Answer Set 4
Physics 204B
Riley 24.21 Consider
H (ln z)2
1+z 2
dz over the path specified. The only singularity inside the contour
2
3
is at z = i, so the integral becomes 2πi (ln2ii) = − π4 . On the large semicircle, parametrize
2
with z = Reiθ . The integrand magnitude approaches (lnRR) , where one factor in the denominator is cancelled by the R in dz (which corresponds to the increasing length of the path as R
grows). This magnitude approaches zero for large R, so the contribution of the large semicircle vanishes. Similarly, on the small semicircle parametrize with z = εeiθ , and the integrand
magnitude approaches (ln ε)2 ε, which goes to zero as ε → 0. (Even though the original function blows up at zero, the path is disappearing even faster.) This leaves the two segments
R
R
R
R
3
xeiπ )2 iπ
x)2
x+iπ)2
x)2
along the real axis, − π4 = Rε (ln
e dx + εR (ln
dx = εR (ln1+x
dx + εR (ln
dx =
2
1+x2 ei2π
1+x2
1+x2
2
x)
ln x
2 R 1
2 εR (ln
dx + 2πi εR 1+x
2 dx − π
ε 1+x2 dx. In the final expression, the third term can
1+x2
be evaluated through contour integration with a single semicircle,
or by
remembering the
R
R π/2 sec2 θdθ
1
derivative of arctan, or by trig substitution x = tan θ: 0∞ 1+x
dx
=
= π/2.
2
0
1+tan2 θ
3
With the prefactor, the term is −π /2. Since the first term is real and the second pure
R
R
3
x)2
ln x
dx = π8 and εR 1+x
imaginary, this means that εR (ln
2 dx = 0.
1+x2
R
R
R
Riley 25.22 The C1 integral starts on the positive x-axis, somewhere near the origin, goes
through z = i, and then heads towards the negative x-axis and goes off to −∞. Take
f (t) = t − 1t . Then f 0 (t) = 1 + t12 is zero at ±i. Only i is on curve C1 . Here f 00 (i) = 2/i,
which has argument −π/2. That gives α = π2 − 12 (− π2 ) = 34 π, which is in thercorrect general direction (right-to-left). The saddle contribution is
−1
1 −(ν+1) z2 (i− 1i ) i 34 π
i
e
e
2πi
2π
z
(2)
2
. When
3
simplifying, note that i i = −1, and incorporate this factor of -1 by changing ei 4 π to
e−iπ/4 . qThe remaining i−ν can be written e−iπν/2 . Also multiply by 2 overall, to get
2 i(z−πν/2−π/4)
Hν1 → πz
e
.
−(ν+1) −iz
For C2 , we must end
e .
q up with α = π/4, and i changes to −i in the factors (−i)
2 −i(z−πν/2−π/4)
2
This gives Hν → πz e
.
q
2
Since Jν (z) = 21 (Hν1 (z) + Hν2 (z)), Jν (z) → πz
cos(z − πν
− π4 ).
2
Note that the combined integral about C1 and C2 doesn’t vanish because of the singularity
at the origin. In principle for the asymptotic form of J one could skip the steepest descent
business and instead evaluate the residue of g at the origin...but it’s an ugly essential singularity, so that’s a lot easier said than done. It’s important that the contour has a cusp
at the singularity, approaching and leaving it horizontally. That means that as far as the
residue theorem is concerned, the singularity acts as inside (not on) the contour.
Riley 25.23 Take f (t) = −iz sin t + iνt, so f 0 (t) = −iz cos t + iν vanishes at to = arccos(ν/z).
This means an infinite set of saddle points all on the real axis, but we’ll aim for the one
between −π and 0. Any of the others would involve a distorted path that
q has to pass
2
00
through a large-amplitude region. Then f (to ) = iz sin(arccos(ν/z)) = −iz 1 − νz2 , which
has argument 3π/2. (I use the negative square root because I’m requiring −π < to < 0, and
sine is negative in this region.) This gives α = π/2−(3π/4) = −π/4, which is good because
it involves going from the upper half-plane to the lower half-plane when crossing the saddle.
(If you use −π/2 as the argument of f 00 , you get α = 3π/4 and then have to adjust the
q
direction by π.) Now plug into the saddle contribution formula: eiz sin to +iνto e−iπ/4 √z2π
2 −ν 2 .
√
2
2
For very large z, ν/z
to → π/2. I’ll also replace z − ν by z. The above
q → 0 and
π
2π i(z+ νπ
−
)
2
4 .
expression becomes z e
1. The pole at z = −1 is within the contour. It has order 2, so calculate the residue by
(z a )0 |−1 = a(−1)a−1 = aeiπ(a−1) = −aeiπa . The integral on the big circle vanishes as
R → ∞, and Rthe integral on Rthe small circle around the origin vanishes as δ → 0. This leaves
iπa
∞ ei2πa xa
xa
dx, so the desired integral equals −2πiae
= sinπaπa .
