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Residue Theory
“Residue theory” is basically a theory for computing integrals by looking at certain terms in the
Laurent series of the integrated functions about appropriate points on the complex plane. We
will develop the basic theorem by applying the Cauchy integral theorem and the Cauchy integral
formulas along with Laurent series expansions of functions about the singular points. We will
then apply it to compute many, many integrals that cannot be easily evaluated otherwise. Most
of these integrals will be over subintervals of the real line.
17.1 Basic Residue Theory
The Residue Theorem
Suppose f is a function that, except for isolated singularities, is single-valued and analytic on
some simply-connected region R. Our initial interest is in evaluating the integral
C 0
f (z) dz.
where C 0 is a circle centered at a point z 0 at which f may have a pole or essential singularity.
We will assume the radius of C 0 is small enough that no other singularity of f is on or enclosed
by this circle. As usual, we also assume C 0 is oriented counterclockwise.
In the region right around z 0 , we can express f (z) as a Laurent series
f (z) =
∞ ∑
k=−∞
ak (z − z 0 )
k ,
and, as noted earlier somewhere, this series converges uniformly in a region containing C 0. So
C 0
f (z) dz =
C 0
∞ ∑
k=−∞
ak (z − z 0 )
k dz =
∞ ∑
k=−∞
ak
C 0
(z − z 0 )
k dz.
But we’ve seen ∮
C 0
(z − z 0 )
k dz for k = 0 , ± 1 , ± 2 , ± 3 ,...
3/10/
Chapter & Page: 17–2 Residue Theory
before. You can compute it using the Cauchy integral theorem, the Cauchy integral formulas,
or even (as you did way back in exercise 14.14 on page 14–17) by direct computation after
parameterizing C 0. However you do it, you get, for any integer k ,
C 0
(z − z 0 )
k dz =
0 if k 6 = − 1
i 2 π if k = − 1
So, the above integral of f reduces to
C 0
f (z) dz =
∞ ∑
k=−∞
ak
0 if k 6 = − 1
i 2 π if k = − 1
= i 2 πa− 1.
This shows that the a− 1 coefficient in the Laurent series of a function f about a point z 0
completely determines the value of the integral of f over a sufficiently small circle centered at
z 0. This quantity, a− 1 , is called the residue of f at z 0 , and is denoted by
a− 1 ,z 0 or Resz 0 ( f ) or · · ·
In fact, it seems that every author has their own notation. We will use Resz 0 ( f ).
If, instead of integrating around the small circle C 0 , we were computing
∮
C
f (z) dz
where C is any simple, counterclockwise oriented loop in R that touched no point of singularity
of f but did enclose points z 0 , z 1 , z 2 , … at which f could have singularities, then, a
consequence of Cauchy’s integral theorem (namely, theorem 15.5 on page 15–7) tells us that
∮
C
f (z) dz =
k
Ck
f (z) dz
where each Ck is a counterclockwise oriented circle centered at zk small enough that no other
point of singularity for f is on or enclosed by this circle. Combined with the calculations done
just above, we get the following:
Theorem 17.1 (Residue Theorem)
Let f be a single-valued function on a region R , and let C be a simple loop oriented counter-
clockwise. Assume C encircles a finite set of points {z 0 , z 1 , z 2 ,.. .} at which f might not be
analytic. Assume, further, that f is analytic at every other point on or enclosed by C. Then
∮
C
f (z) dz = i 2 π
k
Resz k ( f ). (17.1)
In practice, most people just write equation (17.1) as
C
f (z) dz = i 2 π ×
[
sum of the enclosed residues
]
The residue theorem can be viewed as a generalization of the Cauchy integral theorem and
the Cauchy integral formulas. In fact, many of the applications you see of the residue theorem
can be done nearly as easily using theorem 15.5 (the consequence of the Cauchy integral theorem
used above) along with the Cauchy integral formulas.
Chapter & Page: 17–4 Residue Theory
17.2 “Simple” Applications
The main application of the residue theorem is to compute integrals we could not compute (or
don’t want to compute) using more elementary means. We will consider some of the common
cases involving single-valued functions not having poles on the curves of integration. Later, we
will add poles and deal with multi-valued functions.
Integrals of the form
2 π
0
f (sin θ, cos θ) dθ
Suppose we have an integral over ( 0 , 2 π) of some formula involving sin(θ) and cos(θ) , say,
∫ (^2) π
0
dθ
2 + cos θ
We can convert this to an integral about the unit circle by using the substitution
z = e
i θ .
