UNIVERSITY OF BRISTOL
Department of Physics
Level 3 Mathematical Methods 33010 (2009/2010)
Definite Integrals by Contour Integration
Many kinds of (real) definite integrals can be found using the results we have found
for contour integrals in the complex plane.
This is because the values of contour
integrals can usually be written down with very little diﬃculty.
We simply have
to locate the poles inside the contour, find the residues at these poles, and then
apply the residue theorem. The more subtle part of the job is to choose a suitable
contour integral i.e. one whose evaluation involves the definite integral required. We
illustrate these steps for a set of five types of definite integral.
Type 1 Integrals
Integrals of trigonometric functions from 0 to 2
π
:
I
=
2
π
0
(trig function)
dθ
By “trig function” we mean a function of cos
θ
and sin
θ
.
The obvious way to turn this into a contour integral is to choose the unit circle as
the contour, in other words to write
z
= exp
iθ
, and integrate with respect to
θ
.
On
the unit circle
, both cos
θ
and sin
θ
can be written as simple algebraic functions of
z
:
cos
θ
=
1
2
(
z
+ 1
/z
)
sin
θ
=
1
2
i
(
z
−
1
/z
)
and making this replacement turns the trigonometric function into an algebraic
function of
z
whose poles can be easily found.
Example:
I
=
2
π
0
dθ
1 +
a
cos
θ
where
−
1
< a <
+1
I
=
1
1 +
a
2
(
z
+
1
z
)
dz
iz
=
2
i
dz
2
z
+
az
2
+
a
The poles of the function being integrated lie at the roots of the equation
az
2
+2
z
+
a
= 0 i.e. at the points
z
±
=
1
a
−
1
±
√
1
−
a
2
1
C
Z
+
Z

Of the poles, only
z
+
lies inside the unit circle, so
I
= 2
πiR
+
where
R
+
is the
residue at
z
+
To find the residue we note that this is a simple pole and if we write
the integrand as
f
(
z
) =
g
(
z
)
/h
(
z
) the residue at
z
+
is:
g
(
z
+
)
h
(
z
+
)
=
2
2
i
(
az
+
+ 1)
=
1
i
√
1
−
a
2
Hence the integral required is 2
π/
√
1
−
a
2
Type 2 Integrals
Integrals such as
I
=
+
∞
−∞
f
(
x
)
dx
or, equivalently, in the case where
f
(
x
) is an even function of
x
I
=
+
∞
0
f
(
x
)
dx
can be found quite easily, by inventing a closed contour in the complex plane which
includes the required integral. The simplest choice is to close the contour by a very
large semicircle in the upper halfplane. Suppose we use the symbol “
R
” for the
radius. The entire contour integral comprises the integral along the real axis from
−
R
to +
R
together with the integral along the semicircular arc. In the limit as
R
→ ∞
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 Fall '19
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