Solution Set V for 18.075-Fall 2004

SOLUTION SET V FOR 18.075–FALL 2004 10. Functions of a Complex Variable 10.15. Indented contours. . 110. By making use of integration around suitable indented contours in the complex plane, evaluate integrals: ⎩ � the following x (a) −� x(xsin 2 +a2 ) dx (a > 0), ⎩� x (b) −� x(�sin 2 −x2 ) dx. Solution. (a) For R > 1 and 0 < � < 1, define the contour C = C 1 + C2 + C3 + C4 , where C1 is the real interval [−R, −�], C2 is the upper half of the circle |z| = � with clockwise orientation, C3 is the real interval [�, R], and C4 is the upper half of the circle |z| = R with counterclockwise orientation. Then ⎪ ⎪ � ⎪ � eiz eix sin x dx = Im lim dz � ImP dx. 2 2 2 2 2 2) R� z(z + a ) x(x + a C +C −� −� x(x + a ) 1 3 ��0 Moreover, by Theorem 2 of classnotes, ⎪ eiz lim dz = 0, 2 R�� C4 z(z + a2 ) and, by Theorem 4 of classnotes, ⎪ eiz eiz �i dz = −�i Res[ , 0] = − 2 . lim 2 2 ��0 C2 z(z 2 + a2 ) z(z + a ) a So, lim R� ��0 ⎪ C1 +C3 eiz dz = z(z 2 + a2 ) lim R� ��0 ⎨ C eiz dz − lim R� z(z 2 + a2 ) ��0 = 2�i Res[ = �i ⎪ C2 +C4 eiz dz z(z 2 + a2 ) eiz �i , ia] − (− 2 ) 2 2 z(z + a ) a 1 − e−a . a2 Hence, ⎪ � −� sin x dx = Im lim R� x(x2 + a2 ) ��0 ⎪ C1 +C3 Date: October 16, 2002. 1 1 − e−a eiz dz = � . z(z 2 + a2 ) a2 2 SOLUTION SET V FOR 18.075–FALL 2004 (b) For large R > 0 and small � > 0, define the contour C = C 1 +C2 +C3 +C4 +C5 +C6 + C7 + C8 , where C1 , C3 , C5 and C7 are the real intervals [−R, −� − �], [−� + �, −�], [�, � − �] and [� + �, R], C2 is the upper half of the circle |z + �| = � with clockwise orientation, C 4 is the upper half of the circle |z| = � with clockwise orientation, C 6 is the upper half of the circle |z − �| = � with clockwise orientation, and C 8 is the upper half of the circle |z| = R with counterclockwise orientation. Then ⎪ ⎪ � ⎪ � eiz eix sin x dx = Im lim dz � Im P dx. 2 2 2 2 2 2) R� z(� − z ) x(� − x +C +C +C C −� −� x(� − x ) 1 3 5 7 ��0 By Theorem 2 of classnotes, ⎪ lim R�� C8 eiz dz = 0, z(� 2 − z 2 ) and, by Theorem 4 of classnotes, ⎪ eiz −1 −i lim dz = −�i Res[ , −�] = , 2 2 2 2 ��0 C2 z(� − z ) z(� − z ) 2� ⎪ eiz eiz −i lim dz = −�i Res[ , 0] = , 2 2 2 2 ��0 C z(� − z ) z(� − z ) � 4 ⎪ eiz eiz −i lim dz = −�i Res[ , �] = . 2 2 2 2 ��0 C6 z(� − z ) z(� − z ) 2� Since eiz z(� 2 −z 2 ) has no singularities in the upper half plane, we get ⎨ eiz dz = 0, 2 2 C z(� − z ) This gives ⎪ lim R� ��0 C1 +C3 +C5 +C7 eiz dz = − lim R� z(� 2 − z 2 ) ��0 ⎪ C2 +C4 +C6 +C8 So ⎪ � −� sin x dx = Im lim R� x(� 2 − x2 ) ��0 ⎪ C1 +C3 +C5 +C7 2 eiz dz = . 2 2 z(� − z ) � 111. Show that P and hence also that ⎪ � −� � ⎧ ��i eitx dx = 0 ⎧ x � −�i P ⎪ � −� (t > 0), (t = 0), (t < 0), cos tx dx = 0, x eiz 2i dz = . 2 2 z(� − z ) � SOLUTION SET V FOR 18.075–FALL 2004 3 and ⎪ � −� � ⎧ �� sin tx dx = 0 ⎧ x � −� (t > 0), (t = 0), (t < 0). Solution. Case 1, t > 0. For large R and small �, define the contour C = C 1 + C2 + C3 + C4 , where C1 and C3 are the real intervals [−R, −�] and [�, R], C 2 is the upper half of the circle |z| = � with clockwise orientation, and C 4 is the upper half of the circle |z| = R with counterclockwise orientation. Then ⎪ � itx ⎪ eitz e dx = lim dz. P R� C1 +C3 z −� x ��0 By Theorem 2 of classnotes, lim ⎪ R�� C4 eitz dz = 0, z and, by Theorem 4 of classnotes, ⎪ eitz eitz lim dz = −�i Res[ , 0] = −�i. ��0 C2 z z Since eitz z has no singularities on the upper half plane, we get ⎨ itz e dz = 0. C z This gives P ⎪ � −� eitx dx = lim R� x ��0 ⎪ C1 +C3 eitz dz = − lim R� z ��0 ⎪ C2 +C4 eitz dz = �i. z Case 2, t < 0. For large R and