Integration

Question 1

Path

Answer 1

A path @ us a continuous function from an interval [a , b] of real number to (C.

Initial point - (@(a))
Final point - (@ (b))

Question 2

Closed path

Answer 2

A path @ : [a, b] -> (C is said to be closed if @(a) = @(b)

Question 3

Simple path

Answer 3

A path @ : [a , b] -> (C is said to be simple if @(s) =/ @(t) when s=/t, except perhaps if s=a and t=b.

Question 4

Join

Answer 4

Suppose that @1 : [a , b] -> (C and @2 : [ b, c ] -> (C are paths such that the final point of @1 is the initial point of @2, that is @1(b) = @2(b)
The join @1 + @2 of @1 and @2 is the path
@1 + @2(t) = { @1 when a=< t =< b
@2 when b=

Question 5

Reverse path

Answer 5

Suppose that @ : [a ,b] -> (C is a path. The reverse path @* : [a, b] -> (C of @ is given by
@(a + b -t) where a=

Question 6

Derivative of @

Answer 6

Suppose that @ : [a , b] -> (C is a path and that @ = c + di, where c, d: [a ,b ] -> |R. Then we define
@’(t) = c’(t) + i d’(t)
provided that c’(t) and d’(t) exist.

Question 7

Smooth path

Answer 7

We say @ is smooth if all derivative of @ exist in [a,b] and @’(t) =/ 0 for all t e (a,b)

Question 8

Contour

Answer 8

A contour is the join of finitely many smooth paths

Question 9

Piecewise differentiable

Answer 9

A function from an interval into |R or (C that is differentiable except at finitely many points.

Question 10

Jordan curve theorem

Answer 10

If @ : [a,b] -> (C is a simple closed path, the the complement of the range {@(t) : t e [a, b]} of @ is the union of two disjoint domains.
One of these, called the interior of @ and written Int(@) is bounded.
The other, called the exterior of @ and written Ext(@), is unbounded.

Question 11

Standard orientation

Answer 11

The standard orientation of @ is such that the interior is always on our left (anticlockwise)

Question 12

Integral of C

Answer 12

Suppose that A, B : [a,b] -> |R are real valued functions, and that C: [a,b] -> (C is given by C=A + iB. We define
Integral (from a to b) C(t) .dt = Integral (from a to b) A(t) .dt + i Integral (from a to b) B(t). dt

Question 13

Path integral / Contour integral

Answer 13

Given a contour @: [a,b] -> (C and a continuous function f defined on the range of @, we define the path integral or contour integral Integral (over @) f(z) by
Integral (over @) f(z) .dz = Integral (from a to b) f(@(t))@’(t) . dt

Question 14

Properties of contour integrals

Answer 14

(1) Integral (over @) ( $1 f1(z) + $2 f2(z) ) = $1 Integral (over @) (f1(z) .dz) + $2 Integral (over @) (f2(z) .dz
(2) Integral (over @*) (f(z)) = Integral (over @) f(-z)
(3) Integral (@1 + @2) (f(z) = Integral (over @1) f(z) + Integral (over @2) f(z)

Question 15

Composition property

Answer 15

If @1 : [a, b] -> (C and @2 : [ c, d ] -> (C are contours and there is a continuous piecewise differentiable function # : [a,b] -> [c,d] such that #(a) = c and #(b) =d, and @1 = @2 . #, then
Integral (over @1) f(z) = Integral (over @2) f(z)

Question 16

Oriented curve

Answer 16

An oriented curve is the range of a path and the associated direction of movement along it, given by €

Question 17

The ML Lemma

Answer 17

Suppose that € is a contour, and that f is continuous function whose domain includes €. Then
| Integral (over €) f(z) . dz | =< ML,
where M is an upper bound for f on € and L is the length of €.

Question 18

Cauchy- Goursat Theorem

Answer 18

Suppose that % is a simply closed domain in (C, that f e H(%) and that @ is a simply closed contour in %.Then
Integral (over @) f(z) . dz =0.

Question 19

Independence of contours of certain contour integrals

Answer 19

Suppose that % is a simply closed connected domain in (C, that f e H(%) and that €1 and €2 are simple curves with the same initial point p and the same final point q. Then

Integral (over €1) f(z) .dz = Integral (over €2) f(z) .d(z)

Question 20

Existence of primitives

Answer 20

Suppose that % is a simply connected domain in (C, and f e H (%). Then there exists a function F e H(%) such that F’ = f and for any simple contour € in %,

Integral (over €) f(z) dz = F(q) -F(p)

where p and q are the initial and final points of €.

Question 21

Existence of primitives continued…

Answer 21

If F1 is another function such that F1’ = f, then F1-F is a constant and

Integral (over €) f(z) dz = F1(q) - F1(p)

Question 22

Primitives

Answer 22

We call the function F such that F’ = f a primitive or an anti - derivate of f.

Question 23

Importance of the Cauchy- Goursat Theorem

Answer 23

enables us to show that a holomorphic function in a domain % has continuous partial derivatives and is infinitely differentiable.

Question 24

Lemma on contour integrals

Answer 24

Suppose that % is a simply connected domain, that zo e %, and that f e H(%/{zo}). If €1 and €2 are simple closed contours with the standard orientation and zo e Int(€1) n Int ( €2) then

Integral (over €1) f(z) dz = Integral (over €2) f(z) dz

Question 25

Cauchy’s Integral Formula

Answer 25

Suppose that % is a simply connected domain in (C, the f e H(%), that € is a simple closed contour in % and that w e Int(€). Then

f(w) = 1 / 2Pii Integral (over €) f(z) / (z - w) .dz

Enables us to compute integrals without actually integrating.

Question 26

Cauchy’s Integral Formula for power series

Answer 26

Suppose that f e H(B(zo, R)) and that € is a simple closed contour such that zo e Int (€). Then

f(w) = SUM (from n=0, to inf) cn(w-zo)^n for all w e B(zo, R)

where cn = 1/ 2Pii Integral (over €) f(z) / (z - zo)^(n+1) .dz
=( f^(n) (zo) )/ n!