It is a good illustration of special techniques for evaluating definite integrals. The sine integral, an antiderivative of the sinc function, is not an elementary function. However the improper definite integral can be determined in several ways: the Laplace transform, double integration, differentiating under the integral sign, contour integration, and the Dirichlet kernel.
In what follows, one needs the result which is the Laplace transform of the function (see the section 'Differentiating under the integral sign' for a derivation) as well as a version of Abel's theorem (a consequence of the final value theorem for the Laplace transform).
Therefore,
Double integration
Evaluating the Dirichlet integral using the Laplace transform is equivalent to calculating the same double definite integral by changing the order of integration, namely,
Differentiation under the integral sign (Feynman's trick)
First rewrite the integral as a function of the additional variable namely, the Laplace transform of So let
Now, using Euler's formula one can express the sine function in terms of complex exponentials:
Therefore,
Integrating with respect to gives
where is a constant of integration to be determined. Since using the principal value. This means that for
Finally, by continuity at we have as before.
Complex contour integration
Consider
As a function of the complex variable it has a simple pole at the origin, which prevents the application of Jordan's lemma, whose other hypotheses are satisfied.
The pole has been moved to the negative imaginary axis, so can be integrated along the semicircle of radius centered at extending in the positive imaginary direction, and closed along the real axis. One then takes the limit
The complex integral is zero by the residue theorem, as there are no poles inside the integration path :
The second term vanishes as 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 and with one finds
where denotes the Cauchy principal value. Back to the above original calculation, one can write
By taking the imaginary part on both sides and noting that the function is even, we get
Finally,
Alternatively, choose as the integration contour for the union of upper half-plane semicircles of radii and together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of and on the other hand, as and the integral's imaginary part converges to (here is any branch of logarithm on upper half-plane), leading to
Now, as and the term on the left converges with no problem. See the list of limits of trigonometric functions. We now show that 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,
Therefore,
Splitting the integral into pieces, we have
for some constant This shows that the integral is absolutely integrable, which implies the original integral exists, and switching from to was in fact justified, and the proof is complete.
^Bartle, Robert G. (10 June 1996). "Return to the Riemann Integral" (PDF). The American Mathematical Monthly. 103 (8): 625–632. doi:10.2307/2974874. JSTOR 2974874.
^Bartle, Robert G.; Sherbert, Donald R. (2011). "Chapter 10: The Generalized Riemann Integral". Introduction to Real Analysis. John Wiley & Sons. pp. 311. ISBN978-0-471-43331-6.
^Zill, Dennis G.; Wright, Warren S. (2013). "Chapter 7: The Laplace Transform". Differential Equations with Boundary-Value Problems. Cengage Learning. pp. 274-5. ISBN978-1-111-82706-9.
^Appel, Walter. Mathematics for Physics and Physicists. Princeton University Press, 2007, p. 226. ISBN978-0-691-13102-3.
^Chen, Guo (26 June 2009). A Treatment of the Dirichlet Integral Via the Methods of Real Analysis (PDF) (Report).
dirichlet, integral, confused, with, dirichlet, energy, mathematics, there, several, integrals, known, after, german, mathematician, peter, gustav, lejeune, dirichlet, which, improper, integral, sinc, function, over, positive, real, line, peter, gustav, lejeun. Not to be confused with Dirichlet energy 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 Peter Gustav Lejeune Dirichlet 0 sin x x d x p 2 displaystyle int 0 infty frac sin x x dx frac pi 2 This integral is not absolutely convergent meaning sin x x displaystyle left frac sin x x right is not Lebesgue integrable because the Dirichlet integral is infinite in the sense of Lebesgue integration It is however finite in the sense of the improper Riemann integral or the generalized Riemann or Henstock Kurzweil integral 1 2 This can be seen by using Dirichlet s test for improper integrals It is a good illustration of special techniques for evaluating definite integrals The sine integral an antiderivative of the sinc function is not an elementary function However the improper definite integral can be determined in several ways the Laplace transform double integration differentiating under the integral sign contour integration and the Dirichlet kernel Contents 1 Evaluation 1 1 Laplace transform 1 2 Double