PHY6095/PHZ6166: homework assignment #2 SOLUTIONS

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1 PHY695/PHZ666: homework assignment #2 SOLUTIONS

2 2. Problem a The first step is to re-scale the integration variable where c b/a. (y x/a x(x + b e x/a ( y(y + c e y (2 i a b c. Assume that typical y are c. Trying to put c in the integral leads to a logarithmic divergence at y y e y. This is not bad as the assumption y c breaks at y c and the integral needs to cut off at y c. Since the integral is only logarithmically divergent, ambiguity in the cutoff choice (y c, y c/2, etc is not relevant within the Main Log Approximation (MLA. Continuing along these lines c y exp( y. The exponential factor cuts off the integral at y. Again, a particular choice of cutoff in not important within the MLA, and we can drop e y and cut the integral off at y c y ln c ln a b. The result can also be obtained beyond the MLA. To this end, introduce a new variable z y and integrate by parts Integrate by parts: dz ( 2 2 ln z + z z2 + c e z2 2 + c e z2 (3 +4 ( dz ln z + z 2 + c ze z2 ln c + 4 ( dz ln z + z 2 + c ze z2. (4 In the remaining integral one can let c, because the integral is convergent. Upon that, the integral reduces to a pure number 4 dz ln (2z ze z2 where γ is the Euler constant Thus πγ (5 ln b a πγ 4 ln a.29b (6 ii a b c. Neglecting compared y compared to c in the denominator gives c y e y 2 dze z2 c πa b. (7 In the last integral, y and thus neglecting y compared to c was justified. b

3 3 i a b x (x 2 + b 2 sin (x/a (8 (y x/b ( b 2 y (y 2 + sin y b (9 a b 2 y (y 2 + y b a π ( 2 ab ii a b. x (x 2 + b 2 sin (x/a ( (y x/a a 2 y (y 2 + b 2 /a 2 sin y (2 Neglect y 2 compared to b 2 /a 2 in the denominator. The remaining integral is dominated by y. Thus Problem 2. a a. where c 3. More precisely, y sin y π/2 (3 I π a 2 a 2 2 b 2 π 2b 2. (4 x I(a x 2 + a 2 + (5 x (y x/a (6 y a y 2 (7 + y + / a ( y a y 2 + y + / (8 ay ( y a y ay ya (9 (ay 3/2 c a c 2 a + c 3 a 3/2 a 2 c c 2 c 3 y 2π y 2 + y 2 + y, (2 2, (2 y 2 + π 2, (22 2π y 2 c. (23 + y 2 The last term diverges at small y. When expanding the denominator, we have assumed that ay. This assumption breaks down when y /a. Cutting of the divergence at y /a, we obtain the fourth term

4 4 within the MLA Thus, for a, a 2 /a a 2 /a y 2 + y y (24 ln a a 2. (25 2π 2 π 2π a 2 a + ln a 2 a3/2 a 2. (26 a. Within MLA, I(a a x ln a. b Function J(a is defined as J(a Find the first two terms of the expansion of J(a for a. Integrating by parts, J(a ln(y + a 2 e y y + a 2 e y. + ln(y + a 2 e y 2 ln a + ln(y + a 2 e y The integral term is convergent for a, so can put a in this term which gives the next-to-leading term Thus Problem 3. a Expand the function x/ sinh x for x : ln ye y γ. J(a 2 ln a γ +... x cos ωx. (27 sinh x x sinh x 6 x x4 + O ( x 5 (28 We see that the function itself and all its even derivative are finite at x whereas odd derivatives vanish. Now integrate by parts sin ωx x ω sinh x (29 d ( x sin ωx (3 ω sinh x cos ωx d ( x ω 2 x (3 sinh x ( ω 2 d2 x 2 cos ωx (32 sinh x... (33

5 Even derivatives always come in combinations with sin ωx, which vanishes at x, whereas odd derivatives vanish themselves. Thus the integral does not have a power-law expansion in /ω. On the other hand, the integrand is analytic on real axis. To get rid of the factor of x, re-write the intergal as ω sin ωx sinh x 2 ω Im sinh x 2 ω J(ω The poles of the integrand are at z iπn, n, ±, ±2... To take into account only one pole, choose the contour as shown in Fig., where semi-circles circumvent poles at z and z iπ. The vertical segments of the countour (B and D can be moved to ± and, because the integrand does decay along the real axis, the integrals over parts B and D go to zero. The integral over A is equal to the original integral J in the limit of r. On part D, z x + iπ, sinh z sinh x, and By Cauchy theorem, D dz eiωz sinh z e πω e iωx e iωx e πω J. J( + e πω + +, C C 2 5 where C,2 are integrals over semi-circles C,2. On C, z re iφ, dz ire iφ dφ, sinh z re iφ, e iωz, so C ir π dφe iφ iπ. reiφ On C 2, z iπ + re iφ, dz ire iφ dφ, sinh z re iφ, e iωz e πω, so π ire πω dφe iφ re iφ iπe πω. Therefore, C 2 and J iπ e πω + e πω ImJ π e πω + e πω. For ω, e πω and the fraction in the previous equation can be expanded in e πω. Neglecting terms of order e 2πω e πω, we obtain ImJ π ( 2e πω and 2 ω J(ω π2 e πω. b x2 + x 2 e x sin ωx. (34 The integrand is a non-analytic function on the entire real axis, so we should be able to obtain the result via integration by parts f( ω + ω f (x cos ωx f( ω f ( ω

