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Solution of Homework Assigned on February 24, 2009
due March 3, 2009
Problem 1 (from Stein & Shakarchi, p.104, #6). Show that
∫ ∞
−∞
dx
(1 + x^2 )
n+1 =
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
· π.
Solution of Problem 1. The poles of
f (z) =
(1 + z^2 )
n+
are at z = i and z = −i, both of order n + 1. By integrating f (z) dz over
the boundary of the upper half-disk of radius R centered at 0 and letting
R → ∞, we obtain for n ≥ 0
∫ (^) ∞
x=−∞
dx
(1 + x^2 )
n+
= 2πi resif,
where
resif =
n!
d
n
dz
n
(z − i)
n+1 1
(1 + z^2 )
n+
z=i
n!
d
n
dz
n
(z + i)
n+
z=i
n (n + 1)(n + 2) · · · (2n)
n!
(z + i)
2 n+
z=i
n (n + 1)(n + 2) · · · (2n)
n!
(2i)
2 n+
−i
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
Hence (^) ∫ ∞
−∞
dx
(1 + x^2 )
n+
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
· π.
Problem 2 (from Stein & Shakarchi, p.104, #7). Prove that
∫ (^2) π
0
dθ
(a + cos θ)
2
2 πa
(a
2 − 1)
3 2
whenever a > 1.
Solution of Problem 2. Use the parametrization a = e
iθ for 0 ≤ θ ≤ 2 π
so that
cos θ =
z +
z
, dθ =
dz
iz
Hence
∫ (^2) π
θ=
dθ
(a + cos θ)
2
= −i
|z|=
dz
z
a +
1 2
z +
1 z
= −i
|z|=
4 zdz
(2az + z^2 + 1)
2
= −i
|z|=
4 zdz ( (z + a)
2 − (a
2 − 1)
= − 4 i
|z|=
zdz ( z + a +
a
2 − 1
z + a −
a
2 − 1
For the meromorphic function
f (z) =
z ( z + a +
a
2 − 1
z + a −
a
2 − 1
the only poles are at −a −
a
2 − 1 and −a +
a
2 − 1, both double poles.
Since a > 1, clearly the pole −a −
a^2 − 1 is outside the unit circle. The
point −a +
a
2 − 1 has absolute value < 1, because we can look at the right-
angled triangle with two sides of length 1 and
a^2 − 1 and with hypothenuse
a. The difference of the length of the hypothenuse a and the length of one
side abutting the right angle is less than the length of the other side which
is 1. The residue of f at the double pole −a +
a
2 − 1 is computed by
res −a+
√ a^2 − 1
f =
d
dz
z + a −
a^2 − 1
f (z)
z=−a+
√ a^2 − 1
d
dz
z ( z + a +
a
2 − 1
z=−a+
√ a^2 − 1
−z + a +
a^2 − 1 ( z + a +
a
2 − 1
z=−a+
√ a^2 − 1
a
4 (a
2 − 1)
3 2
log z
z + ai
z=ai
log a +
π 2
i
2 ai
The theory of residues yields
∫ (^) ∞
x=
log x dx
x
2
2
x=−∞
(log(−x) + πi) dx
x
2
2
= 2πi resaif =
π
log a +
π 2
i
a
Taking the real part of both sides, we get
∫ (^) ∞
x=
log x dx
x
2
2
x=−∞
log(−x) dx
x
2
2
π log a
a
Changing the sign of the variable x from x to −x in the second integral on
the left-hand side, we get
x=
log x dx
x
2
2
π log a
a
and (^) ∫ ∞
x=
log x dx
x
2
2
π log a
2 a
Problem 4 (from Stein & Shakarchi, p.105, #12). Suppose u is not an integer.
Prove that ∞ ∑
−∞
(u + n)
2
π
2
(sin πu)
2
by integrating
f (z) =
π cot (πz)
(u + z)
2
over the circle |z| = RN = N +
1 2
(N an integer and |N | ≥ |u|), adding the
residues of f inside the circle, and letting N tend to infinity.
Solution of Problem 4. For fixed u which is not an integer, the function
f (z) =
π cot (πz)
(u + z)
2
has simple poles at z equal to an integer n and a double pole at z = −u.
The residues are computed as follow.
res−u f = lim z→−u
d
dz
(π cot (πz)) = −
π
2
sin
2 (πu)
resn f = lim z→n
(z − n)
π cot πz
(u + z)
2
= lim z→n
(z − n)
π cos (πz)
sin (πz) (u + z)
2
= lim z→n
π(z − n)
sin (π (z − n))
cos (π (z − n))
(u + z)
2 =^
(u + n)
2
We have to check that (^) ∫
|z|=N + (^12)
f (z)dz → 0
as N → ∞. There exists a positive integer N 0 such that for N ≥ N 0 the two
conditions |z| = N +
1 2
and |Im z| ≤ 1 imply that the distance from Rez to
the nearest integer is ≥
1 4
. Since cot (πz) has period 1 and its poles occur at
z equal to an integer, the function cot (πz) is uniformly bounded on the set
of all z satisfying |Im z| ≤ 1 with the property that the distance from Rez to
the nearest integer is ≥
1 4
. When z = x + iy with y ≥ 1, we have
cot (πz) = i
e
iz
−iz
eiz^ − e−iz^
= i
e
2 iz
e^2 iz^ − 1
and
|cot (πz)| ≤
1 + e
− 2 y
1 − e−^2 y^
1 + e
− 2
1 − e−^2
Likewise, when z = x + iy with y ≤ −1, we have
cot (πz) = i
e
iz
−iz
e
iz − e
−iz
= i
1 + e
− 2 iz
1 − e
− 2 iz
and
|cot (πz)| ≤
1 + e
2 y
1 − e^2 y^
1 + e
− 2
1 − e−^2
We conclude that
M := sup N ≥N 0
sup
|z|=N +
1 2
|cot (πz)| < ∞.
Thus (^) ∣ ∣ ∣ ∣ ∣
|z|=N + (^12)
f (z)dz
M
N +
1 2
− |u|
2
2 π
N +
for N ≥ N 0 and N ≥ |u|. This implies that
∫
|z|=N + (^12)
f (z)dz → 0 as N → ∞.