integration 1 3 Differentiation under the integral sign Feynman s trick 1 4 Complex contour integration 1 5 Dirichlet kernel 2 See also 3 References 4 External linksEvaluation EditLaplace transform Edit Let f t displaystyle f t be a function defined whenever t 0 displaystyle t geq 0 Then its Laplace transform is given byL f t F s 0 e s t f t d t 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 isL f t t s F u d u displaystyle mathcal L left frac f t t right int s infty F u du provided lim t 0 f t t displaystyle lim t to 0 frac f t t exists In what follows one needs the result L sin t 1 s 2 1 displaystyle mathcal L sin t frac 1 s 2 1 which is the Laplace transform of the function sin t displaystyle sin t see the section Differentiating under the integral sign for a derivation as well as a version of Abel s theorem a consequence of the final value theorem for the Laplace transform Therefore 0 sin t t d t lim s 0 0 e s t sin t t d t lim s 0 L sin t t lim s 0 s d u u 2 1 lim s 0 arctan u s lim s 0 p 2 arctan s p 2 displaystyle begin aligned int 0 infty frac sin t t dt amp lim s to 0 int 0 infty e st frac sin t t dt lim s to 0 mathcal L left frac sin t t right 6pt amp lim s to 0 int s infty frac du u 2 1 lim s to 0 arctan u Biggr s infty 6pt amp lim s to 0 left frac pi 2 arctan s right frac pi 2 end aligned Double integration Edit Evaluating the Dirichlet integral using the Laplace transform is equivalent to calculating the same double definite integral by changing the order of integration namely I 1 0 0 e s t sin t d t d s I 2 0 0 e s t sin t d s d t 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 I 1 0 1 s 2 1 d s p 2 I 2 0 sin t t d t provided s gt 0 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 gt 0 Differentiation under the integral sign Feynman s trick Edit First rewrite the integral as a function of the additional variable s displaystyle s namely the Laplace transform of sin t t displaystyle frac sin t t So letf s 0 e s t sin t t d t displaystyle f s int 0 infty e st frac sin t t dt In order to evaluate the Dirichlet integral we need to determine f 0 displaystyle f 0 The continuity of f displaystyle f can be justified by applying the dominated convergence theorem after integration by parts Differentiate with respect to s gt 0 displaystyle s gt 0 and apply the Leibniz rule for differentiating under the integral sign to obtaind f d s d d s 0 e s t sin t t d t 0 s e s t sin t t d t 0 e s t sin t d t displaystyle begin aligned frac df ds amp frac d ds int 0 infty e st frac sin t t dt int 0 infty frac partial partial s e st frac sin t t dt 6pt amp int 0 infty e st sin t dt end aligned Now using Euler s formula e i t cos t i sin t displaystyle e it cos t i sin t one can express the sine function in terms of complex exponentials sin t 1 2 i e i t e i t displaystyle sin t frac 1 2i left e it e it right Therefore d f d s 0 e s t sin t d t 0 e s t e i t e i t 2 i d t 1 2 i 0 e t s i e t s i d t 1 2 i 1 s i e t s i 1 s i e t s i 0 1 2 i 0 1 s i 1 s i 1 2 i 1 s i 1 s i 1 2 i s i s i s 2 1 1 s 2 1 displaystyle begin aligned frac df ds amp int 0 infty e st sin t dt int 0 infty e st frac e it e it 2i dt 6pt amp frac 1 2i int 0 infty left e t s i e t s i right dt 6pt amp frac 1 2i left frac 1 s i e t s i frac 1 s i e t s i right 0 infty 6pt amp frac 1 2i left 0 left frac 1 s i frac 1 s i right right frac 1 2i left frac 1 s i frac 1 s i right 6pt amp frac 1 2i left frac s i s i s 2 1 right frac 1 s 2 1 end aligned Integrating with respect to s displaystyle s givesf s d s s 2 1 A arctan s displaystyle f s int frac ds s 2 1 A arctan s where A displaystyle A is a constant of integration to be determined Since lim s f s 0 displaystyle lim s to infty f s 0 A lim s arctan s p 2 displaystyle A lim s to infty arctan s frac pi 2 using the principal value This means that for s gt 0 displaystyle s gt 0 f s p 2 arctan s displaystyle f s frac pi 2 arctan s Finally by continuity at s 0 displaystyle s 0 we have f 0 p 2 arctan 0 p 2 displaystyle f 0 frac pi 2 arctan 0 frac pi 2 as before Complex contour integration Edit Considerf z e i z z displaystyle f z frac e iz z As a function of the complex variable z 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 g z e i z z i e displaystyle g z frac e iz z i varepsilon The pole has been moved to the negative imaginary axis so g z displaystyle g z can be integrated along the semicircle g displaystyle gamma of radius R displaystyle R centered at z 0 displaystyle z 0 extending in the positive imaginary direction and closed along the real axis One then takes the limit e 0 displaystyle varepsilon to 0 The complex integral is zero by the residue theorem as there are no poles inside the integration path g displaystyle gamma 0 g g z d z R R e i x x i e d x 0 p e i R e i 8 8 R e i 8 i e i R d 8 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 R 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 