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18.112 Functions of a Complex Variable
Fall 2008
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Lecture 15: Contour Integration and Applications
(Text 154-161)
Remarks on Lecture 15
In parts 4 and 5 (p. 154-160) some clarification of the use of the logarithm are
called for.
Example 4 p.159
The relation
(−z)2α = e2πiα z 2α
which is crucial for proof deserves explanation.
We consider the function
l
logθ z = log |z| + iargθ z
in the region C − lθ (the plane
with the ray lθ removed) where
the angle is fixed by
Fig. 15-1
θ < argθ z < θ + 2π.
In the problem of computing
∞
xα R(x) dx
0
we consider
log− π (z)
2
1
in the plane C with the negative imaginary axis removed and us the Residue theorem
on the contour in Fig. 4.13. As in the text we arrive at the integral
∞
∞(
)
z 2α+1 + (−z)2α+1 R(z 2 ) dz.
z 2α+1 R(z 2 ) dz =
−∞
0
On the right z belongs to (0, ∞) and
(
π π )
−π
3π
log− π2 (z) = log |z| +
−
+
i,
< arg− π2 z <
,
2
2
2
2
π 3π
i
log− π2 (−z) = log |z| + − +
2
2
= log− π2 (z) + iπ,
z > 0.
Thus for z > 0,
(2α+1) log−
π (−z)
(−z)2α+1 = e
2
= e(2α+1)(log |z|+iπ)
= −e2αiπ z 2α+1 ,
so the last integrals combine to
∞
2αiπ
(1 − e
z 2α+1 R(z 2 ) dz.
)
0
For z > 0 we have from the above
log− π2 (z) = log |z|,
so
1
2πi
∞
∞
1
(1 − e2αiπ )
2πi
1
= − eαπi sin πα
π
z 2α+1 R(z 2 ) dz =
−∞
x2α+1 R(x2 ) dx
0
∞
(1)
2α+1
x
2
R(x ) dx.
0
The left hand side of (1) is the sum of the residues of
z 2α+1 R(z 2 ) = f (z)
in the upper half plane. If
g(z)
,
h(z)
where g and h are holomorphic, g(a) = 0, and h has a simple zero at a, then
R(z 2 ) =
Resz=a f (z) = z (2α+1) (a)
where
g(a)
,
h′ (a)
(2α+1) log−
π (z)
z 2α+1 = e
2
2
.
(2)
Example: Exercise 3(g) p.161
To calculate
Z
∞
Z
∞
1
x
3
0
dx
,
1 + x2
2
we use x = t and arrive at
5
z
3
−∞
dz
1 + z 4
in (1). The poles in the upper half plane are
π
z =
e
i
4
z =
e
i( 4
+ 2
)
.
π
and
π
We use (2) to calculate the residues:
)
5
5 (
1
1
i
π
3
e
4
Res
z=ei
π4
z
3
=
z
π
4
1 +
z
4(e
i
4
)3
π
5
1
log π ei
4
= e
3
−
2
( )
i
π 3
4(e 4
)
5
π
1
= e
3
(i 4
)
i
π 3
4(e
4
)
1
π
= e−i 3
,
4
and
(
3π )
5
5
1
1
i 4
3
Res
z=ei 34
π z
3
e
=
z
3π
1 + z 4
4(ei 4
)3
5
= e
3
log−
π (i(− π2
+ 5
4
π ))
2
1
3π
4(ei 4
)3
1
= e−iπ .
4
Thus (1) gives
so
Z
1
−i π 1 −iπ
1
1
πi
π
∞ 5
dx
e
3
+
e
=
−
e
3
sin
x
3
,
4
4
π
3 0
1 + x4
Z ∞
5
dx
π
x
3
= √ .
4
1+x
2 3
0
3
Example 5 p.160
The last four lines on the page are a bit misleading because the specific logarithm
has already been chosen. So here is a completion of the proof after the equation
Z π
Log(−2ieix sin x) dx = 0.
0
We know (Lecture 2) that
Log(z1 z2 ) = Logz1 + Logz2 ,
if − π < Argz1 + Argz2 < π.
Using this for z = 2 sin x we get
Zπ
Z
log(2 sin x) dx +
0
π
Log(−ieix ) dx = 0.
0
But
πi
,
Logeix = ix (0 < x < π),
2
so since − π2 + x is in (−π, π), (3) implies
Log(−i) = −
Log(−ieix ) = −
πi
+ ix.
2
Now (4) implies the result
Z
π
0
log sin θ dθ = −π log 2.
4
(3)
(4)