Note that z does go around the unit circle in the counterclockwise direction as θ goes from 0
to 2π. Under this substitution, we have
dz = d
[
e
i θ
]
= ie
i θ dθ = i z dθ.
So,
dθ =
1
i z
dz = −i z
− 1 dz.
For the sines and cosines, we have
cos θ =
e
i θ
−i θ
2
z +
(^1) / z
2
z + z
− 1
2
and
sin θ =
e
i θ − e
−i θ
2 i
z −
(^1) / z
2 i
z − z
− 1
2 i
These substitutions convert the original integral to an integral of some function of z over the unit
circle, which can then be evaluated by finding the enclosed residues and applying the residue
theorem.
! ◮ Example 17.1: Let’s evaluate ∫ 2 π
0
dθ
2 + cos θ
Letting C denote the unit circle and applying the substitution z = e
i θ , as described above,
we get
∫ 2 π
0
dθ
2 + cos θ
C
−i z − 1 dz
2 +
1 2
[ z + z − 1
]
C
2 z
2 z
−i z
− 1
2 +
1 2
[ z + z − 1
] dz
C
−i 2
4 z +
[ z 2
] dz = − 2 i
C
1
z 2
dz.
“Simple” Applications Chapter & Page: 17–
Letting
f (z) =
1
z 2
and applying the residue theorem, the above becomes
2 π
0
dθ
2 + cos θ
= − 2 i × i 2 π ×
[
sum of the residues of f (z) in the unit circle
]
= 4 π ×
[
sum of the residues of f (z) in the unit circle
]
To find the necessary residues we must find where the denominator of f (z) vanishes,
z
2
Using the quadratic formula, these points are found to be
z± =
− 4 ±
√ 4 2 − 4
2
So
f (z) =
z −
[
])(
z −
[
]).
Now, since
|z+| =
∣−^2 +
∣ ≈ |−^2 +^1.^7 | =^0.^3
while
|z−| =
Clearly, − 2 +
3 is the only singular point of f (z) enclosed by the unit circle, and the
singularity there is a simple pole. We can rewrite f (z) as
f (z) =
g(z)
z −
[ − 2 +
√ 3
] where g(z) =
1
z −
[ − 2 −
√ 3
].
Thus,
Resz
[ f ] = g(z+) = g(− 2 +
1 [ − 2 +
√ 3
] −
[ − 2 −
√ 3
] =
1
2
√ 3
Plugging this back into the last formula obtained for our integral, we get
2 π
0
dθ
2 + cos θ
= 4 π ×
[
sum of the residues of f (z) in the unit circle
]
= 4 π Resz
[ f ]
= 4 π
1
2
√ 3
2 π √ 3
version: 3/10/
“Simple” Applications Chapter & Page: 17–
counterclockwise, the direction of travel on the interval IR is from x = −R to x = R (the
normal ‘positive’ direction of travel on the X–axis) and, so,
IR
f (z) dz =
R
−R
f (x) d x.
Since we are letting R → ∞ and there are only finitely many singularities, we can always
assume that we’ve taken R large enough for Γ
R to enclose all the singularities of f in the
upper half plane (UHP). Then
i 2 π
[
sum of the residues of f in the UHP
]
Γ
R
f (z) dz
IR
f (z) dz +
C
R
f (z) dz.
Thus,
R
−R
f (x) d x =
IR
f (z) dz
= i 2 π
[
sum of the residues of f in the UHP
]
C
R
f (z) dz. (17.6)
To deal with the integral over C
R , first note that this curve is parameterized by
z = z(θ) = Re
i θ where 0 ≤ θ ≤ π.
So “ dz = d[Re
i θ ] = i Re
i θ dθ ” and
C
R
f (z) dz =
π
0
f
Re
i θ
i Re
i θ dθ.
Now assume assumption (3a) holds, and let
M(R) = the maximum of |z f (z)| when |z| = R.
By definition, then, ∣ ∣Rei^ θ^ f
Re
i θ
∣ ≤ M(R)
Moreover, it is easily verified that assumption (3a) (that z f (z) → 0 as |z| → ∞ ) implies that
M(R) → 0 as R → ∞.
