Mathematical Methods For Physicists Webber/Arfken Selecet Solutions ch. 7

Physics 451

Fall 2004 Homework Assignment #11 — Solutions

Textbook problems: Ch. 7: 7.2.5, 7.2.7, 7.2.14, 7.2.20, 7.2.22 Chapter 7 7.2.5 The unit step function is defined as u(s − a) =

0, 1,

sa

Show that u(s) has the integral representations Z ∞ ixs 1 e a) u(s) = lim dx + →0 2πi −∞ x − i Let us first suppose we may close the contour with a semi-circle in the upper half plane z IR R iε u(s)

C

θ

Always assuming the limit → 0+ , we see that the real integral for u(s) may be promoted to a closed contour integral 1 2πi

I C

eizs dz = u(s) + IR z − i

(1)

where IR denotes the integral along the semi-circle at infinity. We now show that, at least for s > 0, the integral IR vanishes. To do so, we make an explicit variable substitution z = Reiθ , dz = iReiθ dθ so that 1 IR = 2πi

Z 0

π

iθ

1 eisRe iReiθ dθ = iθ Re − i 2π

Z

π

eisR(cos θ+i sin θ) dθ

0

where we have safely taken the limit → 0+ . Expanding out the exponent, we find Z π 1 IR = eisR cos θ e−sR sin θ dθ (2) 2π 0

This vanishes by Jordan’s lemma provided sR sin θ > 0 so that the real exponential is suppressed instead of blowing up (in fact, this is Jordan’s lemma). Since R → ∞ is positive and sin θ > 0 in the upper half plane, this corresponds to the requirement s > 0 (as alluded to above). In this case, since IR = 0, (1) simplifies to I eizs eizs 1 dz = residue of at z = i (s > 0) u(s) = 2πi C z − i z − i The residue at i is simply lim→0+ e−s = 1. Hence we have confirmed that u(s) = 1 for s > 0. For s < 0, on the other hand, Jordan’s lemma makes it clear that we should instead close the contour with a semi-circle in the lower half plane z iε u(s)

C

IR

Since there are no residues inside the contour, we simply obtain u(s) = 0 for s < 0. Although the problem does not discuss the case when s = 0, it is worth considering. In this case, we might as well close the contour in the upper half plane. Then IR can be directly evaluated by inserting s = 0 into (2). The result is simply IR = 12 . Since the contour integral still has the value of 1 (residue at the pole at i), inserting IR = 12 into (1) gives 1 = u(0) +

1 2

⇒

u(0) =

1 2

which is a nice result indicating that the step function is given completely by ( 0, s < a u(s − a) = 12 , s = a 1, s > 1 at least using this definition. Z ∞ ixs 1 1 e b) u(s) = + P dx 2 2πi −∞ x The principal value integral can be evaluated by deforming the contour above and below the pole and then taking the average of the results. For s > 0, this corresponds to something like z

z

IR

IR

or C u(s)

C u(s)

As in part a), the residue of the pole at z = 0 is simply 1. So for the contour on H eizs 1 the left we have 2πi z dz = 1, while for the one on the right (no poles inside) we have 0. The principal value then gives 1/2 (the average of 1 and 0). This indicates that 1 1+0 = 1, (s > 0) u(s) = + 2 2 For s < 0, on the other hand, we close the contour on the lower half plane. Again, we average between the case when the pole is inside and when it is outside the contour. However, it is important to realize that by closing the contour on the lower half plane, we are actually choosing a clockwise (‘wrong direction’) contour. This means the contour integral gives either −1 or 0 depending on whether the pole is inside or outside. The principal value prescription then yields u(s) =

1 −1 + 0 + = 0, 2 2

(s < 0)

