s7.3.ImproperIntegralsFromFourierAnalysisResidue theorycanbe useful in evaluating convergent improper integrals of theform["f(x)sinaxdx or Jf(x)cosaxdx,(7.3.1)where a denotes a positive constant. As in Sec. 7.1, we assume that f(x)= p(x)/q(x),where p(x) and q(x) are polynomials with real coefficients and no factors in common.Alsoq(=) has no real zeros. Integrals of type (7.3.1) occur in the theory and application of the Fourierintegral.The method described in Sec.7.1 and used in Sec.7.2 cannot be applied directly here since(see Sec. 3.6)Isin az '= sin ax + sinh’ ayandI cos ax P= cos? ax + sinh’ ayMore precisely, sinceeay-e-aysinh ay2the moduli Isin az| and Icosaz increase like ey as y tends to infinity. The modificationillustrated in the example below is suggested by thefact that[f(x)cosaxdx +if"f(x)sinaxdx=frf(x)edx,togetherwiththefactthatthe modulusle Heia(+)Heei eyis bounded in the upper half planey ≥ 0.Example.Let us show that[.cos 3x d = 21(7.3.2)(x2+1)2eBecause the integrand is even, it is sufficient to show that the Cauchy principal value of theintegral exists and to find that value.We introduce the function1f(a)=(7.3.3)(22 + 1)2and observe that the product f(z)ei3= is analytic everywhere on and above the real axis exceptat the point z =i. The singularity z =i lies in the interior of the semicircular region whoseboundary consists of the segment - R≤ x≤ R of the real axis and the upper half C of thecircle I=-R(R>1) from ==R to z=-R. Integration of f(=)er3- around thatboundaryyields the equationei3xR(r +1 dx = 2元iB, - Jf(=)ei3=dz(7.3.4)whereB, = Res[f(=)ei3]Sinceer3f(=)ei3==_()whereΦ()(z-i)?(z+i)2the point z=i is evidently a pole oforder m=2 of f(z)ei3=; and
§7.3. Improper Integrals From Fourier Analysis Residue theory can be useful in evaluating convergent improper integrals of the form ∫ ∞ ∞− sin)( axdxxf or , (7.3.1) ∫ ∞ ∞− cos)( axdxxf where denotes a positive constant. As in Sec. 7.1, we assume that , where and are polynomials with real coefficients and no factors in common. Also, has no real zeros. Integrals of type (7.3.1) occur in the theory and application of the Fourier integral. a = xqxpxf )(/)()( xp )( xq )( zq )( The method described in Sec. 7.1 and used in Sec. 7.2 cannot be applied directly here since (see Sec. 3.6) az ax ay 22 2 sin|sin| += sinh and ax ax ay 2 2 2 cos|cos| += sinh . More precisely, since 2 sinh ay ay ee ay − − = , the moduli and increase like as tends to infinity. The modification illustrated in the example below is suggested by the fact that az |sin| az |cos| ay e y ∫ ∫∫ − −− + = R R R R R R iax cos)( sin)( )( dxexfaxdxxfiaxdxxf , together with the fact that the modulus iaz iyxia iaxay ay eeeee + − − |||||| === )( is bounded in the upper half plane y ≥ 0 . Example. Let us show that ∫ ∞ ∞− π = + 22 3 2 )1( 3cos e dx x x . (7.3.2) Because the integrand is even, it is sufficient to show that the Cauchy principal value of the integral exists and to find that value. We introduce the function 22 )1( 1 )( + = z zf (7.3.3) and observe that the product is analytic everywhere on and above the real axis except at the point zi ezf 3 )( z = i . The singularity z = i lies in the interior of the semicircular region whose boundary consists of the segment − ≤ ≤ RxR of the real axis and the upper half of the circle from CR RRz >= )1(|| z = R to z = −R . Integration of around that boundary yields the equation zi ezf 3 )( ∫ ∫ − −π= + R R C zi xi R dzezfiBdx x e 3 22 1 3 2 )( )1( , (7.3.4) where ])([Res 3 1 zi iz ezfB = = . Since 2 3 )( )( )( iz z ezf zi − = φ where 2 3 )( )( iz e z zi + φ = , the point = iz is evidently a pole of order m = 2 of ; and zi ezf 3 )(
1B, =(i)=ie3By equating the real parts on each side of equation (7.3.4), then, we find thats dt--Rel.(a)eidR(7.3.5)J-R (x2 + 1)?e3JFinally, we observe that when z is a point on Cr1If(=)KMRwhere MR=(R2-1)?and that le3"e-3y ≤1 for sucha point. Consequently,[ReJe, (2)e"-d|≤c (a)e=de≤ MrR-→0,(7.3.6)JCas R tends to oo and because of inequalities (7.3.6), we need only let R tend to co inequation (7.3.5) to arrive at the desired result (7.3.2)
1 3 1 )( ie = φ′ iB = . By equating the real parts on each side of equation (7.3.4), then, we find that ∫ ∫ − − π = + R R C zi R dzezf e dx x x 3 22 3 )(Re 2 )1( 3cos . (7.3.5) Finally, we observe that when z is a point on , CR ≤ Mzf R |)(| where 22 )1( 1 − = R M R and that 1|| for such a point. Consequently, 3 3 ≤= zi − y ee )(Re )( 0 3 ≤ 3 →≤ ∫∫ RMdzezfdzezf R C zi C zi R R π , (7.3.6) as R tends to ∞ and because of inequalities (7.3.6), we need only let R tend to in equation (7.3.5) to arrive at the desired result (7.3.2). ∞