Question

Evaluate the integrals by contour integration. $$P V \int_{-\infty}^{\infty} \frac{\sin x d x}{(3 x-\pi)\left(x^{2}+\pi^{2}\right)}$$

Short answer

The integral evaluates to zero.

Step-by-step solution

- Identify the function for integration

The integrand is given by \ \ \( f(x) = \frac{\text{sin}(x)}{(3x - \pi)(x^2 + \pi^2)} \), where \( PV \int_{-\infty}^{\infty} \) indicates the Cauchy Principal Value.

- Analyze the poles

Find the poles of the integrand: \ \ 1. \( x = \frac{\pi}{3} \) from the factor \((3x - \pi)\). \ \ 2. \( x = i\pi \) and \( x = -i\pi \) from the factor \((x^2 + \pi^2)\). \ \ Thus, the poles are at \( \frac{\pi}{3}, i\pi \) and \( -i\pi \).

- Construct the contour

Use a semicircular contour \( C \) in the upper half-plane that closes to the right. This contour bypasses the real axis and includes the singularity at \( x = \frac{\pi}{3} \).

- Apply the Residue Theorem

Using the Residue Theorem, the integral around the contour \( C \) is given by \ \ \( \int_{C} f(z) dz = 2\pi i \sum_{\text{residues~inside}} f(z) \). We will calculate the residues at \( z = i\pi \).

- Calculate the residue at \( z = i\pi \)

The residue at \( z = i\pi \) can be found using: \ \ \( \text{Res} \left( f(z), i\pi \right) = \lim_{z \to i\pi} (z - i\pi) f(z) \).\ \ Substitute the value \( i\pi \) into: \ \ \( f(z) = \frac{\text{sin}(z)}{(3z - \pi)(z^2 + \pi^2)} \).

- Simplify the residue expression

Using \( z \to i\pi \),\( \sin(i\pi) = 0 \Rightarrow \text{Res} \left(f(z), i\pi \right) = 0 \). Thus, \( f(z) \) evaluates to zero at this pole.

- Account for contributions of other poles

Since the other pole at \( x = \frac{\pi}{3} \) lies on the real axis and has symmetric contributions on the whole real axis, the integral evaluates to zero considering the Sine function is odd. Therefore, the value of the principal value integral is zero.

Key concepts

Cauchy Principal Value

The Cauchy Principal Value (often abbreviated as PV) is a method to assign values to certain improper integrals that would otherwise be undefined. In integrals where the integrand has singularities along the path of integration (such as zeros in the denominator), the integral might not converge in the traditional sense. The PV is a way to manage such singularities by balancing the contributions around these points.
For example:
\[PV \int_{-\infty}^{\infty} \frac{\sin (x) dx}{(3x - \pi) (x^2 + \pi^2)}\]
The important idea is to symmetrically approach the singularities from both sides, which effectively cancels out the infinite contributions, making the integral well-defined.

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis that simplifies the computation of contour integrals. The theorem states that if you have a function that is analytic inside and on some simple closed contour, except at a finite number of isolated singularities (poles), then the integral of the function over that contour is given by:
\[\int_{C} f(z) dz = 2\pi i \sum_{\text{residues~inside}} f(z) \]
This means the value of the integral is 2πi times the sum of the residues of the function at the enclosed poles. Residues can be thought of as the coefficients of the 1/(z - z0) term in the Laurent series expansion of the function around the singularity z0. Calculating these residues allows us to transform a potentially complex integral into a simple sum.

Poles of Functions

In complex analysis, poles are specific types of singularities of complex functions where the function's value goes to infinity. A simple pole at z=c is characterized by the fact that as z approaches c, the function behaves like 1/(z - c), multiplied by some analytic function that is not zero at z=c.
In our example:
\[f(x) = \frac{\sin(x)}{(3x - \pi)(x^2 + \pi^2)}\]
We identified the poles at x = π/3, iπ, and -iπ. Noticing these locations helps us understand where the function is not defined and where the residues need to be calculated. Proper identification and treatment of these poles are critical as they define the contributions to the integral around a closed contour.

Semicircular Contour

Using a semicircular contour is a clever technique in contour integration, especially when dealing with integrals over the entire real line. Here’s how it works:
1. We consider the contour composed of two parts: the real axis and a semicircle in the complex plane, usually in the upper or lower half-plane.
2. The radius of the semicircle is chosen to be large enough to enclose the necessary poles.
3. By applying the Residue Theorem on this contour, we can relate the integral along the real axis (our original integral) to an integral over the semicircle, which often vanishes as the radius goes to infinity.
In our example, the semicircular contour in the upper half-plane was used to avoid the real axis singularity at x = π/3 and includes the singularities at iπ and -iπ. This technique efficiently evaluates the given integral by understanding the symmetrical properties and contributions of the function at these poles.