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Computing Definite Integrals
Adrian Down
November 17, 2005
1 Convergence of real-valued functions
Suppose we have a continuous function f : R → C. We would like to know
when the following integral exists.
lim
R 2 → ∞
R 1 → ∞
∫ R
2
−R 1
d(x)dx
As with infinite series, there is no single test. We can establish a sufficient
condition, however.
Fact. If ∃C < ∞ such that
∫ (^) R 2
−R 1
|f (x)|dx ≤ C ∀R 1 , R 2
then the limit (R 1 , R 2 → ∞) exists.
Example.
f (x) =
cos(x)
1 + x
2
|f (x)| ≤
1 + x
2
x ≥ 1 ⇒ |f (x)| ≤ x
− 2
and
1
dx
x
2
Similar arguments hold for x ≤ −1. Thus,
∫ (^) R 2
−R 1
|f (x)|dx < C < ∞
f satisfies the condition above, and so the integral exists.
Note. In the previous examples that we did, we proved directly that the
integral existed. The formulation above is good to use as a preliminary check
when approaching a problem to determine if the integral actually converges.
2
∫ ∞
e
ıx
x
dx
2.1 Limiting Behavior
e
ıx
x
|x|
lim N →∞
∫ N
1
dx
x
lim → 0
ln
As written, this integral is problematic for large positive or negative x, as
well as x = 0.
2.2 Related example
As a warm up, consider the simpler integral,
∫ (^1)
− 1
dx
x
We need to decide how to define this integral. Consider,
lim → 0
<|x|< 1
dx
x
Because the integrand is odd, the contributions on either side of the gap of
width cancel exactly, leaving
lim → 0
Only the imaginary part of the integral remains,
P
−∞
e
ıx
x
dx = ıP
−∞
sin(x)
x
dx
We solve the first integral, although the second is generally more useful for
applications. The second follows immediately from the first.
2.4.1 Define a contour
Make a closed loop and use Cauchy’s Residue theorem. Traverse the real
axis in the usual direction. Take a rectangle with height R and length R.
Make a semi-circular indentation centered at 0 of radius . Call the whole
contour ΓR.
The function to integrate is
f (z) =
e
ız
z
2.4.2 Singularities
The only singularity is at z = 0. We have excluded this singularity from the
inside of the contour, so from Cauchy’s Residue Theorem,
∫
Γ
f = 0 ∀, R
Note. Although the integral of f along the entire path Γ is 0, the integral
that we would like to calculate along the real axis is not 0.
2.4.3 Indentation in the contour
First consider the semi-circle about 0. It will be the most important integral.
On this path, expand e
ız to isolate the singular behavior of the function at
e
ız = 1 + (e
ız − 1)
Using this substitution,
∫
C
e
ız
z
dz =
C
dz
z
C
e
ız − 1
z
dz
The first integral is related to one calculated previously.
C
dz
z
(2ıπ) = −ıπ
Note. The
1 2
appears because we previously integrated around an entire circle,
and it is negative because the circle is traversed in the opposite direction.
Using the Estimation Lemma,
C
e
ız − 1
z
dz
≤ π · max |z|=
e
ız − 1
z
The integrand can be be Taylor expanded about 0.
e
ız− 1
z
= ı +
ı
2
Since the function has an extension which is a convergent series at z = 0,
the singularity is removable. Since the function is differentiable on the rest
of the complex plane, it is bounded everywhere. Thus the integral tends to
0 as → 0.
We will see that the only contribution to the integral comes from this
indentation in the contour, so
∫
C
e
ız
z
dz → −ıπ
2.4.4 Other concours in ΓR
Now we show that the integral along all of the other paths is 0.
Horizontal path From the estimation lemma applied to the top of the
rectangle γ 2 ,
γ 2
e
ız
z
≤ L(γ 2 ) · max −R≤x≤R
|e
ıx − R
x + ıR
≤ (2R)
e
−R
R
R→∞
Vertical paths
limits, → 0 and R → ∞.
0 = P
−∞
e
ıx
x
dx − ıπ
⇒ P
−∞
sin(x)
x
dx = π
Note. • This method of integrating around singular points holds for any
singularities on the real axis.
- We could try to go around a path in the lower complex plane. However,
then we would have
e
ız = e
ıx e
−y y < 0
which is not bounded. Thus the desired limits cold not be taken.
3 Summation of infinite series
3.1 Motivating examples
Example. •
∞ ∑
n=
n^2
π
2
sin
2 (w)
∞ ∑
n=−∞
(w − nπ)^2
Note. – The poles of the two expression for the function are the
same, which is reassuring.
- This series is unlike anything we have learned thus far. It is an
infinite series of functions. It is not a power series, nor is it a
Laurent series.
z
e
z − 1
z
∞ ∑
n=
2 z
2
z
2
2 n
2
Note. • We would like to see the relationship between these series that
allows their sums to be calculated.
- The method that we develop considers a particular complex function,
from which a set of identities is developed. It cannot be used to generate
a formula given an arbitrary sum.
3.2 General method by example
g(z) =
cot(πz)
z
2
z
2
cos(πz)
sin(πz)
3.2.1 Singularities
sin(πz) vanishes at the real integers, which includes the singularity of
1 z^2
at
z = 0.
3.2.2 Construct a path
General method Consider a semi-circular path in the upper half plane of
radius r < 1. The path only encloses the singularity at z = 0. The integral
is, by Cauchy’s residue theorem,
∫
γ
g = 2πı · res(g, 0)
If the radius of the path is increased to be 1 < r < 2,
∫
γ
g = 2πı
∑
n=− 1
res(g, n)
Our approach is to let the radius of the path of integration go to ∞. The
integral is then an infinite sum of the values of the residues of g(z) at the
singularities. We hope that the integral goes to 0 in the limit as R → ∞,
which will give an identity in terms of the sum of the residues.
From Cauchy’s Residue theorem,
γN
h = 2πı
+N ∑
n=−N
res(h, n)
We would like to show two things: that the integral on the left is 0, and the
explicit value for the residues.
