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| In [[mathematics]], there are several [[integral]]s known as the '''''Dirichlet integral''''', after the German mathematician [[Peter Gustav Lejeune Dirichlet]].
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| One of those is
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| : <math>\int_0^\infty \frac{\sin \omega}{\omega}\,d\omega = \frac{\pi}{2}</math> | |
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| This integral is not [[absolutely convergent]], and so the integral is ''not even defined'' in the sense of [[Lebesgue integration]], but it ''is'' defined in the sense of the [[Riemann integral]] or the [[Henstock–Kurzweil integral]].<ref>Robert G. Bartle, [http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=2900 Return to the Riemann Integral], ''The American Mathematical Monthly'', vol. 103, 1996, pp. 625-632.</ref> The value of the integral (in the Riemann or Henstock sense) can be derived in various ways. For example, the value can be determined from attempts to evaluate a double [[improper integral]], or by using [[differentiation under the integral sign]].
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| == Evaluation ==
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| === Double improper integral method ===
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| Pre-knowledge of properties of [[Laplace transform#Evaluating improper integrals|Laplace transforms]] allows us to evaluate this Dirichlet integral succinctly in the following manner:
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| : <math>\int_0^\infty\frac{\sin t}{t}\, dt=\int_{0}^{\infty}\mathcal{L}\{\sin t\}(s)\; ds=\int_{0}^{\infty}\frac{1}{s^{2}+1}\, ds=\arctan s\bigg|_{0}^{\infty}=\frac{\pi}{2},
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| </math>
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| where <math>\mathcal{L}\{\sin t\}(s)</math> is the Laplace transform of the function, <math>\sin t</math> that returns the function <math>\tfrac{1}{s^{2}+1}</math>, a function of the Laplace transform variable, ''s''. This is equivalent to attempting to evaluate the same double definite integral in two different ways, by reversal of the [[order of integration (calculus)|order of integration]], ''viz.'',
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| : <math>\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 = \int_0^\infty \sin t{\int _0^\infty e^{-st}\, ds} \, dt\right ),</math>
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| : <math>\left ( I_1=\int_0^\infty {\frac{1}{s^2+1}}\, ds = \frac{\pi}{2}\right ) = \left ( I_2=\int_0^\infty \sin t\, \frac{1}{t} \, dt\right ) \text{, provided } s>0.</math>
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| === Differentiation under the integral sign ===
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| First rewrite the integral as a function of variable <math>\!a</math>. Let
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| : <math>f(a)=\int_0^\infty e^{-a\omega} \frac{\sin \omega}{\omega} d\omega ;</math>
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| then we need to find <math>\!f(0)</math>.
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| Differentiate with respect to <math>\!a</math> and apply the [[Leibniz integral rule]] to obtain:
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| : <math>\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 = -\int_0^\infty e^{-a\omega} \sin \omega \,d\omega = -\mathcal{L}\{\sin \omega\}(a).</math>
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| This integral was evaluated without proof, above, based on Laplace transform tables; we derive it this time. It is made much simpler by recalling Euler's formula,
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| : <math>\! e^{i\omega}=\cos \omega + i\sin \omega ,</math>
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| then,
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| : <math>\Im e^{i\omega}=\sin \omega,</math> where <math>\Im</math> represents the [[imaginary part]].
