Contour Integration Solution For Integral Of 1/(1+cos(u)cos(t))^2

Hey guys! Today, we're diving deep into the fascinating world of complex analysis to tackle a seemingly daunting integral. We're going to use the magic of contour integration to show that:

$$\int_0^{2\pi}\frac{\mathrm dt}{(1+\cos(u)\cos(t))^2}=\frac{2\pi}{\sin^3(u)}$$

for any $u\in(0,\frac{\pi}{2}$). Buckle up, because this is going to be a fun ride!

The Initial Transformation: Setting the Stage

Our journey begins with a clever transformation. First, we recognize that the integral involves trigonometric functions, specifically cosine. To make our lives easier, we'll convert this into a complex integral using the substitution $z = e^{it}$. This is a standard trick in contour integration, and it's super powerful.

Recall Euler's formula:

$$e^{it} = \cos(t) + i\sin(t)$$

From this, we can deduce that:

$$\cos(t) = \frac{e^{it} + e^{-it}}{2} = \frac{z + z^{-1}}{2}$$

and

$$dt = \frac{dz}{iz}$$

These substitutions are key to transforming our real integral into a complex one. Plugging these into our original integral, we get:

$$\int_0^{2\pi}\frac{\mathrm dt}{(1+\cos(u)\cos(t))^2} = \oint_C \frac{1}{\left(1 + \cos(u)\frac{z + z^{-1}}{2}\right)^2} \frac{dz}{iz}$$

where $C$ is the unit circle in the complex plane, parameterized by $z = e^{it}$ for $0 \le t \le 2\pi$. This contour is perfect for our situation because it nicely captures the periodicity of the trigonometric functions.

Simplifying the Complex Integral: Taming the Beast

Now, let's simplify the complex integral. Our goal is to get it into a form where we can easily apply the Residue Theorem. This involves some algebraic manipulation, but don't worry, we'll take it step by step.

First, let's focus on the denominator inside the integral:

$$\left(1 + \cos(u)\frac{z + z^{-1}}{2}\right)^2 = \left(1 + \frac{\cos(u)}{2}\left(z + \frac{1}{z}\right)\right)^2$$

To get rid of the fraction inside the parentheses, we can multiply the inside by $\frac{z}{z}$:

$$\left(\frac{2z + \cos(u)(z^2 + 1)}{2z}\right)^2 = \frac{(2z + \cos(u)(z^2 + 1))^2}{4z^2}$$

Plugging this back into our integral, we have:

$$\oint_C \frac{1}{\frac{(2z + \cos(u)(z^2 + 1))^2}{4z^2}} \frac{dz}{iz} = \oint_C \frac{4z^2}{(2z + \cos(u)(z^2 + 1))^2} \frac{dz}{iz}$$

We can simplify this further by canceling a $z$ and rearranging:

$$\oint_C \frac{4z}{i(\cos(u)z^2 + 2z + \cos(u))^2} dz = -4i \oint_C \frac{z}{(\cos(u)z^2 + 2z + \cos(u))^2} dz$$

Now, we have a much cleaner complex integral to work with. The next step is to find the poles of the integrand.

Finding the Poles: Unearthing the Singularities

The poles of our integrand are the values of $z$ for which the denominator is zero. So, we need to solve the equation:

$$(\cos(u)z^2 + 2z + \cos(u))^2 = 0$$

This is equivalent to solving the quadratic equation:

$$\cos(u)z^2 + 2z + \cos(u) = 0$$

We can use the quadratic formula to find the roots:

$$z = \frac{-2 \pm \sqrt{2^2 - 4\cos^2(u)}}{2\cos(u)} = \frac{-1 \pm \sqrt{1 - \cos^2(u)}}{\cos(u)}$$

Since $1 - \cos^2(u) = \sin^2(u)$, we have:

$$z = \frac{-1 \pm \sin(u)}{\cos(u)}$$

So, our poles are:

$$z_1 = \frac{-1 + \sin(u)}{\cos(u)} \quad \text{and} \quad z_2 = \frac{-1 - \sin(u)}{\cos(u)}$$

Now, we need to determine which of these poles lie inside our contour $C$, the unit circle. Let's analyze their magnitudes.

Identifying Poles Inside the Unit Circle: The Residency Requirement

Let's look at the magnitude of $z_1$:

$$|z_1| = \left|\frac{-1 + \sin(u)}{\cos(u)}\right|$$

Since $u \in (0, \frac{\pi}{2})$, we have $0 < \sin(u) < 1$ and $\cos(u) > 0$. Thus, $-1 < -1 + \sin(u) < 0$, and since $\cos(u) > 0$, we get:

$$|z_1| = \frac{1 - \sin(u)}{\cos(u)}$$

Multiplying the top and bottom by $1 + \sin(u)$, we have:

$$|z_1| = \frac{1 - \sin^2(u)}{\cos(u)(1 + \sin(u))} = \frac{\cos^2(u)}{\cos(u)(1 + \sin(u))} = \frac{\cos(u)}{1 + \sin(u)}$$

Since $0 < \sin(u) < 1$ and $0 < \cos(u) < 1$, we have $1 < 1 + \sin(u)$, so

$$|z_1| = \frac{\cos(u)}{1 + \sin(u)} < 1$$

Therefore, $z_1$ lies inside the unit circle.

