Lecture Notes: Logarithm and Integrals in Complex Analysis

Introduction

• Welcome back, audience.
• Presenter is a math enthusiast exploring logarithmic functions with imaginary components.

Logarithm of Complex Function

• Focus on integrating the logarithm of the function:
  [ \log(1 - e^{i\Theta}) ]
  • Where ( i ) is the imaginary unit.

Series Expansion of Logarithm

• Series expansion:
  [ \log(1 - Z) = -\sum_{k=1}^{\infty} \frac{Z^k}{k} ]
• Replacing ( Z ) with ( e^{i\Theta} ):
  [ \log(1 - e^{i\Theta}) = -\sum_{k=1}^{\infty} \frac{e^{ik\Theta}}{k} ]

Integration Process

• Integrate from 0 to x with respect to ( \Theta ):
  [ \int_0^x \log(1 - e^{i\Theta}) d\Theta ]
  • This leads to
    [ I = -\sum_{k=1}^{\infty} \frac{1}{k} \int_0^x e^{ik\Theta} d\Theta ]

Integral Evaluation

• Evaluating the integral ( \int e^{ik\Theta} d\Theta ):
  • Resulting in ( \frac{e^{ik\Theta}}{ik} ) with limits from 0 to x.
• Resulting expression:
  [ I = -\sum_{k=1}^{\infty} \frac{1}{k^2} (e^{ikx} - 1) ]

Relationship to Zeta Function

• Recognizing ( \sum_{k=1}^{\infty} \frac{1}{k^2} ) is ( \zeta(2) = \frac{\pi^2}{6} ).
• Thus, express as:
  [ I = i \left( -\sum_{k=1}^{\infty} \frac{e^{ikx}}{k^2} + \zeta(2) \right) ]

Polylogarithm Function

• Introduction of the polylogarithm function:
  • Series expansion:
    [ \text{Li}s(z) = \sum{k=1}^{\infty} \frac{z^k}{k^s} ]
• For simplicity, using ( s = 2 ) and ( z = e^{ix} ).

Different Approaches

• Switching method:
  • Expanding the complex exponential using Euler's formula: [ e^{i\Theta} = \cos(\Theta) + i\sin(\Theta) ]

Logarithmic Properties

• Breaking down integrals:
  • [ I = \int_0^x \log(2\sin(\Theta/2)) d\Theta + \int_0^x \log(\sin(\Theta/2) - i\cos(\Theta/2)) d\Theta ]
• Introduces the Clausen function:
  [ I = -\text{Cl}_2(x) + i\int_0^x \left(\frac{\pi}{2} - \frac{\Theta^2}{4}\right) d\Theta ]

Final Expressions

• Final forms of the integral:
  [ I = -\text{Cl}_2(x) - i \left( \frac{\pi}{2}x - \frac{x^2}{4} \right) ]

Conclusion

• Equating formulas leads to beautiful results involving the Clausen function and the zeta function.
• Exploration of series expansions provides insight into complex integrals.

Closing Remarks

• Encourage audience engagement: Like and subscribe for more math content.