Solve $\int_0^\infty\frac{1-\cos X}{x^2(x^2+1)}\,dx$ With Complex Analysis

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Hey everyone! Today, we're diving deep into a fascinating problem: evaluating the integral $\int_0^\infty\frac{1-\cos x}{x^2(x^2+1)}\,dx$ using the powerful tools of complex analysis. If you've ever been intrigued by the elegance of complex integration, residues, and Cauchy's Theorem, you're in the right place. We'll break down each step, making it super clear and easy to follow. So, grab your thinking caps, and let's get started!

Understanding the Integral and Why Complex Analysis?

Before we jump into the solution, let's first understand why this integral is interesting and why complex analysis is a great approach. The integral $\int_0^\infty\frac{1-\cos x}{x^2(x^2+1)}\,dx$ might look intimidating at first glance. It involves a trigonometric function ($\cos x$) divided by a rational function ($x^2(x^2+1)$), and we're integrating over an infinite interval. Direct integration techniques from real calculus can be quite challenging here. This is where complex analysis shines. By extending our analysis to the complex plane, we can leverage powerful theorems like Cauchy's Residue Theorem to simplify the problem.

Complex analysis provides a unique perspective by allowing us to consider functions of complex variables. This opens up a whole new world of techniques, including contour integration and residue calculus, which can transform seemingly difficult real integrals into manageable problems. The core idea is to replace the real integral with a contour integral in the complex plane and then use the residues of the function's singularities to evaluate the integral. This approach is not just a clever trick; it’s a testament to the deep connections between real and complex analysis. So, complex analysis offers a robust and elegant way to tackle integrals that are cumbersome to solve using real calculus methods alone. By choosing an appropriate contour and applying residue theory, we can convert this integral into a sum of residues, making the calculation straightforward. This method highlights the power of extending problems into the complex plane to leverage its unique properties.

Setting Up the Complex Integral

Okay, let’s dive into the nitty-gritty. Our first step is to convert the real integral into a complex one. We do this by considering a complex function that closely relates to our integrand. A common trick is to replace $\cos x$ with $e^{iz}$, where $z$ is a complex variable. So, we define our complex function as:

$$f(z) = \frac{1 - e^{iz}}{z^2(z^2 + 1)}$$

Why $e^{iz}$? Because of Euler's formula, $e^{iz} = \cos z + i\sin z$. The real part of $e^{iz}$ is $\cos z$, which is directly related to our original integrand. By using $e^{iz}$, we can work in the complex plane and later take the real part of our result to get the value of the original integral. This is a clever way to leverage the power of complex exponentials.

Now, we need to choose a suitable contour in the complex plane. A semi-circular contour in the upper half-plane is often a good choice for integrals over $(0, \infty)$. Let's denote this contour by $C$. It consists of two parts:

1. A semi-circle $C_R$ with radius $R$ centered at the origin, lying in the upper half-plane (i.e., $\text{Im}(z) \geq 0$).
2. The real interval $[-R, R]$ along the real axis.

As $R$ approaches infinity, the semi-circular arc will enclose more and more of the upper half-plane, and the integral along the real interval will approach our desired integral. This setup is crucial because it allows us to use Cauchy's Residue Theorem, which relates the integral of a function around a closed contour to the residues of the function's singularities inside the contour. Choosing the right contour is an art in itself, and the semi-circular contour is a classic choice for integrals involving trigonometric functions over infinite intervals.

Identifying Singularities and Calculating Residues

Next up, we need to find the singularities of our function $f(z)$. Singularities are points where the function becomes undefined, usually due to division by zero. For $f(z) = \frac{1 - e^{iz}}{z^2(z^2 + 1)}$, the singularities occur when the denominator is zero:

$$z^2(z^2 + 1) = 0$$

This gives us singularities at $z = 0$ and $z = \pm i$. However, we're only interested in singularities inside our contour $C$, which lies in the upper half-plane. So, we focus on $z = 0$ and $z = i$.

Now, let’s calculate the residues at these points. The residue of a function at a singularity is a measure of the function's behavior near that point. It’s a crucial concept in complex analysis because it directly relates to the value of the contour integral through Cauchy's Residue Theorem.

