Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt

We could evaluate the real integral by putting a branch cut on the positive real axis with 0 < argz < 2π and integrating f z on the contour in Figure13.12.Integrating on this contour wou

Solution 13.28

Convergence If xaf (x)  xα as x → 0 for some α > −1 then the integral

Z 1 0

xaf (x) dxwill converge absolutely If xaf (x)  xβ as x → ∞ for some β < −1 then the integral

Z ∞ 1

xaf (x)will converge absolutely These are sufficient conditions for the absolute convergence of

Z ∞ 0

xaf (x) dx

Contour Integration We put a branch cut on the positive real axis and choose 0 < arg(z) < 2π We considerthe integral of zaf (z) on the contour in Figure ?? Let the singularities of f (z) occur at z1, , zn By the residuetheorem,

On the circle of radius , the integrand is o(−1) Since the length of C is 2π, the integral on C vanishes as

 → 0 On the circle of radius R, the integrand is o(R−1) Since the length of CR is 2πR, the integral on CR vanishes

Z ∞ 0

n

X

k=1

Res (zaf (z), zk) ,Differentiating the solution of Exercise 13.26 m times with respect to a yields

Z ∞ 0

Solution 13.30

Taking the limit as a → 0 ∈ Z in the solution of Exercise 13.26yields

Z ∞ 0

f (x) dx = ı2π lim

a→0

 Pn k=1Res (zaf (z), zk)

1 − eı2πa



The numerator vanishes because the sum of all residues of znf (z) is zero Thus we can use L’Hospital’s rule

Z ∞ 0

f (x) dx = ı2π lim

a→0

 Pn k=1Res (zaf (z) log z, zk)

−ı2π eı2πa



Z ∞ 0

This suggests that we could have derived the result directly by considering the integral of f (z) log z on the contour

in Figure ?? We put a branch cut on the positive real axis and choose the branch arg z = 0 Recall that we haveassumed that f (z) has only isolated singularities and no singularities on the positive real axis, [0, ∞) By the residuetheorem,

By assuming that f (z)  zα as z → 0 where α > −1 the integral on C will vanish as  → 0 By assuming that

f (z)  zβ as z → ∞ where β < −1 the integral on CR will vanish as R → ∞ The value of the integrand below thebranch cut, z = x eı2π is f (x)(log x + ı2π) Taking the limit as  → 0 and R → ∞, we have

Z ∞ 0

If f (z)  zα as z → 0 for some α > −1 then the integral on C will vanish as  → 0 f (z)  zβ as z → ∞ for some

β < −1 then the integral on CR will vanish as R → ∞ Below the branch cut the integrand is f (x)(log x + ı2π)2

f (x) log x dx + 4π2

Z ∞ 0

has first order poles at z = (±1 ± ı)/√

2 and a branch point at z = 0 We could evaluate the real integral by putting

a branch cut on the positive real axis with 0 < arg(z) < 2π and integrating f (z) on the contour in Figure13.12.Integrating on this contour would work because the value of the integrand below the branch cut is a constant timesthe value of the integrand above the branch cut After demonstrating that the integrals along C and CR vanish in thelimits as  → 0 and R → ∞ we would see that the value of the integral is a constant times the sum of the residues atthe four poles However, this is not the only, (and not the best), contour that can be used to evaluate the real integral.Consider the value of the integral on the line arg(z) = θ

f (r eıθ) = r

aeıaθ

1 + r4eı4θ

Cε

Figure 13.12: Possible path of integration for f (z) = 1+zza4

If θ is a integer multiple of π/2 then the integrand is a constant multiple of

f (x) = r

a

1 + r4.Thus any of the contours in Figure 13.13 can be used to evaluate the real integral The only difference is how manyresidues we have to calculate Thus we choose the first contour in Figure 13.13 We put a branch cut on the negativereal axis and choose the branch −π < arg(z) < π to satisfy f (1) = 1

We evaluate the integral along C with the Residue Theorem

C C

C C

C C

Figure 13.13: Possible Paths of Integration for f (z) = 1+zza4

The integral on C vanishes as  → 0 We demonstrate this with the maximum modulus integral bound

Z

C 

za

1 + z4 dz

≤ π

2 maxz∈C 

za

1 + z4

≤ πR2

Rαeπ|β|/2

R4− 1

→ 0 as R → ∞

The value of the integrand on the positive imaginary axis, z = x eıπ/2, is

xa

2(e−ıπ(a+1)/4− eıπ(a+1)/4)

Z ∞ 0

xa

1 + x4 dx = π

4 csc

 π(a + 1)4

C ρ

z1/2log z(z + 1)2 dz

≤ 2πρ max

C ρ

...

z1 /2< /small>log z(z + 1)2< /small> dz

≤ 2? ?ρ max

C ρ

z1 /2< /small>log z(z + 1)2< /small>

= 2? ?ρρ... class="page_container" data-page="8">

The value of the integrand on the positive imaginary axis, z = x eıπ /2< /small>, is

xa

2( e−ıπ(a+1)/4− eıπ(a+1)/4)... πR

2 maxz∈C R

za

1 + z4

≤ πR2

Rαeπ|β| /2< /small>

R4−