So
lim R→∞
C
R
f (z) dz
= lim R→∞
π
0
f
Re
i θ
i Re
i θ dθ
≤ lim R→∞
π
0
f
Re
i θ
i Re
i θ
dθ
≤ lim R→∞
π
0
M(R) dθ = lim R→∞
M(R) π = 0.
version: 3/10/
Chapter & Page: 17–8 Residue Theory
Thus,
lim R→∞
C
R
f (z) dz = 0. (17.7)
Under assumption (3b),
∫
C
R
f (z) dz =
C
R
g(z)e
i αz dz
for some α > 0. Now,
∣ ∣ e
i αz
e
i α(x+i y)
e
i αx e
−αy
= e
−αy ,
which goes to zero very quickly as y → ∞. So it is certainly reasonable to expect equation
(17.7) to hold under assumption (3b). And it does — but a rigorous verification requires a little
more space than is appropriate here. Anyone interested can find the details in the proof of lemma
17.2 on page 17–14.
So, after letting R → ∞ , formula (17.6) for the integral of f (x) on (−R, R) becomes
∫ ∞
−∞
f (x) d x = i 2 π
[
sum of the residues of f in the UHP
]
On the other hand, if it is assumption (3c) that holds, then
∫
C
R
f (z) dz =
C
R
g(z)e
−i αz dz
for some α > 0. In this case, though,
∣ ∣ e
−i αz
e
−i α(x+i y)
e
−i αx e
αy
= e
αy ,
which rapidly blows up as y gets large. So it is not reasonable to expect equation (17.7) to hold
here. Instead, take
ΓR = Γ
− R
= −IR + C
− R
where C
− R is the semicirle in the lower half plane of radius R and centered at 0 (see figure
17.1b). (Observe that, this time, the direction of travel on IR is opposite to the normal ‘positive’
direction of travel on the X–axis.) Again, since we are letting R → ∞ and there are only
finitely many singularities, we can assume that we’ve taken R large enough for Γ
− R to enclose
all the singularities of f in the lower half plane (LHP). Then
i 2 π
[
sum of the residues of f in the LHP
]
Γ
− R
f (z) dz
−IR
f (z) dz +
C − R
f (z) dz
IR
f (z) dz +
C − R
f (z) dz.
Thus,
∫ R
−R
f (x) d x = −
−IR
f (z) dz
= −i 2 π
[
sum of the residues of f in the UHP
]
C
R
f (z) dz
Chapter & Page: 17–10 Residue Theory
Standard “Simple” Tricks
Using Real and Imaginary Parts
It is often helpful to observe that one integral of interest may be the real or imaginary part of
another integral that may, possibly, be easier to evaluate. Do remember that
∫ (^) b
a
Re[ f (x)] d x = Re
[∫
b
a
f (x) d x
]
and
∫ (^) b
a
Im[ f (x)] d x = Im
[∫
b
a
f (x) d x
]
(If this isn’t obvious, spend a minute to (re)derive it.) In this regard, it is especially useful to
observe that
cos(X) = Re
[
e
i X
]
and sin(X) = Im
[
e
i X
]
! ◮ Example 17.3: Consider ∫ ∞
−∞
cos( 2 π x)
1 + x 2
d x
Using the above observations and our answer from the previous exercise,
∞
−∞
cos( 2 π x)
1 + x 2
d x =
∞
−∞
Re
[
e i 2 π x
1 + x 2
]
d x
= Re
[∫
∞
−∞
e i 2 π x
1 + x 2
d x
]
= Re
[
πe
− 2 π
]
= πe
− 2 π .
Clever Choice of Curve and Function
The main “trick” to applying residue theory in computing integrals of real interest (i.e., integrals
that actually do arise in applications) as well as other integrals you may encounter (e.g., other
integrals in assigned homework and tests) is to make clever choices for the functions and the
curves so that you really can extract the value of desired integral from the integral over the closed
curve used. Some suggestions, such as were given on page 17–6 for computing certain integrals
on (−∞, ∞) , can be given. In general, though, choosing the right curves and functions is a
cross between an art and a skill that one just has to develop.
A good example of using both clever choices of functions and curves, and in using real and
imaginary parts, is given in computing the Fresnel integrals (which arise in optics).
! ◮ Example 17.4: Consider the Fresnel integrals
∞
0
cos
x
2
d x and
∞
0
sin
x
2
d x.
