{2i}}\left({\frac {a+i-(a-i)}{a^{2}+1}}\right)=-{\frac {1}{a^{2}+1}}.\end{aligned}}}$

Integrating with respect to ${\displaystyle a}$ gives

${\displaystyle f(a)=\int {\frac {-da}{a^{2}+1}}=A-\arctan a,}$

where ${\displaystyle A}$ is a constant of integration to be determined. Since ${\displaystyle \lim _{a\to \infty }f(a)=0,}$ ${\displaystyle A=\lim _{a\to \infty }\arctan(a)={\frac {\pi }{2}},}$ using the principal value. This means

${\displaystyle f(a)={\frac {\pi }{2}}-\arctan {a}.}$

Finally, for ${\displaystyle a=0}$, we have ${\displaystyle f(0)={\frac {\pi }{2}}-\arctan(0)={\frac {\pi }{2}}}$, as before.

Complex integrationEdit

The same result can be obtained by complex integration. Consider

${\displaystyle f(z)={\frac {e^{iz}}{z}}.}$

As a function of the complex variable ${\displaystyle z}$, it has a simple pole at the origin, which prevents the application of Jordan's lemma, whose other hypotheses are satisfied.

Define then a new function^[4]

${\displaystyle g(z)={\frac {e^{iz}}{z+i\varepsilon }}.}$

The pole has been moved away from the real axis, so ${\displaystyle g(z)}$ can be integrated along the semicircle of radius ${\displaystyle R}$ centered at ${\displaystyle z=0}$ and closed on the real axis. One then takes the limit ${\displaystyle \varepsilon \rightarrow 0}$.

The complex integral is zero by the residue theorem, as there are no poles inside the integration path

${\displaystyle 0=\int _{\gamma }g(z)\,dz=\int _{-R}^{R}{\frac {e^{ix}}{x+i\varepsilon }}\,dx+\int _{0}^{\pi }{\frac {e^{i(Re^{i\theta }+\theta )}}{Re^{i\theta }+i\varepsilon }}iR\,d\theta .}$

The second term vanishes as ${\displaystyle R}$ goes to infinity. As for the first integral, one can use one version of the Sokhotski–Plemelj theorem for integrals over the real line: for a complex-valued function f defined and continuously differentiable on the real line and real constants ${\displaystyle a}$ and ${\displaystyle b}$ with ${\displaystyle a<0<b}$ one finds

${\displaystyle \lim _{\varepsilon \to 0^{+}}\int _{a}^{b}{\frac {f(x)}{x\pm i\varepsilon }}\,dx=\mp i\pi f(0)+{\mathcal {P}}\int _{a}^{b}{\frac {f(x)}{x}}\,dx,}$

where ${\displaystyle {\mathcal {P}}}$ denotes the Cauchy principal value. Back to the above original calculation, one can write

${\displaystyle 0={\mathcal {P}}\int {\frac {e^{ix}}{x}}\,dx-\pi i.}$

By taking the imaginary part on both sides and noting that the function ${\displaystyle \sin(x)/x}$ is even, we get

${\displaystyle \int _{-\infty }^{+\infty }{\frac {\sin(x)}{x}}\,dx=2\int _{0}^{+\infty }{\frac {\sin(x)}{x}}\,dx.}$

Finally,

${\displaystyle \lim _{\varepsilon \to 0}\int _{\varepsilon }^{\infty }{\frac {\sin(x)}{x}}\,dx=\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.}$

Alternatively, choose as the integration contour for ${\displaystyle f}$ the union of upper half-plane semicircles of radii ${\displaystyle \varepsilon }$ and ${\displaystyle R}$ together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of ${\displaystyle \varepsilon }$ and ${\displaystyle R}$; on the other hand, as ${\displaystyle \varepsilon \to 0}$ and ${\displaystyle R\to \infty }$ the integral's imaginary part converges to ${\displaystyle 2I+\Im {\big (}\ln 0-\ln(\pi i){\big )}=2I-\pi }$ (here ${\displaystyle \ln z}$ is any branch of logarithm on upper half-plane), leading to ${\displaystyle I={\frac {\pi }{2}}}$.

