Question

Find the principal values of the following integrals. (a) \(\int_{-\infty}^{\infty} \frac{\sin x d x}{\left(x^{2}+4\right)(x-1)}\). (b) \(\int_{-\infty}^{\infty} \frac{\cos a x}{1+x^{3}} d x\) where \(a \geq 0\). (c) \(\int_{-\infty}^{\infty} \frac{x \cos x}{x^{2}-5 x+6} d x\). (d) \(\int_{-\infty}^{\infty} \frac{1-\cos x}{x^{2}} d x\).

Short answer

The answers to the integrals are: (a) \(\pi e^{-2}\), (b) \(\frac{2 \pi}{\sqrt{3}} e^{-a/2} \sin (\frac{\sqrt{3}}{2} a)\), (c) \(0\), and (d) \(-\pi\).

Step-by-step solution

General Step

All the questions are related to improper integral, so they follow a general structure. The first step is always to break down the function and see if common integrals techniques, like substitution or integration by parts, can be used. If the function is too complex or does not have an elementary primitive function, then use the residue theorem combined with contour integration.

Step 1(a)

For the integral \(\int_{-\infty}^{\infty} \frac{\sin x}{\left(x^{2}+4\right)(x-1)} dx\) we will be using the Residue Theorem from Complex Analysis. The singularities of the integrand are at \(x = 1, x = 2i, x = -2i\). By applying the Residue Theorem, one can calculate that the principal value of the integral is \(\pi e^{-2}\).

Step 1(b)

For the integral \(\int_{-\infty}^{\infty} \frac{\cos a x}{1+x^{3}} d x\) where \(a \geq 0\), again applying the Residue Theorem from Complex Analysis is the benefitial strategy. Singularities of the integrand are at \(x = \omega, \omega^{2}, -1\), where \(\omega = e^{2 \pi i/3}\). With this in mind, the principal value of the integral is found to be \(\frac{2 \pi}{\sqrt{3}} e^{-a/2} \sin (\frac{\sqrt{3}}{2} a)\).

Step 1(c)

For the integral \(\int_{-\infty}^{\infty} \frac{x \cos x}{x^{2}-5 x+6} d x\), we will take a similar approach. The integrand has singularities at \(x = 2, x = 3\). Again applying the Residue Theorem, the principal value of the integral can be shown to be 0.

Step 1(d)

For the integral \(\int_{-\infty}^{\infty} \frac{1-\cos x}{x^{2}} d x\), it is not possible to apply the Residue Theorem directly since \(x=0\) doesn't have a valid Laurent series. By expanding the integrand at \(x=0\) and integrating term-by-term, the result will be \(-\pi\).

Key concepts

Improper Integrals

Improper integrals are a fascinating topic in calculus that involve limits. They typically arise when you try to evaluate integrals with infinite intervals or when the integrand becomes infinite within the given limits.
To solve these integrals, let's break it down:

• If the integral is from \( -\infty \) to \( \infty \), it means evaluating the area under the curve over the entire x-axis.
• Sometimes, these integrals converge to a finite value, indicating the total area stays limited despite an infinite range.
• If a function has discontinuity or undefined regions within its limits, it's termed as having improper points.

Leveraging these concepts ensures that the evaluation yields results that physically or mathematically make sense, often using limits to provide a finite area or outcome for integrals previously deemed impossible.

Residue Theorem

The residue theorem is a powerful tool in complex analysis used to evaluate integrals, especially those that are otherwise difficult or impossible to solve using standard integration methods.
Here's the core of how and why it works:

• The theorem revolves around the concept of residues, which are values derived from singularities (points where a function doesn't behave nicely) in a complex function.
• Singularities can be thought of as 'poles' or points of no return for a function. To find the integral of such a function, one can look at these points and calculate their residues.
• By summing up these residues within a specific contour, you can find the integral over a real interval. This is where nature's beauty of complex integration shines through.

In essence, the residue theorem acts as a shortcut, transforming a complex path of integration into something manageable and often elegant by following the leads set by singularities.

Contour Integration

Contour integration is a central concept in the realm of complex analysis that complements the residue theorem perfectly. It involves constructing a path or 'contour' in the complex plane and then evaluating the integral along this path.
Here’s a closer look:

• The idea is to extend real-valued functions into the complex plane, where certain properties and values of integrals become apparent.
• To do this, you construct a path around singularities (like circles or combinations of lines and curves) in a manner that respects the rules of analytic functions.
• This technique transforms an otherwise challenging integral into a simpler one by making use of symmetry and regularity in the complex plane.

Contour integration often feels like weaving through a maze, but it’s regularized by calculus laws and guided by the properties of analytic functions. As a result, complex integrals often become approachable or even trivial.