If we wanted to be careful, we could also work this out for s = 0 to find the same answer u(0) = 12 . 7.2.7 Generalizing Example 7.2.1, show that Z 2π Z 2π dθ dθ 2π = = 2 a ± b cos θ a ± b sin θ (a − b2 )1/2 0 0 for a > |b|. What happens if |b| > |a|? Since this integral is over a complete period, we note that we would get the same answer whether we integrate cos θ or sin θ. Furthermore, it does not matter whether we integrate a + b cos θ or a − b cos θ. This can be proven directly by considering the substitutions θ → π2 − θ or θ → θ + π into the integral. In any case, this means we only need to consider Z 2π dθ I= a + b cos θ 0 where we assume a > b > 0. For these types of trig integrals, we make the substitutions z = eiθ ,

dz = ieiθ dθ = izdθ,

cos θ =

z + z −1 2

to change the real integral into a contour integral on the unit circle |z| = 1 I I dz −2i dz I= = b 2 −1 b C z + 2a )) C iz(a + 2 (z + z b z+1 Since the contour is already closed, we do not need to worry about finding a way to close the contour. All we need is to identify the poles inside the contour and

their residues. To do this, we solve the quadratic equation in the denominator to obtain I dz −2i I= b C (z − z+ )(z − z− ) where a z± = − ± b

r

a2 −1 b2

(3)

Since we have assumed a > b > 0, the two zeros of the denominator, z− and z+ are located as follows z

C z−

z+

In particular, it is not hard to check that the pole at z+ lies inside the circle of unit radius. As a result −2i 1 I = (2πi) at z = z+ residue of b (z − z+ )(z − z− ) 4π 1 4π 2π = = p =√ b (z+ − z− ) a2 − b2 2b a2 /b2 − 1 Note that, for a < 0, the integrand would always be negative. In this case, I would be negative. Thus the complete answer is I=√

2π sign(a) a2 − b2

For |b| > |a|, the integrand would blow up when θ = − cos−1 (a/b) so the integral is not defined. What happens in this case is that, on the complex plane, the two poles z+ and z− , which still solve (3), move off the real axis but stay on the unit circle contour itself. z z+

C

z−

So the complex integral is just as bad as the real integral. This is an example where we could consider using a principal value prescription to make the integral well defined.

7.2.14 Show that (a > 0) Z ∞ cos x π dx = e−a a) 2 2 a −∞ x + a How is the right-hand side modified if cos x is replaced by cos kx? For these types of integrals with sin or cos in the numerator, it is best to consider sin x or cos x as the imaginary or real part of the complex exponential eix . In this case, we write Z ∞ Z ∞ eix cos x dx = < dx I= 2 2 2 2 −∞ x + a −∞ x + a Using Jordan’s lemma, we may close the contour using a semi-circle in the upper half plane. z IR ia C

Since IR = 0 (by Jordan’s lemma), we have simply I I e−a eiz π eiz I=< dz = < 2πi dz = < = e−a 2 2 z +a (z − ia)(z + ia) 2ia a (for a positive). If cos x is replaced by cos kx, we would write the numerator as 0 we would close the contour in the upper half plane as before. In addition, the exponential factor in the residue would be e−ka , so for cos kx, the integral would be (π/a)e−ka . For k < 0, on the other hand, we could close the contour in the lower half plane. However, it is actually easier to see that cos(−kx) = cos kx, so the answer should be independent of the sign of k. Hence Z ∞ cos kx π −|ka| dx = e 2 2 |a| −∞ x + a is valid for any sign of k and a. Z ∞ x sin x b) dx = πe−a 2 2 −∞ x + a How is the right-hand side modified if sin x is replaced by sin kx? As above, we write sin x = =eix . Closing the contour in the same manner, and using Jordan’s lemma to argue that IR = 0, we obtain I I Z ∞ zeiz zeiz x sin x dx = = dz = = dz 2 2 z 2 + a2 (z − ia)(z + ia) −∞ x + a iae−a = = 2πi = πe−a 2ia

If sin x is replaced by sin kx, the residue would get modified so that e−a is replaced by e−ka . As a result Z ∞ x sin kx dx = πe−|ka| 2 + a2 x −∞ The reason for the absolute value is the same as for part a) above. 7.2.20 Show that

Z 0

∞

(x2

dx π = 3, 2 2 +a ) 4a

a>0

This problem involves a double pole at z = ia. Choosing to close the contour in the upper half plane z IR ia C

we obtain Z

∞

I= 0

1 dx = (x2 + a2 )2 2

Z

∞

−∞

1 dx = (x2 + a2 )2 2

I C

dz (z 2 + a2 )2

(4)

= πi (residue at z = ia) Although this is a double pole, it may still have a residue. To see this, imagine expanding the integrand in a power series near the pole at z = ia (z 2