Rather than use cos
x
2
and sin
x
2
directly, it is clever to use e
i x 2 and the fact that
e
i x 2 = cos
x
2
x
2
So, if we can evaluate ∫ ∞
0
e
i x 2 d x ,
“Simple” Applications Chapter & Page: 17–
R
Re
i π/ 4
AR
lR ΓR
π/ 4
X
Y
(^0) IR
Figure 17.2: The closed curve for computing the Fresnel integrals.
then we can get the two integrals we want via
∫ ∞
0
cos
x
2
d x =
∞
0
Re
[
e
i x
2 ]^
d x = Re
[ ∫
∞
0
e
i x 2 d x
]
and ∫ ∞
0
sin
x
2
d x =
∞
0
Im
[
e
i x^2
]
d x = Im
[∫
∞
0
e
i x^2 d x
]
Now, to compute ∫ ∞
0
e
i x 2 d x ,
we naturally choose the function f (z) = e
i z 2
. The clever choice for the closed curve is
sketched in figure 17.2. It is
ΓR = IR + AR − lR
where, for any choice of R > 0 ,
IR = the straight line on the real line from z = 0 to z = R
= the interval on R from x = 0 to x = R ,
AR = the circular arc centered at 0 starting at z = R and going to z = Re
i π/ 4 ,
and
lR = the straight line from z = 0 to z = Re
i π/ 4 .
Notice that the chosen function, f (z) = e
i z 2 , is analytic on the entire complex plane. So
there are no residues, and, for each R > 0 , we have
ΓR
e
i z^2 dz =
IR
e
i z^2 dz +
AR
e
i z^2 dz −
lR
e
i z^2 dz
R
x= 0
e
i x^2 d x +
AR
e
i z^2 dz −
lR
e
i z^2 dz.
So, ∫ R
x= 0
e
i x 2 d x =
lR
e
i z 2 dz −
AR
e
i z 2 dz.
Letting R → ∞ , this becomes
∫ ∞
0
e
i x 2 d x = lim R→∞
lR
e
i z 2 dz − lim R→∞
AR
e
i z 2 dz. (17.10)
version: 3/10/
“Simple” Applications Chapter & Page: 17–
Gathering together what we’ve derived (equations (17.10), (17.11) and (17.12)), we finally
get
∞
0
e
i x 2 d x = lim R→∞
lR
e
i z 2 dz − lim R→∞
AR
e
i z 2 dz
= e
i π/ 4
√ π
2
[
cos
π
4
π
4
)] √
π
2
[
1 √ 2
1 √ 2
] √
π
2
1
2
π
2
1
2
π
2
Consequently,
∞
0
cos
x
2
d x = Re
[ ∫
∞
0
e
i x 2 d x
]
1
2
π
2
and ∫ (^) ∞
0
sin
x
2
d x = Im
[∫
∞
0
e
i x 2 d x
]
1
2
π
2
? ◮ Exercise 17.1: Consider the last example.
a: Why was the curve ΓR = IR + AR − lR , as illustrated in figure 17.2, such a clever
choice of curves? (Consider what happened to the function e
i z 2 on IR .)
b: Why would the curve ΓR = IR + C
R illustrated in figure 17.1a not be a clever choice
for computing the Fresnel integrals?
In the last example, as in a previous example, the closed curve constructed so the residue
theorem can be applied included a piece of a circle of radius R , say, the AR in our last example.
Our formula for the desired integral then includes a term of the form
lim R→∞
AR
f (z) dz ,
and it is important that this limit be zero (or some other computable finite number). Otherwise,
either this general approach fall apart, or we can show that the desired integral does not converge.
Keep in mind that the length of AR increases as R increases; so the vanishing of the
integrand is not enough to ensure the vanishing of the above limit. To take into account the
increasing length of the curve, it is often a good idea to parameterize it using angular measurement,
z = z(θ) = Re
i θ
with θ limited to some fixed interval (α, β). Then, as we’ve seen in examples,
AR
f (z) dz
β
α
f
Re
i θ
i Re
i θ dθ
β
α
f
Re
i θ
i Re
i θ
dθ =
β
α
f
Re
i θ
R dθ.
version: 3/10/
Chapter & Page: 17–14 Residue Theory
With luck, you will be able to look at the inside of the last integral and tell whether the desired
limit is zero or whether the limit does not exist.
Of course, you may ask, why not just compute the limit by bringing the limit inside the
integral,
lim R→∞
β
α
f
Re
i θ
R dθ =
β
α
lim R→∞
f
Re
i θ
R dθ?
Because this last equation is not always valid. There are choices for f such that
lim R→∞
∫ (^) β
α
∣ (^) f
Re
i θ
∣ (^) R dθ 6 =
∫ (^) β
α
lim R→∞
∣ (^) f
Re
i θ
∣ (^) R dθ!
For example, f (z) = e
−z with (α, β) = ( 0 ,
π/ 2 )^.
