Problem 33 - Advanced Engineering Mathematics

Question

$$\begin{aligned} &\int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}}=\frac{1}{2} \int_{0}^{2 \pi} \frac{d \theta}{(a+\cos \theta)^{2}}=\frac{2}{i} \oint_{C} \frac{z}{\left(z^{2}+2 a z+1\right)^{2}} d z \quad(C \text { is }|z|=1)=\frac{2}{i} \oint_{C} \frac{z}{\left(z-r_{1}\right)^{2}\left(z-r_{2}\right)^{2}} d z\\\ &\text { where } r_{1}=-a+\sqrt{a^{2}-1}, r_{2}=-a-\sqrt{a^{2}-1} . \text { Now }\\\ &\oint_{C} \frac{z}{\left(z-r_{1}\right)^{2}\left(z-r_{2}\right)^{2}} d z=2 \pi i \operatorname{Res}\left(f(z), r_{1}\right)=2 \pi i \frac{a}{4(\sqrt{a^{2}-1})^{3}}=\frac{a \pi}{2(\sqrt{a^{2}-1})^{3}} i \end{aligned}$$ Thus $$\int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}}=\frac{2}{i} \cdot \frac{a \pi}{2(\sqrt{a^{2}-1})^{3}} i=\frac{a \pi}{(\sqrt{a^{2}-1})^{3}}$$ When \(a=2\) we obtain $$\begin{aligned} &\int_{0}^{\pi} \frac{d \theta}{(2+\cos \theta)^{2}}=\frac{2 \pi}{(\sqrt{3})^{3}}\\\ &\text { and so } \int_{0}^{2 \pi} \frac{d \theta}{(2+\cos \theta)^{2}}=\frac{4 \pi}{3 \sqrt{3}} \end{aligned}$$

Step-by-Step Solution

Answer

The integral value is \( \frac{2 \pi}{3 \sqrt{3}} \) when \( a = 2 \).

Step 1: Transform the Integral

Start by expressing the integral from 0 to \( \pi \) in a more general form. \[\int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}} = \frac{1}{2} \int_{0}^{2\pi} \frac{d \theta}{(a+\cos \theta)^{2}}\]This allows us to relate it to a contour integral in the complex plane.

Step 2: Convert to Contour Integral

Recognize that the original integral can be converted to a contour integral using the substitution \( z = e^{i\theta} \), where \( \cos \theta = \frac{z + z^{-1}}{2} \), and change the variable in the integral.\[\int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}} = \frac{2}{i} \oint_{C} \frac{z}{\left(z^{2}+2az+1\right)^{2}} \ dz\]This contour integral takes place around the unit circle \( |z| = 1 \).

Step 3: Factor the Denominator

The denominator of the integrand can be factored as follows:\[z^2 + 2az + 1 = (z - r_1)(z - r_2)\]where \( r_1 = -a + \sqrt{a^2 - 1} \) and \( r_2 = -a - \sqrt{a^2 - 1} \). This simplifies the contour integral as:\[\oint_{C} \frac{z}{\left(z-r_{1}\right)^{2}\left(z-r_{2}\right)^{2}} \ dz\]

Step 4: Evaluate the Residue

Identify the poles and use the residue theorem to evaluate the contour integral. The pole inside the unit circle \( |z| = 1 \) is \( r_1 \).Compute the residue at \( r_1 \):\[\text{Res}(f(z), r_1) = \frac{a}{4(\sqrt{a^2 - 1})^3}\]Thus, we have:\[\oint_{C} \frac{z}{\left(z-r_{1}\right)^{2}\left(z-r_{2}\right)^{2}} \ dz = 2 \pi i \frac{a}{4(\sqrt{a^2 - 1})^3} = \frac{a \pi}{2(\sqrt{a^2 - 1})^3} i\]

Step 5: Solve the Original Integral

Substitute the residue result back into the integral expression and multiply by the factor \( \frac{2}{i} \):\[\int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}} = \frac{2}{i} \cdot \frac{a \pi}{2(\sqrt{a^2 - 1})^3} i = \frac{a \pi}{(\sqrt{a^2 - 1})^3}\]

Step 6: Substitute Specific Value

To find a specific value of the integral, substitute \( a = 2 \) into the result:\[\int_{0}^{\pi} \frac{d \theta}{(2+\cos \theta)^{2}} = \frac{2 \pi}{(\sqrt{3})^{3}}\]And for the integral from 0 to \( 2\pi \):\[\int_{0}^{2\pi} \frac{d \theta}{(2+\cos \theta)^{2}} = \frac{4 \pi}{3 \sqrt{3}}\]

Key Concepts

Contour Integrals

Contour integrals are a fascinating concept in complex analysis, involving the integration of a complex function along a specific path in the complex plane. When we talk about integrating around a contour, we refer to paths or curves over which an integral is taken. In the exercise given, the contour is the unit circle with the equation \(|z| = 1\).
The purpose of changing a real integral into a contour integral is to utilize the properties of complex functions that make calculation easier, such as the Residue Theorem, which we'll discuss next.
When converting from real to complex integrals, variables are substituted. Here, the substitution \(z = e^{i\theta}\) transforms the integral from trigonometric to complex using the complex plane. You might picture the point \(z\) moving along the unit circle as \(\theta\) changes from 0 to \(2\pi\). This circle is our path of integration.
Understanding contour integration is not only beneficial for evaluating integrals but also provides insight into the profound connectivity between complex and real analysis. It opens up a rich toolbox for tackling otherwise challenging or unsolvable real integrals.

Residue Theorem

The Residue Theorem is one of the central tools in complex analysis for evaluating contour integrals. It simplifies complex integration by relating it to the sum of residues at the poles inside a closed contour.
Residues can be thought of as the coefficients of the \((z - r)^{-1}\) term in the Laurent series expansion of a complex function. These play a critical role in determining the contribution of poles to the integral.
In the given problem, we identify the poles by factoring the denominator \(z^2 + 2az + 1\) into \( (z - r_1)(z - r_2) \). We then find that the pole \(r_1\) is inside the unit circle and use the Residue Theorem to calculate the integral.
It's calculated by evaluating the residue at \(r_1\), which corresponds to: \[\text{Res}(f(z), r_1) = \frac{a}{4(\sqrt{a^2 - 1})^3} \]
Finally, applying the Residue Theorem: \[ \oint_{C} \frac{z}{\left(z-r_{1}\right)^{2}\left(z-r_{2}\right)^{2}} \ dz = 2 \pi i \frac{a}{4(\sqrt{a^2 - 1})^3} \]
Understanding this theorem empowers one to simplify complex integrations significantly, making it fundamental not only in analysis but also in physics and engineering.

Trigonometric Integrals

Trigonometric integrals involve the integration of expressions containing trigonometric functions. They often appear in calculus and engineering problems, and solving them can sometimes be tricky due to their periodic nature.
In the exercise, we start with a trigonometric integral of the form:
\[ \int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}} \]
Trigonometric functions like cosine have periodic patterns, and this pattern influences the integral's evaluation. The integral presented is complex due to the composition \((a + \cos \theta)^2\).
Combining trigonometric and complex analysis by converting cosine to its exponential form through the substitution \(z = e^{i\theta}\) helps rectify the periodic difficulty. The substitution is done by expressing \(\cos \theta\) as \(\frac{z + z^{-1}}{2}\).
Ultimately, solving trigonometric integrals with these methods not only provides exact values for seemingly intractable integrals but also enhances one's toolkit for mathematical and practical problem-solving in fields such as physics, engineering, and signal processing.