Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

The value of the integral is the sum of the residues fromsingularities inside the contour.. Example 13.2.3 Clearly R−∞∞ x dx diverges, however the Cauchy principal value exists... 13.3 C

The series about z = ∞ for 1/(z + 2) is

an= 1ı2πI

C

f (z)

zn+1dz

and C is a contour going around the origin once in the positive direction We manipulate the coefficient integral intothe desired form.

an = 1ı2πI

C

(z + 1/z)m

zn+1 dz

= 1ı2π

Z 2π 0

(eıθ+ e−ıθ)m

eı(n+1)θ ı eıθ dθ

= 12π

Z 2π 0

cosmθ(cos(nθ) − ı sin(nθ)) dθNote that cosmθ is even and sin(nθ) is odd about θ = π

= 2

m−1

π

Z 2π 0

cosmθ cos(nθ) dθ

Binomial Series Now we find the binomial series expansion of f (z)



z + 1z

m(m − n)/2



zn

The coefficients in the series f (z) =P∞

(cos θ)mcos(nθ) dθ =

( π

2 m−1

m (m−n)/2

f (z) = αz

3+ βz2+ γz + δ(z − ı/2)(z − 2)2

Since f (z) is a rational function we expand it in partial fractions to obtain a form that is convenient to integrate

We use the value of the integrals of f (z) to determine the constants, a, b, c and d

I

|z|=1

a

z − ı/2 +

b

z − 2+

c(z − 2)2 + d



dz = ı2πı2πa = ı2π

a = 1

b = −1Note that by applying the second constraint, we can change the third constraint to

z − ı/2 − 1

z − 2+

c(z − 2)2 + d



dz = 0I

f (z) = 1

z − ı/2 − 1

z − 2+

2 − ı/2(z − 2)2 + d,where d is an arbitrary constant We can also write the function in the form:

f (z) = dz

3+ 15 − ı84(z − ı/2)(z − 2)2.Complete Laurent Series We find the complete Laurent series about z = 0 for each of the terms in the partial

X

n=0

 ı2z

2 − ı/2(z − 2)2 = (2 − ı/2)1

4(1 − z/2)

−2

= 4 − ı8

∞

X

n=0

−2n





−z2

n

, for |z/2| < 1

= 4 − ı8

z2



1 −2z

 

−2z

1/2 < |z| < 2 and 2 < |z| For |z| < 1/2, we have

k3/3k

(k + 1)3/3k+1

= 3 lim

n→∞

k3

(k + 1)3

If you are really observant, you may have noticed that we did something a little funny in evaluating

The principal value of an integral may exist when the integral diverges If the integral does converge, then it isequal to its principal value.

We can use the method of Example 13.4.1 to evaluate the principal value of integrals of functions that vanish fastenough at infinity

Result 13.4.1 Let f (z) be analytic except for isolated singularities, with only first order poles

on the real axis Let CR be the semi-circle from R to −R in the upper half plane If

This result is proved in Exercise 13.13 Of course we can use this result to evaluate the integrals of the form

Z ∞ 0

f (z) dz,where f (x) is an even function

e−R sin θ dθ

Since 2θ/π ≤ sin θ for 0 ≤ θ ≤ π/2,

e−R sin θ ≤ e−R2θ/π for 0 ≤ θ ≤ π/2

Z π/2 0

e−R sin θ dθ ≤

Z π/2 0

e−R2θ/π dθ

=

h

− π2Re

−R2θ/πiπ/2

0

= − π2R(e

−R−1)

≤ π2R

→ 0 as R → ∞

We can use this to prove the following Result 13.5.1 (See Exercise 13.17.)

Result 13.5.1 Jordan’s Lemma.

Z π

0

e−R sin θ dθ < π

R . Suppose that f (z) vanishes as |z| → ∞ If ω is a (positive/negative) real number and CR is

a semi-circle of radius R in the (upper/lower) half plane then the integral

Z

C R

f (z) eıωz dz vanishes as R → ∞.

We can use Jordan’s Lemma and the Residue Theorem to evaluate many Fourier integrals ConsiderR∞

−∞f (x) eıωx dx,where ω is a positive real number Let f (z) be analytic except for isolated singularities, with only first order poles onthe real axis Let C be the contour from −R to R on the real axis and then back to −R along a semi-circle in theupper half plane If R is large enough so that C encloses all the singularities of f (z) in the upper half plane then

Result 13.5.2 Fourier Integrals Let f (z) be analytic except for isolated singularities, with only first order poles on the real axis Suppose that f (z) vanishes as |z| → ∞ If ω is a positive real number then

13.6 Fourier Cosine and Sine Integrals

Fourier cosine and sine integrals have the form,

Z ∞ 0

f (x) cos(ωx) dx and

Z ∞ 0

f (x) sin(ωx) dx

If f (x) is even/odd then we can evaluate the cosine/sine integral with the method we developed for Fourier integrals

Let f (z) be analytic except for isolated singularities, with only first order poles on the real axis Suppose that f (x)

is an even function and that f (z) vanishes as |z| → ∞ We consider real ω > 0

−

Z ∞ 0

−

Z ∞ 0

Result 13.6.1 Fourier Cosine and Sine Integrals Let f (z) be analytic except for isolated singularities, with only first order poles on the real axis Suppose that f (x) is an even function and that f (z) vanishes as |z| → ∞ We consider real ω > 0.

x + 1 dx, 0 < a < 1,where x−a denotes exp(−a ln(x)) We choose the branch of the function

f (z) = z

−a

z + 1 |z| > 0, 0 < arg z < 2πwith a branch cut on the positive real axis

Let C and CR denote the circular arcs of radius  and R where  < 1 < R C is negatively oriented; CR ispositively oriented Consider the closed contour C that is traced by a point moving from C to CR above the branchcut, next around CR, then below the cut to C, and finally around C (See Figure 13.6.)

We write f (z) in polar coordinates

f (z) = exp(−a log z)

exp(−a(log r + iθ))

r eıθ+1

− 48z coth(z)csc(z)2< /sup>csch(z)2< /sup>+ 24 z2< /sup>cot(z)...



24 cot(z) coth(z)csc(z)2< /sup>− 32z coth(z)csc(z)4

− 16z cos(2z) coth(z)csc(z)4+ 22 z2< /sup>cot(z) coth(z)csc(z)4+ 2z2< /sup>cos(3z)... 24 z2< /sup>cot(z) coth(z)csc(z)2< /sup>csch(z)2< /sup>+ 16z2< /sup>csc(z)4csch(z)2< /sup>+ 8z2< /sup>cos(2z)csc(z)4csch(z)2< /sup>

− 32z cot(z)csch(z)4−