Residue at $z = 0$

The singularity at $z = 0$ is a pole of order 2 (because of the $z^2$ term in the denominator). To find the residue at a pole of order $n$, we use the formula:

$$\text{Res}(f, z_0) = \frac{1}{(n-1)!} \lim_{z \to z_0} \frac{d^{n-1}}{dz^{n-1}} [(z - z_0)^n f(z)]$$

For our function, $n = 2$ and $z_0 = 0$, so:

$$\text{Res}(f, 0) = \lim_{z \to 0} \frac{d}{dz} [z^2 f(z)] = \lim_{z \to 0} \frac{d}{dz} \left(\frac{1 - e^{iz}}{z^2 + 1}\right)$$

Taking the derivative using the quotient rule, we get:

$$\text{Res}(f, 0) = \lim_{z \to 0} \frac{(-ie^{iz})(z^2 + 1) - (1 - e^{iz})(2z)}{(z^2 + 1)^2}$$

Plugging in $z = 0$, we find:

$$\text{Res}(f, 0) = \frac{(-i)(1) - (0)(0)}{1} = -i$$

Residue at $z = i$

The singularity at $z = i$ is a simple pole (order 1). The residue at a simple pole is given by:

$$\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z)$$

So, for $z_0 = i$:

$$\text{Res}(f, i) = \lim_{z \to i} (z - i) \frac{1 - e^{iz}}{z^2(z^2 + 1)} = \lim_{z \to i} \frac{1 - e^{iz}}{z^2(z + i)}$$

Plugging in $z = i$, we get:

$$\text{Res}(f, i) = \frac{1 - e^{-1}}{i^2(2i)} = \frac{1 - e^{-1}}{-2i} = \frac{i(1 - e^{-1})}{2}$$

Alright, we've nailed the residues! These values are super important because they'll be the key to unlocking our integral using Cauchy's Residue Theorem. Calculating these residues carefully is crucial, as any mistake here will propagate through the rest of the solution. Now that we have the residues, we’re ready to apply Cauchy's Residue Theorem and see how it all comes together.

Applying Cauchy's Residue Theorem

Now comes the moment we've been waiting for – applying Cauchy's Residue Theorem. This theorem states that the integral of a complex function around a closed contour is equal to $2\pi i$ times the sum of the residues of the function's singularities inside the contour. Mathematically:

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

where the sum is over all singularities $z_k$ inside the contour $C$. In our case, the singularities inside $C$ are $z = 0$ and $z = i$, and we've already calculated their residues.

So, applying the theorem, we get:

$$\oint_C \frac{1 - e^{iz}}{z^2(z^2 + 1)} \, dz = 2\pi i [\text{Res}(f, 0) + \text{Res}(f, i)]$$

Plugging in the values we found earlier:

$$\oint_C f(z) \, dz = 2\pi i \left[-i + \frac{i(1 - e^{-1})}{2}\right] = 2\pi i \left[-i + \frac{i}{2} - \frac{i}{2e}\right]$$

Simplifying, we have:

$$\oint_C f(z) \, dz = 2\pi \left[1 - \frac{1}{2} + \frac{1}{2e}\right] = \pi \left[1 + \frac{1}{e}\right]$$

Now, remember that our contour $C$ consists of the semi-circle $C_R$ and the real interval $[-R, R]$. So, we can write:

$$\oint_C f(z) \, dz = \int_{-R}^{R} f(x) \, dx + \int_{C_R} f(z) \, dz$$

As $R$ approaches infinity, the integral over the real interval $[-R, R]$ approaches the integral we want to evaluate. We need to show that the integral over the semi-circle $C_R$ goes to zero as $R \to \infty$. This is a crucial step because it allows us to isolate the integral along the real axis, which is what we're actually interested in.

Showing the Integral over $C_R$ Vanishes

To show that the integral over the semi-circle $C_R$ vanishes as $R \to \infty$, we use the Estimation Lemma (also known as the ML-estimate). This lemma provides an upper bound for the magnitude of a contour integral. It states that if $|f(z)| \leq M$ on a contour $C$ of length $L$, then:

$$\left| \int_C f(z) \, dz \right| \leq ML$$

In our case, $f(z) = \frac{1 - e^{iz}}{z^2(z^2 + 1)}$, and $C$ is the semi-circle $C_R$. We need to find an upper bound $M$ for $|f(z)|$ on $C_R$. For $z$ on $C_R$, we have $z = Re^{i\theta}$, where $0 \leq \theta \leq \pi$. Thus, $|z| = R$.

