Solve ∫(arctan5x-arctan3x)/x Dx: A Step-by-Step Guide

Hey guys! Let's dive into a fascinating integral problem that often pops up in calculus discussions:

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

This integral might look intimidating at first glance, especially with those arctangent functions hanging out. But don't worry, we'll break it down step-by-step and explore a clever technique to solve it. Many of you may have tried the arctangent subtraction formula or other direct approaches, only to find yourselves in a maze of complexity. The secret weapon here is a technique often called Feynman's trick, or the method of differentiating under the integral sign. This method allows us to introduce a parameter, differentiate the integral with respect to that parameter, solve a simpler integral, and then integrate back to find our original solution. It's like a mathematical magic trick!

The Feynman's Trick: A Magical Approach

So, how does Feynman's trick actually work? 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 as a difference of two such functions:

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

This is a crucial step. By introducing this parameterized integral, we create a pathway to a simpler problem. Now, the magic begins. We differentiate ${I(a)}$ with respect to ${a}$:

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

Here's where the power of Feynman's trick shines. We can interchange the differentiation and integration (under certain conditions, which hold true in this case). This gives us:

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

Now, we just need to differentiate the integrand with respect to ${a}$. The derivative of ${\arctan(ax)}$ with respect to ${a}$ is ${\frac{x}{1 + (ax)^2}}$. So, our integral becomes:

$$\frac{dI}{da} = \int_0^\infty \frac{1}{1 + (ax)^2} dx$$

This integral looks much friendlier, doesn't it? It's a standard integral that we can easily solve. Let's use a simple substitution: let ${u = ax}$, so ${du = a dx}$, and ${dx = \frac{du}{a}}$. The limits of integration remain the same (0 to ${\infty}$). Thus, the integral transforms to:

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

The integral ${\int_0^\infty \frac{1}{1 + u^2} du}$ is a well-known result: it's simply ${\arctan(u)}$ evaluated from 0 to ${\infty}$, which equals ${\frac{\pi}{2}}$. Therefore:

$$\frac{dI}{da} = \frac{1}{a} \cdot \frac{\pi}{2} = \frac{\pi}{2a}$$

We've successfully differentiated under the integral sign and simplified the problem to a manageable integral. The next step is to integrate back with respect to ${a}$ to find ${I(a)}$.

Reconstructing the Integral: Integrating Back

We've found that the derivative of ${I(a)}$ with respect to ${a}$ is ${\frac{\pi}{2a}}$. To find ${I(a)}$, we need to integrate this expression with respect to ${a}$:

$$I(a) = \int \frac{\pi}{2a} da$$

This is a straightforward integration. The integral of ${\frac{1}{a}}$ is ${\ln|a|}$, so we have:

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

where ${C}$ is the constant of integration. Now, we need to determine the value of this constant. To do this, we can use a boundary condition. Let's consider the case when ${a = 0}$. If we plug ${a = 0}$ into our original parameterized integral, we get:

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

So, ${I(0) = 0}$. However, if we plug ${a = 0}$ into our expression for ${I(a)}$, we get ${\frac{\pi}{2} \ln|0| + C}$, which is undefined due to the logarithm of zero. This might seem like a problem, but it highlights a subtlety of this method. We cannot directly use ${a = 0}$ as a boundary condition because of the singularity. Instead, let’s consider what happens as ${a}$ approaches 0. As ${a}$ approaches 0, ${\arctan(ax)}$ also approaches 0, and thus the integral ${I(a)}$ approaches 0. This suggests that the constant of integration ${C}$ should be 0. Alternatively, we can evaluate the integral at another point, say ${a=1}$. This approach often helps to circumvent the issue of singularity at ${a=0}$.

However, there's a more elegant way to bypass this issue. We can recognize that the constant of integration will cancel out when we compute the difference ${I(5) - I(3)}$, which is what we ultimately want. So, we can proceed without explicitly finding ${C}$.

The Grand Finale: Calculating the Definite Integral

Remember, our goal was to find the value of the original integral:

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

We have found that

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

Now, we can plug in ${a = 5}$ and ${a = 3}$:

$$I(5) = \frac{\pi}{2} \ln(5) + C$$

$$I(3) = \frac{\pi}{2} \ln(3) + C$$

Subtracting these, we get:

$$I(5) - I(3) = \frac{\pi}{2} \ln(5) + C - \left(\frac{\pi}{2} \ln(3) + C\right) = \frac{\pi}{2} \ln(5) - \frac{\pi}{2} \ln(3)$$

The constant of integration ${C}$ beautifully cancels out, just as we anticipated! We can simplify this further using the logarithm property ${\ln(a) - \ln(b) = \ln(\frac{a}{b})}$:

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

Therefore, the value of our integral is:

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

Key Takeaways and Generalizations

Isn't that a fantastic result? We've successfully tackled a challenging integral using Feynman's trick. Here are some key takeaways:

• Feynman's trick (differentiation under the integral sign) is a powerful technique for solving integrals that might seem impossible at first. It involves introducing a parameter, differentiating with respect to that parameter, solving a simpler integral, and then integrating back.
• The choice of the parameter is crucial. In this case, using the argument of the arctangent function as a parameter was the key to simplifying the integral.
• Interchanging differentiation and integration is valid under certain conditions (uniform convergence), which are often met in well-behaved integrals like this one.
• Don't be afraid to introduce auxiliary functions to help solve the problem. The function ${I(a)}$ was instrumental in breaking down the integral.

This method can be generalized to a broader class of integrals. For example, consider the integral:

$$\int_0^\infty \frac{\arctan(ax) - \arctan(bx)}{x} dx$$

Following the same steps, you can show that this integral evaluates to:

$$\frac{\pi}{2} \ln\left(\frac{a}{b}\right)$$

This is a more general result that encompasses our original problem as a special case (when ${a = 5}$ and ${b = 3}$).

Conclusion: Mastering the Art of Integration

Integration can be a challenging but rewarding area of calculus. Problems like this one demonstrate the beauty and power of advanced techniques like Feynman's trick. By mastering these methods, you can tackle a wide range of integrals that might otherwise seem insurmountable. So keep practicing, keep exploring, and never stop learning! This particular example highlights not only the utility of Feynman's technique but also the importance of strategic problem-solving in calculus. We didn't just brute-force our way through the integral; we introduced a clever parameterization that transformed a complex problem into a series of simpler, manageable steps. This is a common theme in advanced mathematics: the key to solving a hard problem often lies in finding the right perspective or transformation. The arctangent function, with its derivative being a simple rational function, is a frequent player in such problems, making the method of differentiating under the integral sign exceptionally effective. Remember, the journey of solving an integral is often as valuable as the solution itself. Each step, from choosing the right technique to performing the calculations, enhances your understanding and builds your problem-solving skills. So, embrace the challenge, and happy integrating!