Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt

Figure 13.8: The real and imaginary part of the integrand for several values of α.Note that the integral exists for all nonzero real α and that < 1... Solution 13.12 Let Ci be the contou

Since the nearest singularity is at z = −ı, the Laurent series will converge in the annulus 0 < |z − ı| < 2.

1 − ı(z − ı)/2

= 14

∞

X

n=0

 ı(z − ı)2

n

= 14

(z − zj)

P (z) =

cPn k=1

Qn j=1 j6=k

(z − zj)

cQn k=1(z − zk)

(z − zj)

Qn j=1(z − zj)

Γ

P0(z)

P (z) dz =

1ı2πZ

the number of roots, counting multiplicities, that lie inside the contour Γ.

1ı2πZ

Γ

P0(z)

P (z) dz =

1ı2π [log P (z)]C

= 1ı2π

the contour occurs at z = 0 We evaluate the integral with the residue theorem.

I

C

ezcos z(z − π) sin zdz = ı2π Res



ezcos z(z − π) sin z, z = 0



= ı2π lim

z=0z e

zcos z(z − π) sin z

= −ı2 lim

z=0

zsin z

= −ı2 lim

z=0

1cos z

= −ı2I

Res



ezcos z(z − π) sin z, z = 0



= −1π

We find the residue at z = −π with the residue formula

Res

 ezcos z(z − π) sin z, z = −π



= lim

z→−π(z + π) e

zcos z(z − π) sin z

= e

−π(−1)

−2π z→−πlim

z + πsin z

= e

−π

2π z→−πlim

1cos z

= −e

−π

2π

We find the residue at z = π by finding the first few terms in the Laurent series of the integrand.

ezcos z(z − π) sin z =

Res



ezcos z(z − π) sin z, z = π



= eπ.The integral is

I

C

ezcos z(z − π) sin zdz = ı2π

Res



ezcos z(z − π) sin z, z = −π

+ Res



ezcos z(z − π) sin z, z = 0



+ Res



ezcos z(z − π) sin z, z = π

Now consider the integral

Z 1

−1

1

x − ıαdxwhere α ∈ R, α 6= 0 Since the integrand is bounded, the integral exists

α

x2+ α2 dx

= ı2

Z 1/α 0

Figure 13.8: The real and imaginary part of the integrand for several values of α.

Note that the integral exists for all nonzero real α and that

<

1

Now note that when α = 0, the integrand is real Of course the integral doesn’t converge for this case, but if wecould assign some value to

maximum modulus of (z−z0 )f (z)−f 0

z−z 0 on C By using the maximum modulus integral bound, we have

Z

C 

(z − z0)f (z) − f0

z − z0 dz

≤ (β − α)M

→ 0 as  → 0+.Thus we see that

ıf0dθ

= ı(β − α)f0

≡ ı(β − α) Res(f (z), z0)This proves the result

Solution 13.12

Let Ci be the contour that is indented with circular arcs or radius  at each of the first order poles on C so as to enclosethese poles Let A1, , An be these circular arcs of radius  centered at the points ζ1, , ζn Let Cp be the contour,(not necessarily connected), obtained by subtracting each of the Aj’s from Ci

Since the curve is C1, (or continuously differentiable), at each of the first order poles on C, the Aj’s becomessemi-circles as  → 0+ Thus

Z

A j

f (z) dz = ıπ Res(f (z), ζj) for j = 1, , n

CONTINUEFigure 13.9: The Indented Contour.

The principal value of the integral along C is

−Z

C

1

z − 1dzwhere C is the unit circle Let Cp be the circular arc of radius 1 that starts and ends a distance of  from z = 1 Let

C be the negative, circular arc of radius  with center at z = 1 that joins the endpoints of Cp Let Ci, be the union of

Cp and C (Cp stands for Principal value Contour; Ci stands for Indented Contour.) Ci is an indented contour thatavoids the first order pole at z = 1 Figure 13.9 shows the three contours

Note that the principal value of the integral is

−Z

We can calculate the integral along Ci with Cauchy’s theorem The integrand is analytic inside the contour.

Z

C i

1

z − 1dz = 0

We can calculate the integral along C using Result 13.3.1 Note that as  → 0+, the contour becomes a semi-circle,

a circular arc of π radians in the negative direction

x2(x2+ 1)(x2+ 4)dx =

12

Z ∞

−∞

x2(x2+ 1)(x2+ 4)dxNext we close the path of integration in the upper half plane Consider the integral along the boundary of the

domain 0 < r < R, 0 < θ < π.

12Z

C

z2

(z2+ 1)(z2+ 4)dz =

12Z

C

z2

(z − ı)(z + ı)(z − ı2)(z + ı2)dz

= ı2π12

Res



= π6Let CR be the circular arc portion of the contour RC =R−RR +RC

R We show that the integral along CRvanishes

as R → ∞ with the maximum modulus bound

z=−b+ıa

= πaLet CR be the circular arc portion of the contour RC =R−RR +RC

R We show that the integral along CRvanishes

as R → ∞ with the maximum modulus bound

Z

C R

dz(z + b)2+ a2

≤ πR max

z∈C R

...

x2< /sup>(x2< /small>+ 1)(x2< /small>+ 4)dx =

12

Z ∞

−∞

x2< /sup>(x2< /small>+ 1)(x2< /small>+... < π.

12Z

C

z2< /small>

(z2< /small>+ 1)(z2< /small>+ 4)dz =

12Z

C

z2< /small>

(z... Res

2z−1

z2< /small> + z + 1, z = 0

+ ? ?2? ? Res 2z

!

ıπ lim

z→0

2

z2< /small>+