6 6 FIG.. Contour of integration for Problem 2a. where f(x Obviously, f( and f ( 2. Thus c Re-write the integral as x2 + x 2 e x x 2 x 3 2 x x5 + O ( x 6. (35 I 2 ω 3 + O(ω 5. (36 2 cos(ωx. (37 (x 2 2/3 + exp(iωx (x 2 + 2/3. The integrand has a branch cut at x i, where x 2 +. Choose the contour as shown in Fig. 2. By Cauchy theorem, I + + +, C BC BC2 where C is a circle around i and BC,2 are the branch cuts. On the circle, z i + re iφ, dz ire iφ dφ, (z 2 + 2/3 r 2/3 and C r/3 r. On BC, z is+δ with δ +, and (z 2 + 2/3 ( s 2 ++iδ 2/3 (s 2 2/3 e 2iπ/3. On BC2, z is δ, and (z 2 + 2/3 (s 2 2/3 e 2iπ/3. Therefore, BC e 2iπ/3 idse ωs (s 2 ie 2iπ/3 2/3 dse ωs (s 2 2/3

7 7 y C δ i r x and FIG. 2. Contour of integration for Problem 2b. BC2 e 2iπ/3 3 2 idse ωs (s 2 2/3 dse ωs (s 2 2/3. In the limit of ω, s tries to be as small as possible, therefore, we change the variable as s + t with t, and replace (s 2 2/3 2 2/3 t 2/3. Thus 3 dte ωt 3Γ( 3 e ω e ω 25/3 t 2/3 2 5/3 ω. /3 Problem 4 Γ(z + x z e x (38 exp ( x + z ln x. (39 f(x z ln x x (4 f z/x (4 x z, f(x z ln z z (42 f (x /z (43 f (x 2/z 2 (44 f (IV (x 6/z 3 (45

8 8 Expanding the argument of the exponential, one gets Γ(z + z z e z exp ( x + z ln x (46 exp( 2z [x z]2 + 3z 2 [x z]3 (47 4z 3 [x z]4. (48 Keeping only the 2nd order term in the exponential gives the result we obtained in class Γ 2 (x z z e z 2πz. To evaluate the effect of the 3rd order term, expand the exponential to leading order in this term. Rescaling the integration variable as x x/ z and changing the variable to y x 2 /2, we obtain: Γ 3 (z + zz e z 3z 2 exp( 2z [x z]2 (x z zz e z ye y 3 zz+ e z/2. (49 z/2 Up to third-order terms, we have Γ(z + Γ 2 + Γ 3 z z e z ( 2πz + z 3/2 e z/2. (5 3 2π Notice that the 3rd order correction is exponentially small and will be neglected in what follows. The 4th correction is obtained by expanding the exponential st order in the 4th-order derivative and to second order in the 3rd-orderderivative (these two terms give the same order in z Recalling that [ ( Γ(z + z z e z exp ] 2z x2 + x3 3z 2 x4 4z 3 ( z z e z e x2 /2z x4 4z 3 + [ ] x z 2 z z e z[ + 8z 4 e x2 /2z 4z 3 x 4 e x2 /2z x 6 e x2 /2z ] (5 2πzz z e z zz e z (2z 5/2 + zz e z (2z 7/2 8z 4 4z 3 x 4 e x2 3 π, 4 x 6 e x2 5 π, 8 x 4 e x2 (52 x 6 e x2 (53 we get Γ(z + ( 2πzz z e z + 2z

9 9 or Γ(z z Γ(z + (54 ( 2πz z /2 e z +. (55 2z Note that the combined correction due to fourth derivative and (third derivative 2 is larger than due to the third derivative alone. Problem 5. a i a the n term dominates and G(a n (n 2 + a 2 2 G(a a 4 ii a. Large number of terms (n a contributes to the sum which therefore can be approximated by an integral n (n 2 + a 2 2 dn (n 2 + a 2 2 a 3 π 4 a 3 (x (the absolute value of a occurs because the integrand is an even function of a. Both of this approximations are good only in giving us the leading terms. To construct regular series, use the Poisson formula. Re-arranging the sum using the formula valid for an even function f(n n and applying the Poisson formular, we obtain n G(a 2 (n 2 + a 2 2 f(n f( + 2 ( k π 2 a 3 π 2 a 3 n k [ + 2 f(n, n (n 2 + a a 4 ei2πkn dn (n 2 + a 2 2 ( + 2π k a e 2π k a ] ( + 2πk a e 2πk a k (a π 2 a 3 [ + 2 ( + 2π a e 2π a ] G(a π 4 a 3 + 2a 4 + π2 2a 2 e 2π a +... In fact, the series over k can be summed exactly (choosing a > for brevity ( + 2πka e 2πka k.

10 [ ] [ ] d 2πa e 2πka d (2πa l e 2πa + 2πae2πa (e 2πa 2 G(a 2a 4 + π [ 4a 3 coth πa + πa ] sinh 2 πa For small a,the expression in the square brackets can be expanded [...] 2 π a + ( 2 45 π3 a 3 + O ( a 5, thus b i a K(a exp( a. ii a G(a 2a 4 + π [ ( ] 2 2 4a 3 πa + 45 π3 a 3... a 4 + π K(a n 2 exp( an 6. n n 2 exp( an 6 n π a dnn 2 exp( an 6 Again, higher-order terms can be obtained via the Poisson formula. dn 3 exp( an 6 exp( ay 2