Limit Of Integral: Proving Convergence To Pi

Hey guys! Let's dive into a fascinating problem from real analysis. We're going to explore how to prove that the integral $\int_{-1}^1\arccos |\arctan x|^\alpha dx$ converges to $\pi$ as $\alpha$ approaches infinity. This is a cool example that combines trigonometric functions, absolute values, and limits, so buckle up!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what the problem is asking. We have an integral where the integrand is $\arccos(|\arctan x|^\alpha)$, and we're integrating with respect to $x$ from -1 to 1. The parameter $\alpha$ is a real number, and we want to see what happens to the value of this integral as $\alpha$ gets incredibly large. Our goal is to show that the integral's value gets closer and closer to $\pi$. To really break this down, we need to consider the behavior of each part of the integrand as $\alpha$ changes.

• $\arctan x$: This is the inverse tangent function. As $x$ varies from -1 to 1, $\arctan x$ varies from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$.
• $|\arctan x|$: Taking the absolute value means we're only considering the magnitude, so $|\arctan x|$ ranges from 0 to $\frac{\pi}{4}$. This part is crucial because it ensures we're dealing with non-negative values.
• $|\arctan x|^\alpha$: This is where $\alpha$ comes into play. When $\alpha$ is large, raising a number between 0 and 1 to the power of $\alpha$ makes it much smaller. This shrinking effect is key to understanding the limit.
• $\arccos(|\arctan x|^\alpha)$: Finally, we take the arccosine. Remember that $\arccos$ is the inverse cosine function, and it's defined for values between -1 and 1. As $|\arctan x|^\alpha$ gets closer to 0, $\arccos(|\arctan x|^\alpha)$ gets closer to $\arccos(0)$, which is $\frac{\pi}{2}$. However, our main focus is on understanding how this whole expression behaves within the integral as $\alpha$ tends to infinity. We need to see how this convergence to $\pi$ occurs across the interval of integration, not just at a single point. The interplay between the shrinking magnitude of $|\arctan x|^\alpha$ and the behavior of the arccosine function is the heart of the problem. By carefully analyzing these components, we can build a rigorous argument to prove the limit.

Strategy for the Proof

Okay, so how do we actually prove this? Here's the plan:

1. Pointwise Limit: First, we'll look at what happens to the integrand, $\arccos(|\arctan x|^\alpha)$, for a fixed $x$ as $\alpha \to \infty$. We'll find the pointwise limit of the integrand.
2. Dominated Convergence Theorem (DCT): To justify bringing the limit inside the integral, we'll use the Dominated Convergence Theorem. This theorem is a powerful tool in real analysis that allows us to interchange limits and integrals under certain conditions. We'll need to find a function that "dominates" our integrand, meaning it's always greater than or equal to the absolute value of our integrand, and is also integrable.
3. Evaluate the Limit of the Integral: Once we've justified using the DCT, we can take the limit inside the integral and evaluate the resulting integral, which should give us $\pi$.

Step 1: Pointwise Limit

Let's find the pointwise limit of $f(x, \alpha) = \arccos(|\arctan x|^\alpha)$ as $\alpha \to \infty$. We need to consider different cases for $x$:

• Case 1: $x = 0$

  If $x = 0$, then $\arctan x = 0$, so $|\arctan x|^\alpha = 0^\alpha = 0$ for any $\alpha > 0$. Thus, $\arccos(|\arctan 0|^\alpha) = \arccos(0) = \frac{\pi}{2}$. This is a straightforward case where the function value is constant regardless of $\alpha$.

