Question

using Cauchy's Residue Theorem and the so-alled pacman inte 6. (10 pts) Evaluate gration contour.

Answer 1

• $\oint$_cf(x) dx = 2$\pi$i { Residues of f(x_n) } , where n is number of poles

If f has a removable singularity at x = x₀, then the residue is equal to zero. If f has a single pole at x = x₀ , then

• Res f(x₀) =  
  lim
  ₀ (x-x₀)*f(x)

and if f has a pole of order k at x = x₀, then

• Res f(x₀) = $\frac{1}{(k-1)!}$   
  lim
  ₀
  dak
  {(x-x₀)^k * f(x)}  , k ∈ {1, 2, 3, . . .}

In the given problem,

$\int_{0 }^{\infty }$
CAT 8)
​​ =   $\int_{0 }^{\infty }$$\frac{dx}{x^{1/2}(x-2(1+i))(x-2(1-i))}$

Let f(x) = $\frac{1}{x^{1/2}(x^{2}-4x+8)}$   ,

now let us find non analytic points i.e, x²-4x+8 = 0

$\Rightarrow$(x-2(1+i))(x-2(1-i)) = 0

$\Rightarrow$x₀ = 2(1+i) or x₁ = 2(1-i)

The residues are,

R0 = Res f(x₀) =  
lim
₀ (x-x₀)*f(x) =  
lim
$\frac{(x-2(1+i))}{x^{1/2}(x-2(1+i))(x-2(1-i))}$

=  
lim
  
r1/2(r - 2(1- i))

  =
4i2(1 + i)

R₁ = Res f(x₁) =  
lim
₁(x-x₁)*f(x) =  $\lim_{x\rightarrow 2(1-i)}$$\frac{(x-2(1-i))}{x^{1/2}(x-2(1+i))(x-2(1-i))}$

=  $\lim_{x\rightarrow 2(1-i)}$  $\frac{1}{x^{1/2}(x-2(1+i))}$

  = $\frac{-1}{4i \sqrt{2(1-i)}}$

$\int_{0 }^{\infty }$f(x) dx = 2$\pi$i { Residues of f(x_n) } , where n is number of poles

$\int_{0 }^{\infty }$
CAT 8)
= 2$\pi$i {R₀ + R₁}

=   2$\pi$i * { $\frac{1}{4i \sqrt{2(1+i)}}-\frac{1}{4i \sqrt{2(1-i)}}$ }

= 2$\pi$i * {$\frac{\sqrt{1-i}-\sqrt{1+i}}{4\sqrt{2}i * \sqrt{(1+i)(1-i)}}$} =   $\frac{2\pi i}{8i} * (\sqrt{1-i} - \sqrt{1+i})$

(Since
1 i)(1
= $\sqrt{2}$ )

$\int_{0 }^{\infty }$
CAT 8)
=   $\frac{\pi }{4}$
1 i)(1