−2πiaeiπa = 0∞ (x+1)
2 dx − 0
(x+1)2
1−ei2πa
2. a) As |z| → ∞ the integrand looks like z −5/3 . The circumference of the big circle is proportional to |z|, so the integral on the large circle vanishes. As |z| → 0 the integrand looks
like z 1/3 , and the small circle’s circumference
also vanishes,
so that integral vanishes.
R
R
i2π/3 x1/3 dx
x1/3 dx
This leaves the two horizontal segments, 0∞ (1+x)(2+x)
− 0∞ e(1+x)(2+x)
. The contour
integral also equals 2πi times the sum of the residues at the two enclosed poles, z = −1
and z = −2. (Note that (−1)1/3 = (eiπ )1/3 = eiπ/3 . This is the only acceptable way
to write −1 in polar form with 0 < θ < 2π, which is the branch the contour covers.)
R
iπ/3
1/3 eiπ/3 )
1/3 −1)
2π
x1/3 dx
= 2πi(e(1−e−2
= π(2
=√
(21/3 − 1).
This gives 0∞ (1+x)(2+x)
i2π/3 )
sin π/3
3
b)
R∞
cos x2 dx = Re[ 0∞ eix dx]. Integrate eiz dz on a wedge of angle π4 . There are no
singularities
within the
contour, so the contour
integral vanishes. Parametrizing each
R π/4
R
R
2 iπ/2
2
2 i2θ
side, 0 = 0R eix dx + 0 eiR e iReiθ dθ − 0R eix e eiπ/4 dx. In the limit as R → ∞,
the integral on the large arc vanishes because the real part of the exponential goes to
0 faster than the extra factor ofR R blows up. Finally,
the Rintegral along the
√
√ slanted
2
π
∞
2
√
√π . (From
side of the wedge becomes eiπ/4 0∞ e−x dx = 1+i
.
Hence
cos
x
dx
=
0
2 2
2 2
R∞
the imaginary part, it turns out that 0 sin x2 dx has exactly the same value.)
R
2
2
H
0
1
5 i5θ
iθ
1
5
1
5
3. a) Let z = Reiθ . Then the integrand becomes eis( 5 R e +Re ) = eis( 5 R cos 5θ+R cos θ) e−s( 5 R sin 5θ+R sin θ) .
As R gets huge the sign of sin 5θ determines whether the exponential is huge or nearly
zero. The trouble is that sin 5θ has both positive and negative portions in both halfplanes, and thanks to the positive portions neither integral will converge.
b) Use function f (z) = i( 51 z 5 + z). Then f 0 (z) = i(z 4 + 1), and the second derivative is
f 00 (z) = 4iz 3 . The zeroes of f 0 and other relevant calculations for each one are shown
in the table. Note that some of these quantities could be written differently; for
example, in the first line I could have used −3π/4 instead of 5π/4 for the argument of
the second derivative. This might change α by π, but for the time being that doesn’t
matter because we’re just looking at the angle the path takes in going over the saddle.
arg(f 00 (zo )) α
f (zo )
zero of f 0 , zo f 00 (zo )
iπ/4
i3π/4
e
4ie
5π/4
−π/8 i( 15 ei5π/4 + eiπ/4 )
4ieiπ/4
3π/4
π/8
i( 15 e−iπ/4 + ei3π/4 )
ei3π/4
−iπ/4
−i3π/4
e
4ie
π/4
3π/8 i( 15 eiπ/4 + e−i3π/4 )
e−iπ/4
4ie−i3π/4 −π/4
5π/8 i( 15 e−i5π/4 + e−iπ/4 )
The good approximation is with a path through the two saddles in the upper halfplane. It goes over the left saddle at the fairly shallow angle of π/8, then over the
right saddle at −π/8. Avoiding either of these saddles requires passing through a
higher-amplitude region to one side or the other. Similarly, getting from the negative
real axis to the saddle at ei5π/4 requires crossing a path of increasing amplitude that
begins at the saddle and initially goes at an angle of 7π/8. The integrand amplitude
along the real axis is 1, so for large enough s the increasing-amplitude path from the
lower saddle always stays below the axis. The same problem arises in getting from
the saddle at ei7π/4 to the positive real axis.
c) Calculate contributions from the two saddle points separately and add them together.
√
The left saddle gives
1 −iπ/4 +ei3π/4 )
eiπ/8
2πeis( 5 e
√
2 s
=
1 1
i
1
i
π is( 5 ( √2 − √2 )+(− √2 + √2 )) iπ/8
e
e
2s
√
i5π/4
1
+eiπ/4 ) −iπ/8
is( e
e
saddle gives 2πe 5 2√s
q
4 1
4 i
π is(− 5 √2 + 5 √2 ) iπ/8
e
e , and similarly the right
2s
q
q
4 √i
4 √1
+
)
is(
π
π −as
−iπ/8
5 2
5 2
e
.
Adding
these
gives
e
e (2 cos(as − π8 ))
2s
2s
1
π). Here a = 5√4 2 .
8
=
q
=
q
2π −as
e
s
=
cos(as −