Appendices to this Section
Two issues are addressed here. One is how to rigorously verify that certain integrals over arcs of
radii R vanish as R goes to infinity. The other is how to evaluate
∞
0
e
−s 2 ds. These appendices
are included for the sake of completeness. You should be acquainted with the general results,
but don’t worry about reproducing the sort of analysis given here.
The Vanishing of Certain Integrals as R → ∞
Often, when using residues to compute integrals with exponentials, it is necessary to verify that
certain integrals over arcs of radii R vanish as R goes to infinity. Sometimes this is easy to show;
sometimes it is not. To illustrate how we might verify the cases involving complex exponentials,
we will rigorously verify the following lemma. It is the lemma needed to rigorously verify
equation (17.7) on page 17–8.
Lemma 17.
Let α > 0 , and assume g is any “reasonably smooth” function on the complex plane
2 satisfying
g(z) → 0 as |z| → ∞.
Then
lim R→∞
C
R
g(z)e
i αz dz = 0
where C
R is the upper half of the circle of radius R about the origin.
PROOF: Keep in mind that C
R
is getting longer as R gets larger. So the simple fact that
the integrand gets smaller as R gets larger is not enough to ensure that the above limit is zero.
To help take into account the increasing length of the curve and to help convert the integral
to something a little more easily to deal with, we make use of the fact that this curve can be
parameterized by
z = z(θ) = Re
i θ = R [cos(θ) + i sin(θ)] where 0 ≤ θ ≤ π.
For convenience, let
M(R) = maximum value of g(z(θ)) when 0 ≤ θ ≤ π.
2 it suffices to assume g(z) is continuous on the region where |z| > R 0 for some finite real value R 0.
Chapter & Page: 17–16 Residue Theory
Thus,
lim R→∞
C
R
g(z)e
i αz dz
< lim R→∞
4 M(R)
α
4 · 0
α
which, of course, means that
lim R→∞
C
R
g(z)e
i αz dz = 0 ,
as claimed.
? ◮ Exercise 17.2: Using the ideas behind inequalities (17.14) and (17.15), show that
π/ 4
0
e
−R 2 sin( 2 θ ) R dθ ≤
π/ 4
0
e
−R 2 θ R dθ =
1
R
→ 0 as R → ∞.
Integral of the Basic Gaussian
While there is no simple formula for the indefinite integral of e
−x 2 , the value of the definite
integral ∫ ∞
0
e
−s 2 ds
is easily computed via a clever trick.
To begin, observe that, by symmetry,
∞
0
e
−s 2 ds =
1
2
I
where
I =
∞
−∞
e
−s 2 ds =
∞
−∞
e
−x 2 d x =
∞
−∞
e
−y 2 dy.
The “clever trick” is based on the observation that I
2 , the product of I with itself, can be
expressed as a double integral over the entire XY –plane,
I
2
∞
−∞
e
−x 2 d x
∞
−∞
e
−y 2 dy
∞
−∞
∞
−∞
e
−x 2 d x
e
−y 2 dy
∞
−∞
∞
−∞
e
−x 2 e
−y 2 d x
dy =
∞
−∞
∞
−∞
e
−(x 2 +y 2 ) d x dy.
This double integral is easily computed using polar coordinates (r, θ) where
x = r cos(θ) and y = r sin(θ).
Recall that
x
2
2 = r
2 and d x dy = r dr dθ.
Integrals Over Branch Cuts of Multi-Valued Functions Chapter & Page: 17–
So, converting to polar coordinates and using elementary integration techniques,
I
2
2 π
0
∞
0
e
−r 2 r dr dθ =
2 π
0
1
2
dθ = π.
Taking the square root gives
I = ±
π.
Because e
−s^2 is a positive function, its integral must be positive. Thus,
∞
−∞
e
−s 2 ds = I =
π
and (^) ∫ ∞
0
e
−s^2 ds =
1
2
I =
1
2
π.
17.3 Integrals Over Branch Cuts of Multi-Valued
Functions
If you need to compute ∫ β
α
f (x) d x
when f (z) is a multi-valued function, then a clever choice for the closed curve may include
using the interval (α, β) as a branch cut for f , and letting it serve as two parts of the curve
enclosing the residues of f. Just what I mean will be a lot clearer if we do one example.
But first, go back and re-read the discussion of multi-valued functions starting on page 14–5.
Also, re-read the bit about Square Roots and Such starting on page 14–8.
! ◮ Example 17.5: Consider evaluating
∞
0
√ x
1 + x 2
d x.
