Problem set 6, TMA4175
Karl Yngve Lervåg
February, 2008
Problem 1. (Ex 9.15 from Nevanlinna)
Investigate the zeros and singularities of the function
2
f (z) = e−1/z sin(1/z).
We have that
⇔
f (z) = 0
1
sin( ) = 0
z
⇔
z=
1
,
nπ
where n = ±1, ±2, ±3, . . . . Also, we have that
2
f (z) = e−1/z sin(1/z)
∞
∞
X
(−1)n X
(−1)n
=
z 2n n!
z 2n+1 (2n + 1)!
n=0
n=0
1
1
1
1
= 1 − 2 + 4 − ···
−
+ ···
z
2z
z 6z 3
1
7
= − 3 + ···
z 6z
and we see that there is an essential singularity at z = 0.
1
Problem 2. (Ex 10.2 from Nevanlinna)
Using the residue theorem evaluate the following integrals:
(a)
I
γ
ez
dz,
z
where γ is the circle |z| = 1.
(b)
I
tan(πz)dz,
γ
where γ is the circle |z| = n and n is a positive integer.
(a) We have one singularity at z = 0 in the given circle, a simple pole with
residue 1, thus
I z
e
dz = 2πi.
γ z
(b) tan(πz) has exactly 2n singularities in the circle |z| = n. For the simple
pole at zk = π/2 + k, k = 0, ±1, ±2, . . . , we have that
sin(πzk )
1
sin(πz)
Resz=zk
=
=− ,
cos(πz
π sin(πzk )
π
which gives
I
tan(πz)dz = 2πi ·
γ
2
−2n
= −4ni.
π
Problem 3. (Ex 10.3 from Nevanlinna)
Evaluate the integral
I iz
e
dz,
γ z
where γ is the curve consisting of the semicircles |z| = R, Im z > 0 and
|z| = r (< R), Im z > 0, together with the segments r ≤ z ≤ R and
−R ≤ z ≤ −r. Let R → ∞ and r → 0, and find the value of
Z ∞
sin x
dx.
x
0
Consider the integral
I
I=
γ
eiz
dz,
z
where γ is the closed curve consisting of the segments given in the exercise
text. Since there are no singularities inside the contour, we have that I = 0.
Consider the integral over the semicircle γ1 : |z| = R, Imz > 0,
Z
Z π
eiz iR(cos θ+i sin θ) ≤
i
e
dθ
γ1 z
0
Z π/2
≤2
e−R sin θ dθ
0
Z
π/2
e−R2θ/π dθ
π 1 − e−R
.
=
2R
≤2
0
When R → ∞, this goes to 0. For the semicircle γ2 : |z| = r, Imz > 0, we
have
Z 0
Z
eiz
= lim i
eir(cos θ+i sin θ) dθ
lim
r→0 γ z
r→0
−π
2
Z 0
=
1dθ
−π
= −iπ.
Also,
Z
lim
−r
r→0,R→∞ −R
eix
dx +
x
Z
R
eix
dx = lim
→0
x
r
Z
|x|>
thus
Z
0
∞
sin x
dx = π/2.
x
3
cos x
dx + 2i
x
Z
x>
sin x
dx,
x
Problem 4. (Ex 10.4 from Nevanlinna)
By using the well-known formula of integral calculus,
√
Z ∞
π
−x2
e dx =
,
2
0
evaluate the so-called Fresnel integrals
r
Z ∞
Z ∞
1 π
2
2
.
cos(x )dx =
sin(x )dx =
2 2
0
0
2
We consider the integral of the function w(z) = eiz along a closed curve γ
consisting of the two segments γ1 and γ2 joining the origin to the points R
and Reiπ/4 respectively, and the minor arc γ3 of the circle |z| = R between
these points, and we see that
I
Z
Z
Z
w(z)dz =
w(z)dz +
w(z)dz +
w(z)dz = 0.
γ
γ1
γ2
γ3
First,
Z
Z π/4
2
w(z)dz ≤
e−R sin θ dθ
γ3
0
Z π/4
2
e−R 2θ/π dθ
≤
0
π −R2 /2
=
1
−
e
2R2
which vanishes when R → ∞. Also,
r
r
Z R
Z
1 π
1 π
−r 2 iπ/4
w(z)dz = −
e e
dr → −
+i
,
2 2
2 2
γ2
0
when R → ∞, so
Z
Z
R
γ1
2
eix dx
w(z)dz =
0
Z
R
Z
2
=
cos(x )dx + i
0
r
r
1 π
1 π
→
+i
,
2 2
2 2
when R → ∞, hence
Z ∞
2
Z
∞
0
0
4
sin(x2 )dx
0
1
sin x dx =
2
2
cos x dx =
R
r
π
.
2
Problem 5. (Ex 10.5 from Nevanlinna)
Derive the formula
Z +∞
dx
1 · 3 · 5 · · · (2n − 1)
=π
2
n+1
2 · 4 · 6 · · · (2n)
−∞ (1 + x )
(n ≥ 1).
Consider the function w(z) = 1/(1 + z 2 ), and integrate it along the upper
half disk of radius R. We see that
Z π
Z R
I
iReiθ dθ
dx
dz
+
= 2πiRi
=
2
n+1
2
2iθ
n+1
(1 + z 2 )n+1
0 (1 + R e )
−R (1 + x )
where the second term on the left-hand side clearly goes to 0 when R goes
to infinity. We need to find the residue Ri at z = i, a pole of order n + 1.
We have that
#n+1
n+1 "
1
1
=
(z − i)(z + i)
2i(z − i)(1 + z−i
2i )
!n+1
n+1
1
1
=
2i(z − i)
1 + z−i
2i
n+1 X
∞ n+j
−(z − i) j
1
,
=
2i(z − i)
j
2i
j=0
where we find the residue when j = n, n ≥ 1,
n+1 1
2n
−(z − i) n
Ri = (z − i)
2i(z − i)
n
2i
n+1 n
1
2n
−1
=
2i
n
2i
2n+1 1
2n
= (−1)n
2i
n
1
1
1 · 2 · 3 · · · 2n
=
2n 2n
2i (1 · 2 · · · n)(1 · 2 · · · n)
1 1 · 3 · 5 · · · (2n − 1)
=
.
2i
2 · 4 · 6 · · · (2n)
In the end, this leaves us with
Z
lim
R
R→∞ −R
dx
1 · 3 · 5 · · · (2n − 1)
= 2πiRi = π
.
(1 + x2 )n+1
2 · 4 · 6 · · · (2n)
5
Problem 6. (Ex 10.6 from Nevanlinna)
Derive the formula
Z ∞
cos(mx)
π
dx = e−m
2
1+x
2
0
(m > 0).
We now consider the function w(z) = eimz /(1 + z 2 ) with m > 0 and the
closed curve consisting of the segment from −R to R and the top half circle
connecting the two points. There is one simple pole at z = i, so
I
e−m
eimz
dz = 2πi
= πe−m .
2
1+z
2i
It is easy to show that the integral over the half circle vanishes when R → ∞,
indeed
Z
Z π imR cos θ−mR sin θ
e
iθ w(z)dz ≤
iRe dθ
1 + R2 e2iθ
γ
0
Z π
R
≤
e−mR sin θ dθ
2
0 R −1
πR
< 2
,
R −1
which goes to zero when R → ∞. Therefore
Z ∞
cos(mx) + i sin(mx)
dx = πe−m ,
2
1
+
x
−∞
and hence
Z
0
∞
cos(mx)
π
dx = e−m .
2
1+x
2
6