Question

By considering the real part of $$ \int \frac{-i z^{n-1} d z}{1-a\left(z+z^{-1}\right)+a^{2}} $$ where \(z=\exp i \theta\) and \(n\) is a non-negative integer, evaluate $$ \int_{0}^{\pi} \frac{\cos n \theta}{1-2 a \cos \theta+a^{2}} d \theta $$ for \(a\) real and \(>1\)

Short answer

The final result is 0 for the integral.

Step-by-step solution

Substitution

Let us use the substitution given in the problem: \[ z = e^{i\theta} \] With this, we have \[ dz = ie^{i\theta} d\theta = iz d\theta \] Also, note that \( z^{-1} = e^{-i\theta} \).

Change of Variables

Now let's change the variable in the integral: \[ \theta = 0 \text{ to } \theta = \pi \] becomes a closed contour on the unit circle for \(z\). The integral becomes: \[ \begin{aligned} \text{Given integral:} & \int_{0}^{\pi} \frac{\cos n \theta}{1-2 a \cos \theta+a^{2}} d \theta \ & = \int_{C} \frac{-i z^{n-1} dz}{1 - a \left( z+z^{-1} \right) + a^2} \end{aligned} \]

Simplification

We recognize that \[ 1 - a \left( z + z^{-1} \right) + a^2 = 1 - a(z + z^{-1}) + a^2 \]which can be rewritten as a perfect square form: \[(z - a)(z^{-1} - a^{ -1}) \].

Integral on Closed Contour

Since the integrand is analytic inside and on the unit circle, except for poles, we apply the residue theorem. Identify poles inside the unit circle: For \( (z-a) \text{, the poles are } z = a \). Since \( a \text{ is greater than } 1\), the poles are outside the unit circle.

Residue Calculation

Since the function has no poles inside the unit circle, the integral evaluates to 0 excluding all poles. Thus: \[ \text{Real part of } \int_{C} \frac{-i z^{n-1} dz}{(z-a)(z^{-1}-a^{-1})} = 0 \].

Final Result

Since the integral evaluates to 0 in the complex plane, the desired real part over [0, π] is also 0. Therefore: \[ \int_{0}^{\pi} \frac{\cos n \theta}{1-2 a \cos \theta + a^2} d\theta = 0 \].

Key concepts

integration on closed contour

In complex analysis, the concept of integration on a closed contour is critical. A closed contour is essentially a path in the complex plane that starts and ends at the same point, forming a loop. When performing integration on such contours, a significant result used is the Cauchy-Goursat theorem which states that if a function is analytic inside and on a simple closed contour, then the integral over that contour is zero. This facilitates calculating complex integrals, especially when combined with the residue theorem.

In our exercise, we transformed the real integral into a complex one by substituting a complex exponential function. The result is a closed contour integral along the unit circle in the complex plane. When integrating functions with closed contours, we examine the behavior of the function within and on the boundary of that contour.

residue theorem

The residue theorem is a powerful tool in complex analysis, often used to evaluate complex integrals, especially integrals around closed contours. The theorem states that if we have a function that is analytic inside and on some closed contour, except for isolated singularities (poles), the integral around that contour is equal to \(2πi\) times the sum of residues of the function inside the contour.

In this problem, we encountered integrals involving functions that might have singularities. We determined which singularities (if any) lay inside the unit circle. Given that the poles were outside the unit circle (since \(a > 1\)), it was concluded that the integral over the closed contour was zero as the residues inside the contour summed to zero.

change of variables

Changing variables is a common strategy to simplify and solve integrals, particularly in complex analysis. By making a substitution, we can often transform a difficult integral into one easier to evaluate. Consider our substitution in this problem where \(z = e^{i\theta}\). This transformed the variable from a real domain (from \(0\) to \(π\)) into a complex one, corresponding to a closed contour traversal along the unit circle.

This method lets us use tools from complex analysis like the residue theorem, which are not directly applicable in the real domain. Through substitution, we can also translate properties and behaviors from one domain to another, facilitating solutions to complex integrals.

complex substitution

Complex substitution is a specific case of variable change, where a real variable is replaced with a complex one. This technique helps in solving integrals involving trigonometric functions by leveraging the properties of complex exponentials. In our exercise, we used the substitution \(z = e^{i\theta}\), transforming real integration limits to the boundaries of the unit circle in the complex plane.

By recognizing that \(z^{-1} = e^{-i\theta}\), it allowed further simplification and setup for applying residues and contour integration methods. Complex substitution provides a bridge between real and complex analysis, often simplifying complicated integral problems and allowing for advanced analytical techniques to be applied.