Download Complex Analysis 7, Exercises Solution - Mathematics and more Exercises Complex Numbers Theory in PDF only on Docsity!
Solution of Homework Assigned on February 19, 2009
due February 24, 2009
(numbering of problems continued from
the last assignment with the same due date)
Problem 4 (from Stein & Shakarchi, p.103, #2). Evaluate the integral
−∞
dx
1 + x^4
Where are the poles of
1 1+z^4
Solution of Problem 4. Let f (z) =
1 z^4 +
. The zeroes of z
4
quartic roots of −1 which are
e
πi (^4) , e
3 πi (^4) , e
5 πi (^4) , e
7 πi (^4).
They are the four poles of f (z), all simple. The first two poles
e
πi (^4) , e
3 πi 4
lie on the upper half-plane. We now integrate f (z)dz over the boundary of
the upper half-disk of radius R centered at the origin and let R → ∞. The
integral ∫
CR
f (z)dz
over the upper half-circle CR of radius R centered at 0 in the counter-
clockwise sense goes to 0 as R → ∞, because
CR
f (z)dz
πR
R^4 − 1
for R > 1.
According to l’Hˆopital’s rule the residue resa f of f at one of the four simple
poles a is given by
resa f = lim z→a
z − a
z^4 + 1
4 a^3
Hence the two residues at the two simple poles in the upper half-plane are
res e
πi 4
f =
4 e
3 πi 4
e
− 3 πi 4
− 1 − i
res e
π 3 i 4
f =
4 e
9 πi 4
e
− 9 πi 4
e
−πi 4
1 − i
By applying the theory of residues to the integration of f (z)dz over the
boundary of the upper half-disk of radius R centered at the origin and letting
R → ∞, we get
∫ (^) ∞
x=−∞
dx
1 + x
4
= 2πi
res e
πi 4
f + Res e
3 πi 4
f
π √ 2
Problem 5 (from Stein & Shakarchi, p.103, #4). Show that
∫ ∞
−∞
x sin x
x
2
2
dx = πe
−a
for all a > 0.
Solution of Problem 5. Let
f (z) =
ze
iz
z
2
2
The poles of f (z) are the zeroes of z
2
2 which are ai and −ai. Only ai
is in the upper half-plane. We now integrate f (z)dz over the boundary of
the upper half-disk of radius R centered at the origin and let R → ∞. The
integral (^) ∫
CR
f (z)dz
over the upper half-circle CR of radius R centered at 0 in the counter-
clockwise sense goes to 0 as R → ∞, because integration by parts by in-
tegrating e
iz yields
CR
ze
iz
z^2 + a^2
dz =
ze
iz
i (z^2 + a^2 )
]z=−R
z=R
CR
e
iz
d
dz
z
i (z^2 + a^2 )
dz
ze
iz
i (z^2 + a^2 )
]z=−R
z=R
CR
e
iz a
2 − z
2
i (z
2
2 )
2
dz
and because |e
iz | ≤ 1 for z ∈ CR and
∣ ∣ ∣ ∣ ∣
ze
iz
i (z^2 + a^2 )
]z=−R
z=R
2 R
R^2 − a^2
for R > a
Let
f (z) =
e
− 2 πiξz
(z^2 + 1)
2
We are going to integrate f (z)dz over the boundary of the lower half-disk of
radius R centered at the origin and let R → ∞. The reason why the lower
half-disk is chosen, instead of the upper half-disk, is that there is a negative
sign in the exponent of the numerator e
− 2 πiξz of the expression defining f (z).
The poles of f (z) are the zeroes of z
2
- 1 which are i and −i. Only −i is
in the lower half-plane and it is a double pole of f (z). The integral
CR
f (z)dz
over the lower half-circle CR of radius R centered at 0 in the counter-clockwise
sense goes to 0 as R → ∞, because
∣e−^2 πiξz^
∣ (^) ≤ 1 for Im z ≤ 0 and ξ ≥ 0
and (^) ∣
∣ ∣ ∣
CR
e
− 2 πiξz
(z^2 + 1)
2
dz
πR
(R^2 − 1)
2
for R > 1.
The residue res−i f of f at the double pole −i of f is given by
res−i f = lim z→ai
d
dz
(z + i)
2 e
− 2 πiξz
(z^2 + 1)
2
d
dz
e
− 2 πiξz
(z − i)
2
z=−i
− 2 πiξ
e
− 2 πiξz
(z − i)
2
e
− 2 πiξz
(z − i)
3
z=−i
= − 2 πiξ
e
− 2 πξ
(− 2 i)
2
e
− 2 πξ
(− 2 i)
3
= −i
(1 + 2πξ)e
− 2 πξ
By applying the theory of residues to the integration of f (z)dz over the
boundary of the lower half-disk of radius R centered at the origin and letting
R → ∞, we get
−∞
e
− 2 πixξ
(1 + x^2 )
2
dx = 2πi res−if =
π
(2π|ξ| + 1) e
− 2 π|ξ| .