Now, let's find an upper bound for $|f(z)|$:

$$|f(z)| = \left| \frac{1 - e^{iz}}{z^2(z^2 + 1)} \right| \leq \frac{|1 - e^{iz}|}{|z^2(z^2 + 1)|}$$

We know that $|e^{iz}| = |e^{i(x + iy)}| = |e^{-y + ix}| = e^{-y}$. Since we're in the upper half-plane, $y \geq 0$, so $e^{-y} \leq 1$. Thus, $|e^{iz}| \leq 1$.

Using the triangle inequality, $|1 - e^{iz}| \leq |1| + |e^{iz}| \leq 1 + 1 = 2$. Also,

$$|z^2(z^2 + 1)| = |z|^2 |z^2 + 1| \geq R^2 (|z|^2 - 1) = R^2(R^2 - 1)$$

So,

$$|f(z)| \leq \frac{2}{R^2(R^2 - 1)}$$

The length of the semi-circle $C_R$ is $L = \pi R$. Applying the Estimation Lemma:

$$\left| \int_{C_R} f(z) \, dz \right| \leq \frac{2}{R^2(R^2 - 1)} \cdot \pi R = \frac{2\pi}{R(R^2 - 1)}$$

As $R \to \infty$, this bound goes to zero:

$$\lim_{R \to \infty} \frac{2\pi}{R(R^2 - 1)} = 0$$

Therefore, the integral over the semi-circle $C_R$ vanishes as $R$ approaches infinity. This step is critical because it confirms that the contour integral is indeed equal to the integral along the real axis, which is the one we're trying to compute. This meticulous application of the Estimation Lemma ensures our result's validity.

Final Calculation and Result

Now that we know the integral over $C_R$ vanishes, we can write:

$$\oint_C f(z) \, dz = \int_{-\infty}^{\infty} \frac{1 - e^{ix}}{x^2(x^2 + 1)} \, dx = \pi \left[1 + \frac{1}{e}\right]$$

We're interested in the integral of the real part of $f(x)$, which is:

$$\int_{-\infty}^{\infty} \frac{1 - \cos x}{x^2(x^2 + 1)} \, dx$$

Since the integrand is an even function, we have:

$$\int_{0}^{\infty} \frac{1 - \cos x}{x^2(x^2 + 1)} \, dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1 - \cos x}{x^2(x^2 + 1)} \, dx$$

Taking the real part of our contour integral result:

$$\frac{1}{2} \int_{-\infty}^{\infty} \frac{1 - \cos x}{x^2(x^2 + 1)} \, dx = \frac{1}{2} \Re \left[ \pi \left(1 + \frac{1}{e}\right) \right] = \frac{\pi}{2} \left(1 + \frac{1}{e}\right)$$

So, finally, we get:

$$\int_{0}^{\infty} \frac{1 - \cos x}{x^2(x^2 + 1)} \, dx = \frac{\pi}{2} \left(1 + \frac{1}{e}\right)$$

Conclusion

Woohoo! We made it! Using the magic of complex integration, we've successfully evaluated the integral $\int_0^\infty\frac{1-\cos x}{x^2(x^2+1)}\,dx$. This journey took us through setting up the complex integral, identifying singularities, calculating residues, applying Cauchy's Residue Theorem, and showing that the integral over the semi-circle vanishes. Each step was crucial, and together, they demonstrate the power and elegance of complex analysis.

Complex analysis might seem intimidating at first, but as you can see, it's a collection of clever techniques that can transform difficult problems into manageable ones. We started with an integral that looked tricky using real calculus methods, but by extending our view to the complex plane, we were able to leverage powerful theorems and arrive at a beautiful solution. Remember, practice makes perfect, so keep exploring these techniques and tackling new problems. You'll be amazed at what you can accomplish with complex analysis!

So, guys, give yourselves a pat on the back for sticking through this lengthy solution. I hope this has helped you understand the process of complex integration a bit better. Until next time, happy integrating!