Now let's consider $z_2$:

$$|z_2| = \left|\frac{-1 - \sin(u)}{\cos(u)}\right| = \frac{1 + \sin(u)}{\cos(u)}$$

Since $0 < \sin(u) < 1$ and $0 < \cos(u) < 1$, we have:

$$|z_2| = \frac{1 + \sin(u)}{\cos(u)} > 1$$

Thus, $z_2$ lies outside the unit circle.

So, only $z_1 = \frac{-1 + \sin(u)}{\cos(u)}$ lies inside our contour $C$. This is a crucial piece of information for applying the Residue Theorem.

Calculating the Residue: The Heart of the Matter

The Residue Theorem states that:

$$\oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)$$

where the sum is over all poles $z_k$ of $f$ inside the contour $C$. In our case, we have only one pole inside the contour, $z_1$. Also, note that since $(\cos(u)z^2 + 2z + \cos(u))^2 = \cos^2(u)(z - z_1)^2(z - z_2)^2$, our pole $z_1$ is a pole of order 2.

The formula for the residue of a pole of order 2 is:

$$\text{Res}(f, z_1) = \lim_{z \to z_1} \frac{d}{dz}\left[(z - z_1)^2 f(z)\right]$$

Our function $f(z)$ is:

$$f(z) = \frac{z}{\cos^2(u)(z - z_1)^2(z - z_2)^2}$$

So,

$$(z - z_1)^2 f(z) = \frac{z}{\cos^2(u)(z - z_2)^2}$$

Now we need to take the derivative:

$$\frac{d}{dz} \left[\frac{z}{\cos^2(u)(z - z_2)^2}\right] = \frac{1}{\cos^2(u)} \frac{(z - z_2)^2 - z \cdot 2(z - z_2)}{(z - z_2)^4} = \frac{(z - z_2)(z - z_2 - 2z)}{\cos^2(u)(z - z_2)^4} = \frac{-z - z_2}{\cos^2(u)(z - z_2)^3}$$

Now we take the limit as $z \to z_1$:

$$\text{Res}(f, z_1) = \lim_{z \to z_1} \frac{-z - z_2}{\cos^2(u)(z - z_2)^3} = \frac{-z_1 - z_2}{\cos^2(u)(z_1 - z_2)^3}$$

Recall that

$$z_1 = \frac{-1 + \sin(u)}{\cos(u)} \quad \text{and} \quad z_2 = \frac{-1 - \sin(u)}{\cos(u)}$$

So,

$$-z_1 - z_2 = -\frac{-1 + \sin(u)}{\cos(u)} - \frac{-1 - \sin(u)}{\cos(u)} = \frac{1 - \sin(u) + 1 + \sin(u)}{\cos(u)} = \frac{2}{\cos(u)}$$

and

$$z_1 - z_2 = \frac{-1 + \sin(u)}{\cos(u)} - \frac{-1 - \sin(u)}{\cos(u)} = \frac{-1 + \sin(u) + 1 + \sin(u)}{\cos(u)} = \frac{2\sin(u)}{\cos(u)}$$

Plugging these in:

$$\text{Res}(f, z_1) = \frac{\frac{2}{\cos(u)}}{\cos^2(u)\left(\frac{2\sin(u)}{\cos(u)}\right)^3} = \frac{\frac{2}{\cos(u)}}{\cos^2(u)\frac{8\sin^3(u)}{\cos^3(u)}} = \frac{2\cos^2(u)}{8\sin^3(u)\cos^2(u)} = \frac{1}{4\sin^3(u)}$$

Applying the Residue Theorem and Finishing the Proof: The Grand Finale

Now we can apply the Residue Theorem:

$$\oint_C f(z) dz = 2\pi i \text{Res}(f, z_1) = 2\pi i \frac{1}{4\sin^3(u)} = \frac{\pi i}{2\sin^3(u)}$$

Remember our original integral was:

$$-4i \oint_C \frac{z}{(\cos(u)z^2 + 2z + \cos(u))^2} dz$$

So,

$$\int_0^{2\pi}\frac{\mathrm dt}{(1+\cos(u)\cos(t))^2} = -4i \cdot \frac{\pi i}{2\sin^3(u)} = \frac{-4i^2 \pi}{2\sin^3(u)} = \frac{2\pi}{\sin^3(u)}$$

And there you have it! We've successfully shown, using contour integration, that:

$$\int_0^{2\pi}\frac{\mathrm dt}{(1+\cos(u)\cos(t))^2}=\frac{2\pi}{\sin^3(u)}$$

for any $u\in(0,\frac{\pi}{2}$). What a journey! This problem beautifully illustrates the power and elegance of complex analysis. Keep exploring, guys, and remember that even the most challenging integrals can be conquered with the right tools and techniques!