1 = (z − ia)−2 (z + ia)−2 = (z − ia)−2 [2ia + (z − ia)]−2 + a2 )2 −2 z − ia −2 −2 = (z − ia) (2ia) 1+ 2ia ! 2 −1 z − ia 2 · 3 z − ia = 2 (z − ia)−2 1 − 2 + − ··· 4a 2ia 2 2ia =

−1/4a2 −i/4a3 + + (3/16a4 ) + · · · 2 (z − ia) (z − ia)

Here we have used the binomial expansion for (1 + r)−2 . This shows that, in addition to the obvious double pole, there is a single pole ‘hidden’ on top of it with residue a−1 = −i/4a3 . Alternatively, we could have computed the residue much more quickly by noting that for a double pole in f (z) = 1/(z 2 + a2 )2 , we form the non-singular function g(z) = (z − ia)2 f (z) = 1/(z + ia)2 . The residue is then the derivative d 1 −2 −2 −i 0 a−1 = g (ia) = = = = 3 2 3 3 dz (z + ia) z=ia (z + ia) z=ia (2ia) 4a

In either case, using this residue in (4), we find π −i = 3 I = πi 3 4a 4a or more precisely I = π/4|a|3 , which is valid for either sign of a. It is worth noting that, for this problem, the integrand falls off sufficiently fast at infinity that we could choose to close the contour either in the upper half plane or the lower half plane. Had we worked with the lower half plane, we would have found a pole at −ia with opposite sign for the residue. On the other hand, the clockwise contour would have contributed another minus sign. So overall we would have found the same result either way (which is a good thing). 7.2.22 Show that

Z

∞

Z

2

∞

cos(t )dt = 0

√ π sin(t )dt = √ 2 2 2

0

Again, when we see sin or cos, it is worth considering this as the imaginary or real parts of the complex exponential. Hence we first choose to evaluate the integral Z ∞ 2 I= eit dt 0

Taking the hint into account, we write down a (closed) contour integral using the contour z IR

I2 π/2 I

Thus

I

C

2

eiz dz = I + IR + I2

C 2

We start by evaluating the contour integral on the left. Although eiz has an essential singularity at infinity, that actually lies outside the contour (this is certainly true for any fixed large radius R; it also remains outside the contour in the limit R → ∞). Since there are no poles and no singularities inside the contour, the contour integral vanishes. As a result, 0 = I + IR + I2

⇒

I = −IR − I2

We now show that the integral on IR vanishes as well by a simple modification to Jordan’s lemma. For IR , we let z = Reiθ , so that Z π/2 Z π/2 2 2 iR2 e2iθ iθ IR = e iRe dθ = iR eiR cos 2θ e−R sin 2θ eiθ dθ 0

0

Hence Z

π/2

e−R

|IR | = R

2

sin 2θ

dθ

0

Using the same argument as in Jordan’s lemma, we can show that the integral R π/2 −R2 sin 2θ e dθ may be bounded by 1/R2 . Hence |IR | itself falls off as 1/R, and 0 vanishes when we take R → ∞. As a result, we are left with the observation I = −I2 To examine I2 , we note that the path of integration is a line of constant slope in the complex plane z = `eiπ/2 , dz = eiπ/2 d` Thus

0

Z I2 =

i(`eiπ/2 )2 iπ/2

e

e

iπ/2

∞

Z

∞

2

e−` d`

d` = −e

0

Note that the minus sign came from consideration of the direction of integration along the I2 contour. At this point, complex analysis does not really help us, and we must recall (or look up) the gaussian integral ∞

Z

−`2

e 0

1 d` = 2

Z

∞

√ −`2

e −∞

d` =

π 2

Thus √ iπ/2

I = −I2 = e

√ √ π π π = (cos(π/2) + i sin(π/2)) = (1 + i) √ 2 2 2 2

R∞ 2 Since I = 0 eit dt, we may now take the real (cos) and imaginary (sin) parts of I to obtain √ Z ∞ Z ∞ π 2 2 cos(t )dt = sin(t )dt = √ 2 2 0 0 This is a curious result, as it is not directly obvious why integrating cos(t2 ) and sin(t2 ) would give identical results.

Mathematical Methods For Physicists Webber/Arfken Selecet Solutions ch. 7

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