a displaystyle a and b displaystyle b with a lt 0 lt b displaystyle a lt 0 lt b one findslim e 0 a b f x x i e d x i p f 0 P a b f x x d x 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 P displaystyle mathcal P denotes the Cauchy principal value Back to the above original calculation one can write0 P e i x x d x p i 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 sin x x displaystyle sin x x is even we get sin x x d x 2 0 sin x x d x displaystyle int infty infty frac sin x x dx 2 int 0 infty frac sin x x dx Finally lim e 0 e sin x x d x 0 sin x x d x p 2 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 f displaystyle f the union of upper half plane semicircles of radii e displaystyle varepsilon and R displaystyle R together with two segments of the real line that connect them On one hand the contour integral is zero independently of e displaystyle varepsilon and R displaystyle R on the other hand as e 0 displaystyle varepsilon to 0 and R displaystyle R to infty the integral s imaginary part converges to 2 I ℑ ln 0 ln p i 2 I p displaystyle 2I Im big ln 0 ln pi i big 2I pi here ln z displaystyle ln z is any branch of logarithm on upper half plane leading to I p 2 displaystyle I frac pi 2 Dirichlet kernel Edit Consider the well known formula for the Dirichlet kernel 5 D n x 1 2 k 1 n cos 2 k x sin 2 n 1 x sin x displaystyle D n x 1 2 sum k 1 n cos 2kx frac sin 2n 1 x sin x It immediately follows that 0 p 2 D n x d x p 2 displaystyle int 0 frac pi 2 D n x dx frac pi 2 Definef x 1 x 1 sin x x 0 0 x 0 displaystyle f x begin cases frac 1 x frac 1 sin x amp x neq 0 6pt 0 amp x 0 end cases Clearly f displaystyle f is continuous when x 0 p 2 displaystyle x in 0 pi 2 to see its continuity at 0 apply L Hopital s Rule lim x 0 sin x x x sin x lim x 0 cos x 1 sin x x cos x lim x 0 sin x 2 cos x x sin x 0 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 f displaystyle f fulfills the requirements of the Riemann Lebesgue Lemma This means lim l 0 p 2 f x sin l x d x 0 lim l 0 p 2 sin l x x d x lim l 0 p 2 sin l x sin x d x displaystyle 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 The form of the Riemann Lebesgue Lemma used here is proven in the article cited We would like to compute 0 sin t t d t lim l 0 l p 2 sin t t d t lim l 0 p 2 sin l x x d x lim l 0 p 2 sin l x sin x d x lim n 0 p 2 sin 2 n 1 x sin x d x lim n 0 p 2 D n x d x p 2 displaystyle begin aligned int 0 infty frac sin t t dt amp lim lambda to infty int 0 lambda frac pi 2 frac sin t t dt 6pt amp lim lambda to infty int 0 frac pi 2 frac sin lambda x x dx 6pt amp lim lambda to infty int 0 frac pi 2 frac sin lambda x sin x dx 6pt amp lim n to infty int 0 frac pi 2 frac sin 2n 1 x sin x dx 6pt amp lim n to infty int 0 frac pi 2 D n x dx frac pi 2 end aligned However we must justify switching the real limit in l displaystyle lambda to the integral limit in n displaystyle n which will follow from showing that the limit does exist Using integration by parts we have a b sin x x d x a b d 1 cos x x d x 1 cos x x a b a b 1 cos x x 2 d x 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 a 0 displaystyle a to 0 and b 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 1 cos x x 2 d x 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 1 cos x 1 k 0 1 k 1 x 2 k 2 k k 1 1 k 1 x 2 k 2 k displaystyle 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 Therefore 1 cos x x 2 k 0 x 2 k 2 k 1 k 0 x k k e x 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 1 cos x x 2 d x e 2 x 2 d x e e e x d x e 2 x 2 d x K 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 K gt 0 displaystyle K gt 0 This shows that the integral is absolutely integrable which implies the original integral exists and switching from l displaystyle lambda to n displaystyle n was in fact justified and the proof is complete See also Edit Mathematics portalDirichlet distribution Dirichlet principle Sinc function Fresnel integralReferences Edit Bartle Robert G 10 June 1996 Return to the Riemann Integral PDF The American Mathematical Monthly 103 8 625 632 doi 10 2307 2974874 JSTOR 2974874 Bartle Robert G Sherbert Donald R 2011 Chapter 10 The Generalized Riemann Integral Introduction to Real Analysis John Wiley amp Sons pp 311 ISBN 978 0 471 43331 6 Zill Dennis G Wright Warren S 2013 Chapter 7 The Laplace Transform Differential Equations with Boundary Value Problems Cengage Learning pp 274 5 ISBN 978 1 111 82706 9 Appel Walter Mathematics for Physics and Physicists Princeton University Press 2007 p 226 ISBN 978 0 691 13102 3 Chen Guo 26 June 2009 A Treatment of the Dirichlet Integral Via the Methods of Real Analysis PDF Report R C Daileda Improper Integrals PDF Report External links EditWeisstein Eric W Dirichlet Integrals MathWorld Retrieved from https en wikipedia org w index php title Dirichlet integral amp oldid 1157650915, wikipedia, wiki, book, books, library,