• Case 2: $x \in (-1, 0) \cup (0, 1)$

  For any other $x$ in the interval $(-1, 1)$ but not equal to 0, we have $0 < |\arctan x| < \frac{\pi}{4} < 1$. As $\alpha \to \infty$, $|\arctan x|^\alpha$ will approach 0 because we're raising a number between 0 and 1 to a very large power. Therefore, $\lim_{\alpha \to \infty} |\arctan x|^\alpha = 0$. Consequently, $\lim_{\alpha \to \infty} \arccos(|\arctan x|^\alpha) = \arccos(0) = \frac{\pi}{2}$. This is the crucial part where we see the integrand converging to a constant value for almost all $x$ in the interval. The rate at which $|\arctan x|^\alpha$ approaches 0 depends on both $x$ and $\alpha$, but the limit is always 0 as long as $0 < |\arctan x| < 1$.

So, the pointwise limit is $\frac{\pi}{2}$ for all $x$ in $(-1, 1)$. Mathematically, we can write this as:

$$\lim_{\alpha \to \infty} \arccos(|\arctan x|^\alpha) = \begin{cases} \frac{\pi}{2}, & \text{if } x \in (-1, 1) \\ \end{cases}$$

Step 2: Dominated Convergence Theorem (DCT)

Now, let's use the Dominated Convergence Theorem (DCT). The DCT states that if we have a sequence of functions $f_n(x)$ that converge pointwise to a function $f(x)$, and if there exists an integrable function $g(x)$ such that $|f_n(x)| \leq g(x)$ for all $n$ and $x$, then we can interchange the limit and the integral:

$$\lim_{n \to \infty} \int f_n(x) dx = \int \lim_{n \to \infty} f_n(x) dx$$

In our case, $f(x, \alpha) = \arccos(|\arctan x|^\alpha)$. We need to find a dominating function $g(x)$ such that $|\arccos(|\arctan x|^\alpha)| \leq g(x)$ and $\int_{-1}^1 g(x) dx$ is finite.

Since the range of $\arccos(u)$ is $[0, \pi]$, we know that $0 \leq \arccos(|\arctan x|^\alpha) \leq \pi$. So, we can choose $g(x) = \pi$ as our dominating function. This works because the arccosine function always returns a value between 0 and $\pi$, regardless of the input. Choosing a constant function like $\pi$ simplifies the domination argument significantly.

Now, we need to check that $g(x)$ is integrable over the interval $[-1, 1]$. This is easy since it's just a constant function:

$$\int_{-1}^1 g(x) dx = \int_{-1}^1 \pi dx = \pi [x]_{-1}^1 = \pi(1 - (-1)) = 2\pi$$

Since $2\pi$ is finite, $g(x) = \pi$ is indeed integrable. We've successfully found a dominating function that meets the conditions of the DCT. This is a crucial step because it rigorously justifies our next move of interchanging the limit and the integral.

Step 3: Evaluate the Limit of the Integral

Now that we've satisfied the conditions of the Dominated Convergence Theorem, we can interchange the limit and the integral:

$$\lim_{\alpha \to \infty} \int_{-1}^1 \arccos(|\arctan x|^\alpha) dx = \int_{-1}^1 \lim_{\alpha \to \infty} \arccos(|\arctan x|^\alpha) dx$$

We already found the pointwise limit in Step 1, which is $\frac{\pi}{2}$ for $x \in (-1, 1)$. So, we can substitute this into the integral:

$$\int_{-1}^1 \frac{\pi}{2} dx = \frac{\pi}{2} \int_{-1}^1 dx = \frac{\pi}{2} [x]_{-1}^1 = \frac{\pi}{2} (1 - (-1)) = \frac{\pi}{2} (2) = \pi$$

And there you have it! We've shown that:

$$\lim_{\alpha \to \infty} \int_{-1}^1 \arccos(|\arctan x|^\alpha) dx = \pi$$

Conclusion

We successfully proved that the integral $\int_{-1}^1 \arccos(|\arctan x|^\alpha) dx$ converges to $\pi$ as $\alpha$ approaches infinity. We did this by first finding the pointwise limit of the integrand, then using the Dominated Convergence Theorem to justify interchanging the limit and the integral, and finally evaluating the resulting integral. This problem highlights the power of real analysis tools like the DCT and the importance of understanding the behavior of functions as parameters change. Great job, guys! This was a fun one to tackle together.