small �, define the contour C = C 1 + C2 + C3 + C4 , where C1 and C3 are the real intervals [−R, −�] and [�, R], C 2 is the lower half of the circle |z| = � with counterclockwise orientation, and C 4 is the lower half of the circle |z| = R with clockwise orientation. Then ⎪ � itx ⎪ eitz e P dx = lim dz. R� C1 +C3 z −� x ��0 By Theorem 2 of classnotes, lim ⎪ R�� C4 eitz dz = 0, z and, by Theorem 4 of classnotes, ⎪ eitz eitz lim dz = �i Res[ , 0] = �i. ��0 C z z 2 4 SOLUTION SET V FOR 18.075–FALL 2004 Since eitz z has no singularities in the lower half plane, we get ⎨ itz e dz = 0. C z This gives P ⎪ � −� eitx dx = lim R� x ��0 Case 3, t = 0. Since 1 x ⎪ C1 +C3 eitz dz = − lim R� z ��0 ⎪ C2 +C4 eitz dz = −�i. z is odd, it’s clear that ⎪ � 1 P dx = 0. −� x Combine all three cases, we get P ⎪ � � ⎧�i � dx = 0 ⎧ x � −�i (t > 0), (t = 0), (t < 0), � ⎧ �� sin tx dx = 0 ⎧ x � −� (t > 0), (t = 0), (t < 0). eitx −� Compare the real and imaginary parts on both sides. We get ⎪ � cos tx P dx = 0, x −� and P � ⎪ −� Note that the integral above becomes ⎩� −� sin tx x ⎪ dx actually converges in all three cases. So the last equation � −� � ⎧ �� sin tx dx = 0 ⎧ x � −� (t > 0), (t = 0), (t < 0). 116. Obtain the evaluation ⎪ � dx −� cos ax − cos bx = �(b − a). x2 [Notice that f (z) = (eiaz − eibz )/z 2 has a simple pole at the origin.] By taking a = 0 and b = 2, also deduce the formula ⎪ � sin2 x = �. dx x2 −� SOLUTION SET V FOR 18.075–FALL 2004 5 Solution. If we naively set cos ax = Re(e iax ) and cos bx = Re(eibx ) and take the Re outside the integral sign, then the resulting integral doesn’t make any sense as is: ⎪ � ⎪ � cos ax − cos bx eiax − eibx I= dx → = Re dx = �. x2 x2 −� −� More precisely, I itself is finite, since the integrand in the left-hand side is well-behaved for all x. For example, by expanding the cosines in the integrand near z = 0, we find cos az − cos bz z2 = [1 − (az)2 2! + ... + (−1)n (az)2n (2n)! + . . .] − [1 − + ... + (−1)n (bz)2n (2n)! + . . .] z2 = − a2 − b 2 a2n − b2n 2(n−1) + . . . + (−1)n z + .... 2! (2n)! On the other hand, the integrand on the right-hand side, f (z) � at z = 0. Indeed zf (z) reads as zf (z) = (bz)2 2! eiaz − eibz z [1 + iaz + (iaz)2 2! eiaz −eibz , z2 (iaz)n n! 2 has a simple pole + . . .] − [1 + ibz + (ibz) 2! + . . . + = z n − bn a2 − b 2 a = i(a − b) − z + . . . + in z n−1 + . . . , 2! n! which is analytic at z = 0 (and furthermore lim z�0 [zf (z)] = → 0 since a →= b). Therefore, we read I = ReIp where Ip is the principal-value integral ⎪ � eiax − eibx Ip � P dx . x2 −� + ... + (ibz)n n! + . . .] We calculate Ip by closing the path by a small semicircle C �+ of radius � around z = 0 in the upper half plane, and a large semicircle C R+ of radius R also in the upper half plane. The resulting closed contour does not contain any singularities of the integrand and has to be zero by the Cauchy integral theorem. In addition, by Theorems 4& 2 of classnotes, ⎪ lim dz f (z) = −i� Rez[f (z), 0] = −i� i(a − b) = �(a − b), ��0+ C�+ lim ⎪ R�� CR+ dz f (z) = 0. The last equation was obtained by noticing that a > 0 and b > 0 while 1/z 2 goes to 0 uniformly in |z| = R as R � �. It follows that ⎪ ⎪ Ip + lim dz f (z) + lim dz f (z) = 0 ��0+ C�+ R�� CR+ ≡ Ip = �(b − a) ≡ I = ReIp = �(b − a). In the special case a = 0 and b = 2, the integrand of the original integral becomes 1 − cos 2x 2 sin2 x cos ax − cos bx = = . x2 x2 x2 6 SOLUTION SET V FOR 18.075–FALL 2004 Hence, the result of integration reads as ⎪ � 2 sin2 x dx = 2�, x2 −� which agrees with the desired formula.