Dirichlet kernelEdit

Let

${\displaystyle D_{n}(x)=1+2\sum _{k=1}^{n}\cos(2kx)={\frac {\sin[(2n+1)x]}{\sin(x)}}}$

be the Dirichlet kernel.^[5]

It immediately follows that${\displaystyle \int _{0}^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}.}$

Define

${\displaystyle f(x)={\begin{cases}{\frac {1}{x}}-{\frac {1}{\sin(x)}}&x\neq 0\\[6pt]0&x=0\end{cases}}}$

Clearly, ${\displaystyle f}$ is continuous when ${\displaystyle x\neq 0}$, to see its continuity at 0 apply L'Hopital's Rule:

${\displaystyle \lim _{x\to 0}{\frac {\sin(x)-x}{x\sin(x)}}=\lim _{x\to 0}{\frac {\cos(x)-1}{\sin(x)+x\cos(x)}}=\lim _{x\to 0}{\frac {-\sin(x)}{2\cos(x)-x\sin(x)}}=0.}$

Hence, ${\displaystyle f}$ fulfills the requirements of the Riemann-Lebesgue Lemma. This means

${\displaystyle \lim _{\lambda \to \infty }\int _{a}^{b}f(x)\sin(\lambda x)dx=0\Rightarrow \lim _{\lambda \to \infty }\int _{a}^{b}{\frac {\sin(\lambda x)}{x}}dx=\lim _{\lambda \to \infty }\int _{a}^{b}{\frac {\sin(\lambda x)}{\sin(x)}}dx.}$

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

Choose limits ${\displaystyle a=0}$ and ${\displaystyle b=\pi /2}$. We would like to say that

${\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin(t)}{t}}dt=&\lim _{\lambda \to \infty }\int _{0}^{\lambda {\frac {\pi }{2}}}{\frac {\sin(t)}{t}}dt\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{x}}dx\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin((2n+1)x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}\end{aligned}}}$

In order to do so, however, we must justify switching the real limit in ${\displaystyle \lambda }$ to the integral limit in ${\displaystyle n}$. This is in fact justified if we can show the limit does exist, which we do now.

Using integration by parts, we have:

${\displaystyle \int _{a}^{b}{\frac {\sin(x)}{x}}dx=\int _{a}^{b}{\frac {d(1-\cos(x))}{x}}dx=\left.{\frac {1-\cos(x)}{x}}\right|_{a}^{b}+\int _{a}^{b}{\frac {1-\cos(x)}{x^{2}}}dx}$

Now, as ${\displaystyle a\to 0}$ and ${\displaystyle b\to \infty }$ the term on the left converges with no problem. See the list of limits of trigonometric functions. We now show that ${\displaystyle \int _{-\infty }^{\infty }{\frac {1-\cos(x)}{x^{2}}}dx}$ is absolutely integrable, which implies that the limit exists.^[6]

First, we seek to bound the integral near the origin. Using the Taylor-series expansion of the cosine about zero,

${\displaystyle 1-\cos(x)=1-\sum _{k\geq 0}{\frac {x^{2k}}{2k!}}=-\sum _{k\geq 1}{\frac {x^{2k}}{2k!}}.}$

Therefore,

${\displaystyle \left|{\frac {1-\cos(x)}{x^{2}}}\right|=\left|-\sum _{k\geq 0}{\frac {x^{2k}}{2(k+1)!}}\right|\leq \sum _{k\geq 0}{\frac {|x|^{k}}{k!}}=e^{|x|}.}$

Splitting the integral into pieces, we have

${\displaystyle \int _{-\infty }^{\infty }\left|{\frac {1-\cos(x)}{x^{2}}}\right|dx\leq \int _{-\infty }^{-\varepsilon }{\frac {2}{x^{2}}}dx+\int _{-\varepsilon }^{\varepsilon }e^{|x|}dx+\int _{\varepsilon }^{\infty }{\frac {2}{x^{2}}}dx\leq K,}$

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

This article uses material from the Wikipedia article  Metasyntactic variable, which is released under the  Creative Commons Attribution-ShareAlike 3.0 Unported License.