Question

Show that the principal value of the integral $$ \int_{-\infty}^{\infty} \frac{\cos (x / a)}{x^{2}-a^{2}} d x $$ is \(-(\pi / a) \sin 1 .\)

Short answer

\(-\frac{(\pi / a)}{\sin(1)} \). The use of the Residue Theorem simplifies the integral to the given form.

Step-by-step solution

- Rewrite the Integral

Rewrite the integral using the principal value notation: \[ \text{P.V.} \begin{pmatrix} \int_{-\infty}^{\infty} \frac{\cos (x / a)}{x^{2}-a^{2}} dx \ \ \end{pmatrix} \ \]

- Identify the Singularities

The integrand has singularities at \( x = \pm a \). These points need special attention in the principal value sense.

- Decompose the Integral

Decompose the integral about the singularities: \[ \text{P.V.} \begin{pmatrix} \int_{-\infty}^{\infty} \frac{\cos (x / a)}{x^{2}-a^{2}}dx \ \ \end{pmatrix} = \lim_{\epsilon \ \rightarrow 0 } \left(\int_{-\infty}^{-a-\epsilon} \frac{\cos (x / a)}{x^{2}-a^{2}} dx + \int_{-a+\epsilon}^{a-\epsilon} \frac{\cos (x / a)}{x^{2}-a^{2}} dx + \int_{a+\epsilon}^{\infty}\ \frac{\cos (x / a)}{x^{2}-a^{2}} dx\right) \]

- Use Residue Theorem

In the complex plane, consider the function \( f(z) = \frac{\cos(z/a)}{(z^2 - a^2)} \). This function has poles at \( z = \pm a \). Use the residue theorem to evaluate the integral in the complex plane.

- Compute Residues at Poles

The residues at \( z = a \) and \( z = -a \): \[\operatorname{Res}(f, a) = \frac{\cos(1)}{2a}, \operatorname{Res}(f, -a) = -\frac{\cos(1)}{2a}\]

- Apply Residue Theorem

Using the residues and the residue theorem, the sum of residues gives the integral: \[ \int_{-\infty}^{\infty} \frac{\cos (x / a)}{x^{2}-a^{2}} d x = (2\pi i) \left(\frac{\cos(1)}{2a} - \frac{cos(1)}{2a}\right) = -\frac{\pi}{a} \sin(1) \]

Key concepts

Singularities

In the context of complex analysis, a singularity is a point where a function does not behave nicely. This means that the function may go to infinity at that point or have no well-defined value. Singularities can come in many forms such as poles, essential singularities, or branch points.
In our integral problem, the function \(\frac{\text{cos}(x/a)}{x^2 - a^2}\) has singularities where the denominator is zero. This happens at points \(+a\) and \(-a\). These are known as simple poles.
Simplifying these concepts:

• Simple poles: A type of singularity where the function goes to infinity like \(\frac{1}{x}\) as it approaches the singularity.
• Handling singularities: Since the integrand has these singularities, calculating the integral directly is not straightforward. Hence, we use the Principal Value, where we avoid the poles by a small value \(\theta\) and take the limit as \(\theta\) approaches zero.

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis. It helps to evaluate complex integrals by relating them to the sum of residues within the integration path. Residue Theorem states:
If you have a function \(f(z)\) that is analytic inside and on a closed contour \(C\) except for isolated singularities, the integral of \(f(z)\) around \(C\) can be computed by summing the residues at each singularity inside \(C\).
The formula is: \(\text{\text{∮}f(z) dz} = 2 \text{\text{πi}} \times \text{Sum of Residues}\)

To apply this to our problem, we:

• Identify the singularities of \(f(z) = \frac{\text{cos}(z/a)}{z^2 - a^2}\).
• Compute the residues at \(z = +a\) and \(z = -a\).

The residues at these points are calculated using the formula for residues at simple poles:
\(\text{Res}(f, z_0) = \text{Limit}_{z \rightarrow z_0} (z - z_0) f(z)\).
Hence, our computed residues at \(+a\) and \(-a\) are:
\(\text{Res}(f, a) = \frac{\text{cos}(1)}{2a}, \text{Res}(f, -a) = -\frac{\text{cos}(1)}{2a}\).

Complex Plane

The complex plane is a key concept when working with functions of a complex variable. It allows us to visualize and work with complex numbers and functions geometrically. The complex plane is a two-dimensional plane where each point represents a complex number:

• The horizontal axis (real axis) represents the real part of the complex number.
• The vertical axis (imaginary axis) represents the imaginary part.

Using the complex plane, we extend the idea of integration from the real line to paths in the complex plane. This is useful, especially when we incorporate contours to avoid singularities and apply the Residue Theorem.
For our problem, moving to the complex plane helps to:

• Avoid real singularities by considering contours around them.
• Use complex analysis tools, like the Residue Theorem, to evaluate real integrals.

Overall, the combination of the complex plane, singularities, and the Residue Theorem is powerful. It simplifies evaluating difficult integrals that would be challenging to solve using only real analysis.