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| : <math>\therefore\frac{df}{da}=-\Im\int_0^\infty e^{-a\omega}e^{i\omega}d\omega=\Im\frac{1}{-a+i}=\Im\frac{-a-i}{a^2+1}=\frac{-1}{a^2+1} \text{, given that } a > 0 .</math>
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| Integrating with respect to <math>\!a</math>:
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| : <math>f(a) = \int \frac{-da}{a^2+1} = A - \arctan a,</math>
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| where <math>\! A</math> is a constant to be determined. As,
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| : <math>f(+\infty)=0 \therefore A = \arctan (+\infty) = \frac{\pi}{2} + m\pi,</math> | |
| : <math>\therefore f(0)=\lim _{a \to 0^+} f(a) = \frac{\pi}{2} + m\pi - \arctan 0 = \frac{\pi}{2} + n\pi,</math>
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| for some integers ''m'' & ''n''. It is easy to show that <math>\!n</math> has to be zero, by analyzing easily observed bounds for this integral:
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| : <math>0<\int _0^\infty \frac {\sin x}{x}dx < \int _0^\pi \frac {\sin x}{x}dx < \pi</math>
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| The left and right bounds can be derived by dividing the integrated region [0:∞] to periodic intervals, over which the integrals have zero value. Left bound, ∫[0:∞]sinx/x dx = ∫[0:2π]sinx/x dx + ∫[2π:4π]sinx/x dx + ... > ∫[0:2π]sinx/(2π) dx + ∫[2π:4π]sinx/(4π) dx + ... = 0. Right bound, ∫[0:∞]sinx/x dx = ∫[0:π] sinx/x dx + ∫[π:∞] sinx/x dx < ∫[0:π] sinx/x dx + ∫[π:∞] sinx dx. The second term is zero, which can be proved using the same approach as for the left bound. The first term, ∫[0:π] sinx/x dx < ∫[0:π] dx = π.
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| End of proof.
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| Extending this result further, with the introduction of another variable, first noting that <math>\! {\sin x}/{x}</math> is an even function and therefore
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| : <math>\int_0^\infty \frac{\sin x}{x}\,dx = \int_{-\infty}^0 \frac{\sin x}{x}\,dx = -\int_0^{-\infty} \frac{\sin x}{x}\,dx,</math>
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| then:
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| : <math>\int_0^\infty \frac{\sin b\,\omega}{\omega}\,d\omega = \int_0^{b\,\infty} \frac{\sin b\,\omega}{b\,\omega}\,d(b\,\omega) = \int_0^{\sgn b\times\infty} \frac{\sin x}{x}\,dx = \sgn b \int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}\,\sgn b</math>
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| === Complex integration ===
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| The same result can be obtained via complex integration. Let's consider
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| : <math> f(z)=\frac{e^{iz}}{z} </math> | |
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| As a function of the complex variable z, it has a simple pole at the origin, which prevents the application of [[Jordan's lemma]], whose other hypotheses are satisfied. We shall then define a new function<ref>Appel, Walter. ''Mathematics for Physics and Physicists''. Princeton University Press, 2007, p. 226.</ref> g(z) as follows
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| : <math> g(z)=\frac{e^{iz}}{z +i\epsilon} </math> | |
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| The pole has been moved away from the real axis, so g(z) can be integrated along the semicircle of radius R centered at z=0 and closed on the real axis, then the limit <math>\epsilon \rightarrow 0</math> should be taken.
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| The complex integral is zero by the residue theorem, as there are no poles inside the integration path
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| : <math> 0 = \int_\gamma g(z) dz = \int_{-R}^R \frac{e^{ix}}{x +i\epsilon} dx + \int_{0}^{\pi} \frac{e^{i(Re^{i\theta} + \theta)}}{Re^{i\theta} +i\epsilon} iR d\theta </math>
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| The second term vanishes as R goes to infinity; for arbitrarily small <math> \epsilon </math>, the [[Sokhotski–Plemelj theorem]] applied to the first one yields
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| : <math> 0= \mathrm{P.V.} \int \frac{e^{ix}}{x} dx - \pi i \int_{-\infty}^{\infty}\delta(x) e^{ix} dx </math> | |
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| Where P.V. indicates [[Cauchy principal value]]. By taking the imaginary part on both sides and noting that <math> \mathrm{sinc}(x) </math> is even and by definition <math> \mathrm{sinc}(0)=1 </math>, we get the desired result
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| : <math> \lim_{\epsilon\rightarrow 0}\int_\epsilon^{\infty} \frac{\sin(x)}{x} dx = \int_0^{\infty} \frac{\sin(x)}{x} dx = \frac{\pi}{2} </math>
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| == See also ==
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| * [[Dirichlet principle]]
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| ==Notes==
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| {{Reflist}}
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| == External links ==
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| * {{MathWorld | urlname=DirichletIntegrals | title=Dirichlet Integrals}}
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| [[Category:Calculus]]
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| [[Category:Special functions]]
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| [[Category:Integral calculus]]
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