s ÃR for all s > 0, where ÃR: [0, Ä„] C is the path with [ÃR] = SR defined such that ÃR(¸) = Rei¸ for all ¸ " [0, Ä„]. Proof It follows from the definition of the path integral that
Ä„
R f(z)eisz dz = f(ÃR(¸))eisà (¸)ÃR(¸) d¸ ÃR 0 Ä„
= f(Rei¸)eiRs cos ¸-Rs sin ¸ iRei¸ d¸. 0 Therefore
Ä„
f(z)eisz dz d" R |f(Rei¸)||eiRs cos ¸-Rs sin ¸| d¸
ÃR 0
Ä„ d" RM(R) e-Rs sin ¸ d¸. 0 Now sin ¸ e" 2¸/Ä„ when 0 d" ¸ d" Ä„/2, and therefore Ä„ Ä„ 2 2 -2Rs¸ Ä„ Ä„ e-Rs sin ¸ d¸ d" e d¸ d" . 2Rs 0 0 Also Ä„ Ä„ 2 e-Rs sin ¸ d¸ = e-Rs sin ¸ d¸. Ä„ 0 2 (This follows on making the substitution that replaces ¸ by Ä„ -¸.) Therefore
Ä„ Ä„ e-Rs sin ¸ d¸ d" . Rs 0 It follows that
Ä„M(R)
f(z)eisz dz d" ,
s ÃR as required. 81 Example We shall apply Cauchy s Residue Theorem (Theorem 6.16) and Lemma 8.1 in order to evaluate
" eisx dx x2 + a2 - " when s > 0. Let a be a positive real number, let R be a real number satisfying R > a, and let ÃR: [0, Ä„] C be the path that sends ¸ " [0, Ä„] to Rei¸. (Thus ÃR(¸) traverses a semicircle of radius R in the upper half of the complex plane from R to -R as ¸ increases from 0 to Ä„. Now it follows from the Triangle Inequality that |z2| d" |z2 + a2| + |a2|, and thus |z2 + a2| e" |z|2 - a2 for all complex numbers z, and therefore
1 1
d" z + a2 2 R2 - a2 for all complex numbers z satisfying |z| e" R. It now follows from Lemma 8.1 that
eisz Ä„
dz d"
z2 + a2 s(R2 - a2) ÃR for all real numbers s and R satisfying s > 0 and R > a. Therefore
eisz lim dz = 0. R+ " z2 + a2 ÃR Now the function f has poles at ia and -ia. Moreover
eisz eisz e-sa lim (z - ia) = lim = , zia zia z2 + a2 z + ia 2ia eisz and therefore the meromorphic function that sends z to as a simple z2 + a2 pole at ia with residue e-sa/2ia. Thus if we apply Cauchy s Residue Theorem (Theorem 6.16) in order to evaluate the path integral of this function around the boundary of the set {z " C : |z| d" R and Im[z] e" 0}, we find that
R eisx dx eisz e-sa Ä„e-sa lim dx + lim = 2Ä„i × = R+ " R+ " x2 + a2 z2 + a2 2ia a -R ÃR 82 when s > 0. If we then take the limit of the left hand side of this identity as R + ", we find that
" eisx Ä„e-sa dx = . x2 + a2 a - " when s > 0. This formula does not hold when s d" 0. And indeed, if we take the complex conjugate of the above identity we find that
" e-isx Ä„e-sa dx = . x2 + a2 a - " when s > 0. It follows that
" eisx Ä„e-|s|a dx = . x2 + a2 a - " when |s| = 0. This identity also holds when s = 0, though this does not
follow from the above calculations. Example We can also evaluate the above integral by applying Cauchy s Residue Theorem to the path integral taken around the boundary of a rect- angle in the complex plane with vertices at -R, R, R + iM and -R + iM, where R and M are large positive real numbers. Let a be a real number, and let R and M be real numbers satisfying R > a and M > a. The inequality
1 1
d" z + a2 2 R2 - a2 is satisfied for all complex numbers z for which |z| > a. It follows from this that
eisz eisz - dz - dz z2 + a2 z2 + a2 [-R+iR,R+iR] [-R,-R+iR] when s > 0. It follows that
" e-isx eisz Ä„e-sa dx = lim dz = R+ " x2 + a2 z2 + a2 a - " [-R,R] when s > 0. Example Let Ä… be a real number satisfying 0 < Ä… < 1. We evaluate the integral
" xÄ… dx x(x + 1) 0 through an application of Cauchy s Residue Theorem. Let D = C \ {x " R : x d" 0}, so that D is the open set obtained on removing the negative real axis from the complex plane, let log: D C denote the principal branch of the logarithm that sends rei¸ to log r + i¸ for all real numbers r and ¸ satisfying r > 0 and -Ä„ < ¸ < Ä„, and let zÄ… = exp(Ä… log z) for all z " D. Then the function f zÄ… that sends z " D\{1} to is a meromorphic function on D. The only z(z - 1) pole of this function that lies within the open set D is a simple pole at z = 1 with residue 1. Let R and · be real numbers satisfying R > 1 and 0 < · < 1, 84
and let ¸· " [3Ä„, Ä„] be determined such that -R + i· = R2 + ·2 exp(i¸·). 4 It follows from Cauchy s Residue Theorem (Theorem 6.16) that
f(z) dz - f(z) dz [-R+i·,-·+i·] Ä…·
- f(z) dz + f(z) dz [-R-i·,-·-i·] ²R,· = 2Ä„i, 3 where Ä…·: [-3Ä„, Ä„] C is the path from -· - i· to -· + i· that sends 4 4 " 3 t " [-3Ä„, Ä„] to 2 ·eit, and ²R,·: [-¸·, ¸·] C is the path from -R - i· 4 4
to -R + i· that sends t " [-¸·, ¸·] to R2 + ·2eit. [Thus Ä…·(t) traverses " a three-quarters of a circle of radius 2 · about zero in the anti-clockwise 3 direction as t increases from -3Ä„ to Ä„, and ²R,·(t) traverses most of a 4 4
circle of radius R2 + ·2 about zero as t increases from -¸· to ¸·.] Now the inequality Ä… > 0 ensures that
lim f(z) dz = 0. ·0 Ä…· Also
lim f(z) dz = f(z) dz, ·0 ²R,· ÃR where ÃR: [-Ä„, Ä„] C is the path that sends t " [-Ä„, Ä„] to Reit. It follows that