Naturally, we will let
f (z) =
z
(^1) / 2
1 + z 2
This function has singularities at z = ±i (where the denominator is zero). Also, because of
the z
(^1) / (^2) factor, f (z) is multi-valued with a branch point at z = 0. We will cleverly take the
cut line to be the positive X –axis, and define z
(^1) / (^2) to be given by
z
(^1) / (^2) =
|z| e
i θ/ 2
where θ is the polar angle (argument) of z satisfying
0 < θ < 2 π.
version: 3/10/
Integrals Over Branch Cuts of Multi-Valued Functions Chapter & Page: 17–
The circle Cε isolates the branch point from the region encircled by Γε R. This is a good
idea because “bad things” can happen near branch points. Later, we will check that we can
(or cannot) let ε → 0.
The curves I
ε R and I
− ε R are each, in fact, simply the subinterval (ε, R) on the X –axis
oriented in opposing directions. They are treated, however, as two distinct pieces of Γε R ,
with I
ε R viewed as a lower boundary to the region above it, and I
− ε R viewed as the upper
boundary to the region below it. Indeed, it may be best to first view them as laying a small
distance above and below the X –axis, exactly as sketched in figure 17.4, with this distance
shrunk to zero by the end of the computations. Consequently, the region encircled by Γε R is
the shaded region in figure 17.4.
Applying the residue theorem, we have
i 2 π × [ sum of the resides of f in the shaded region]
Γ
f (z) dz
I
ε R
f (z) dz +
CR
f (z) dz +
I
− ε R
f (z) dz −
Cε
f (z) dz
where ∫
I
ε R
f (z) dz =
R
ε
[
lim z→x Im z> 0
f (z)
]
d x =
R
ε
√ x
1 + x 2
d x
and
I − ε R
f (z) dz =
ε
R
[
lim z→x Im z< 0
f (z)
]
d x = −
R
ε
[
√ x
1 + x 2
]
d x =
R
ε
√ x
1 + x 2
d x.
Thus,
i 2 π × [ sum of the resides of f in the shaded region]
∫ R
ε
√ x
1 + x 2
d x +
CR
f (z) dz −
Cε
f (z) dz ,
and, so,
R
ε
√ x
1 + x 2
d x = iπ × [ sum of the resides of f in the shaded region]
1
2
Cε
f (z) dz −
1
2
CR
f (z) dz.
(Notice that, because of the multi-valueness of f ,
I
ε R
f (z) dz and
I − ε R
f (z) dz
did not cancel each other out even though I
ε R and I
− ε R are the same curve oriented in opposite
directions. That is what will make these computations work.)
Naturally, we want ε → 0 and R → ∞. Now, if |z| = ε and ε < 1 , then
| f (z)| =
z
(^1) / 2
1 + z 2
ε
(^1) / 2
1 − ε 2
version: 3/10/
Chapter & Page: 17–20 Residue Theory
So, using the parametrization z = z(θ) = εe
i θ ,
Cε
f (z) dz
∫ (^2) π
0
f (z(θ)) iεe
i θ dθ
∫ (^2) π
0
| f (z(θ))| ε dθ
2 π
0
ε
(^1) / 2
1 − ε 2
ε dθ =
ε
(^3) / 2
1 − ε 2
2 π → 0 as ε → 0.
By a very similar computations, we get (since R > 1 ),
CR
f (z) dz
2 π
0
R
(^1) / 2
R^2 − 1
R dθ =
R
(^3) / 2
R^2 − 1
2 π → 0 as R → ∞.
Thus, after letting ε → 0 and R → ∞ , equation (17.16) reduces to
∞
0
√ x
1 + x 2
d x = iπ × [ sum of the resides of f in the shaded region] (17.17)
Clearly, the only singularities of
f (z) =
z
(^1) / 2
1 + z 2
are at z = ±i , and
f (z) =
g(z)
z − i
and f (z) =
h(z)
z − (−i)
where
g(z) =
z
(^1) / 2
z + i
and h(z) =
z
(^1) / 2
z − i
So these singularities are simple poles, and
∞
0
√ x
1 + x 2
d x = iπ × [ sum of the resides of f in the shaded region]
= iπ
[
Resi ( f ) + Res−i ( f )
]
= iπ[g(i) + h(−i)]
= iπ
[
i
(^1) / 2
i + i
(−i)
(^1) / 2
−i − i
]
= π
[
e
i π/ 4
2
e
i 3 π/ 4
2
]
π √ 2
