Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

Let f z be a single-valued analytic tion with only isolated singularities and no singularities on the positive, real axis, [0, ∞... 13.8 Exploiting SymmetryWe have already used symmetry

Result 13.7.1 Integrals from Zero to Infinity Let f (z) be a single-valued analytic tion with only isolated singularities and no singularities on the positive, real axis, [0, ∞) Let

func-a 6∈ Z If the integrfunc-als exist then,

Z ∞ 0

where z1, , zn are the singularities of f (z) and there is a branch cut on the positive real axis with 0 < arg(z) < 2π.

13.8 Exploiting Symmetry

We have already used symmetry of the integrand to evaluate certain integrals For f (x) an even function we wereable to evaluate R0∞f (x) dx by extending the range of integration from −∞ to ∞ For

Z ∞ 0

xαf (x) dx

we put a branch cut on the positive real axis and noted that the value of the integrand below the branch cut is aconstant multiple of the value of the function above the branch cut This enabled us to evaluate the real integral withcontour integration In this section we will use other kinds of symmetry to evaluate integrals We will discover thatperiodicity of the integrand will produce this symmetry

We note that zn = rneınθ is periodic in θ with period 2π/n The real and imaginary parts of zn are odd periodic

in θ with period π/n This observation suggests that certain integrals on the positive real axis may be evaluated byclosing the path of integration with a wedge contour

Example 13.8.1 Consider

Z ∞ 0

1

1 + xndx

where n ∈ N, n ≥ 2 We can evaluate this integral using Result 13.7.1.

Z ∞ 0

of r and θ, it is periodic in θ with period 2π/n

contour We evaluate the residue there.

Res

1

Z

C R

1

1 + zndz

≤ 2πR

n z∈CmaxR

1cosh z

≤ π max

y∈[0 π]

2

e±R+ıy+ e∓R−ıy

...

scos2< /small>x cosh2< /sup>y + sin2< /sup>x sinh2< /sup>ysin2< /sup>x cosh2< /sup>y + cos2< /small>x sinh2< /sup>y

≤

scosh2< /sup>ysinh2< /sup>y...

x2

I

C

1cosh z dz = ? ?2? ? Res

1cosh z,

ı? ?2



= ? ?2? ? lim

z→ıπ /2< /small>

z − ıπ/2cosh... integral exists for −1 < a <