Solving ∫₀^∞ (arctan(5x) - Arctan(3x))/x Dx A Comprehensive Guide

Hey guys! Today, we're diving deep into the fascinating world of calculus to tackle a tricky integral problem. Specifically, we're going to explore how to solve the improper integral:

$$\int_0^\infty \frac{\arctan5x-\arctan3x}{x}dx$$

This integral is a classic example that often pops up in calculus discussions, and it beautifully demonstrates the power of techniques like Feynman's trick (differentiation under the integral sign). So, buckle up, and let's get started!

Understanding the Problem

Before we jump into the solution, let's take a moment to understand what makes this integral interesting. First off, it's an improper integral because the upper limit of integration is infinity. This means we can't just plug in the limits and evaluate in the traditional sense. We need to use a limit to handle the infinite bound. Secondly, the integrand involves a difference of arctangent functions divided by x. This form doesn't immediately lend itself to standard integration techniques, which is where the clever tricks come in handy.

Many of you might have initially thought about using the arctangent subtraction formula, but as you've probably noticed, that path can quickly lead to a dead end. That's where the magic of Feynman's technique comes into play. This method, also known as differentiation under the integral sign, allows us to introduce a parameter into the integral, differentiate with respect to that parameter, solve a simpler integral, and then integrate back to find our original solution. It sounds like a mouthful, but trust me, it’s super cool once you see it in action!

The Feynman's Trick: Differentiation Under the Integral Sign

The core idea behind Feynman's trick is to transform a difficult integral into something more manageable by introducing a parameter. Let's define a function I(a) as follows:

$$I(a) = \int_0^\infty \frac{\arctan(ax)}{x} dx$$

Notice that our original integral can be expressed in terms of I(a). If we can find a way to evaluate I(a), we can then use it to solve our problem. Specifically, we want to find:

$$\int_0^\infty \frac{\arctan5x-\arctan3x}{x}dx = I(5) - I(3)$$

Now, here's the brilliant part: we differentiate I(a) with respect to a. This is where the "differentiation under the integral sign" comes in. We're essentially swapping the order of differentiation and integration. The derivative of I(a) with respect to a is:

$$\frac{dI}{da} = \frac{d}{da} \int_0^\infty \frac{\arctan(ax)}{x} dx = \int_0^\infty \frac{\partial}{\partial a} \frac{\arctan(ax)}{x} dx$$

Performing the Differentiation

We need to differentiate the integrand with respect to a. Recall that the derivative of arctan(u) with respect to u is 1/(1 + u²). So, using the chain rule:

$$\frac{\partial}{\partial a} \frac{\arctan(ax)}{x} = \frac{1}{x} \cdot \frac{1}{1 + (ax)^2} \cdot x = \frac{1}{1 + a^2x^2}$$

This simplifies our integral significantly:

$$\frac{dI}{da} = \int_0^\infty \frac{1}{1 + a^2x^2} dx$$

Solving the Simplified Integral

This integral is much more manageable. We can solve it using a simple substitution. Let u = ax, so du = a dx, and dx = du/a. The limits of integration remain the same (0 to ∞). Our integral becomes:

$$\frac{dI}{da} = \int_0^\infty \frac{1}{1 + u^2} \frac{du}{a} = \frac{1}{a} \int_0^\infty \frac{1}{1 + u^2} du$$

The integral of 1/(1 + u²) is a standard result: arctan(u). So we have:

$$\frac{dI}{da} = \frac{1}{a} [\arctan(u)]_0^\infty = \frac{1}{a} (\arctan(\infty) - \arctan(0)) = \frac{1}{a} (\frac{\pi}{2} - 0) = \frac{\pi}{2a}$$

Integrating Back

Now, we have an expression for dI/da. To find I(a), we need to integrate with respect to a:

$$I(a) = \int \frac{\pi}{2a} da = \frac{\pi}{2} \int \frac{1}{a} da = \frac{\pi}{2} \ln|a| + C$$

Here, C is the constant of integration. To determine C, we need a boundary condition. A convenient choice is to consider I(0). When a = 0, we have:

$$I(0) = \int_0^\infty \frac{\arctan(0)}{x} dx = \int_0^\infty 0 dx = 0$$

Plugging a = 0 into our expression for I(a), we get:

$$0 = \frac{\pi}{2} \ln|0| + C$$

This might seem problematic since ln(0) is undefined. However, we should consider the limit as a approaches 0. In this context, the constant C is indeed 0.

Thus, we have:

$$I(a) = \frac{\pi}{2} \ln|a|$$

Finding the Final Solution

Remember that we wanted to find I(5) - I(3). Now that we have a formula for I(a), it's straightforward:

$$I(5) - I(3) = \frac{\pi}{2} \ln(5) - \frac{\pi}{2} \ln(3) = \frac{\pi}{2} (\ln(5) - \ln(3))$$

Using the logarithm property ln(a) - ln(b) = ln(a/ b), we get our final answer:

$$\int_0^\infty \frac{\arctan5x-\arctan3x}{x}dx = \frac{\pi}{2} \ln\frac{5}{3}$$

Alternative Approaches and Considerations

While Feynman's trick is elegant, there are other ways to approach this integral. For instance, we could use contour integration, which involves integrating a complex function along a specific path in the complex plane. This method is powerful but often requires a solid understanding of complex analysis.

Another approach involves using Frullani's integral, which is a general result that can be applied to integrals of this form. Frullani's integral states that if f(x) is a function such that the limits f(0) and f(∞) exist, then:

$$\int_0^\infty \frac{f(ax) - f(bx)}{x} dx = (f(\infty) - f(0)) \ln(\frac{a}{b})$$

In our case, f(x) = arctan(x), a = 5, and b = 3. Since arctan(∞) = π/2 and arctan(0) = 0, we can directly apply Frullani's integral to obtain the same result:

$$\int_0^\infty \frac{\arctan5x-\arctan3x}{x}dx = (\frac{\pi}{2} - 0) \ln(\frac{5}{3}) = \frac{\pi}{2} \ln\frac{5}{3}$$

This confirms our solution using Feynman's trick!

Convergence and Conditions

It's important to note that when dealing with improper integrals, we should always consider convergence. In this case, the integral converges because the integrand behaves nicely as x approaches 0 and ∞. The arctangent functions "tame" the growth of the integrand, ensuring that the integral doesn't diverge.

Conclusion

So there you have it! We've successfully navigated the integral of (arctan(5x) - arctan(3x))/x from 0 to infinity. We primarily used Feynman's trick, which involves differentiating under the integral sign, to transform a challenging integral into a manageable one. We also verified our result using Frullani's integral, showcasing the versatility of different integration techniques. Remember, practice makes perfect, so keep exploring and tackling those integrals, guys! Happy integrating!

Keywords

Improper integral integration, Feynman's trick calculus, differentiation under the integral sign, arctan integral solution, Frullani's integral application, definite integral evaluation, calculus integration techniques, integral convergence analysis, complex analysis contour integration, advanced calculus problems