Solved: The Tricky Integral ∫[-1,1] (1/x)√(1+x)/(1-x)ln(...)

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Hey guys! Today, we're diving headfirst into a fascinating integral that looks intimidating at first glance, but trust me, it's a beautiful mathematical puzzle waiting to be solved. The integral in question is:

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$

This integral combines several interesting elements: a square root, a logarithmic function, and a rational function, all within the bounds of -1 to 1. Let's break it down step by step and explore the techniques we can use to crack this mathematical nut.

Understanding the Challenge

Before we jump into solving, let's appreciate the complexity of this integral. The presence of the square root term, $\sqrt{\frac{1+x}{1-x}}$, immediately suggests that we might encounter some singularities or points where the function behaves wildly, especially near the boundaries of integration, x = -1 and x = 1. The logarithmic term, $\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)$, adds another layer of complexity. Logarithmic functions can be tricky, especially when their arguments approach 1 or 0. Furthermore, the rational function $\frac{1}{x}$ introduces a singularity at x = 0, which lies within our integration interval. This means we'll need to be extra careful when dealing with this point.

The integrand's behavior is crucial to understanding how to approach the integral. The graph provided shows a somewhat symmetrical shape, which hints that there might be some clever substitutions or symmetry arguments we can exploit. We also notice the sharp changes near x = -1, 0, and 1, reinforcing the need for careful handling of these points.

In this comprehensive exploration, we will dissect the integral, employing a combination of analytical techniques, substitutions, and possibly even a touch of complex analysis to arrive at a solution. Our goal is not just to find the answer, but also to understand the underlying principles and strategies that make this solution possible. So, buckle up, and let's embark on this mathematical journey together!

Strategic Approaches to Integral Evaluation

To effectively evaluate the definite integral, $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx$, we need to carefully consider several strategic approaches. Given the complexity of the integrand, a direct analytical solution might be challenging. Therefore, we should explore various techniques that simplify the integral or transform it into a more manageable form. Here are several key strategies we'll consider:

1. Substitution Techniques: Substitution is a powerful tool for simplifying integrals. We should look for suitable substitutions that eliminate the square root or simplify the logarithmic term. For instance, a trigonometric substitution might help with the square root, while a substitution related to the argument of the logarithm could linearize the expression.
2. Symmetry Arguments: Examining the symmetry of the integrand and the interval of integration can significantly simplify the problem. If the integrand is even or odd, we can reduce the integration interval or even show that the integral is zero. The graph suggests potential symmetry, so we'll need to investigate this further.
3. Integration by Parts: Integration by parts is useful when the integrand is a product of two functions. We might be able to apply it by choosing appropriate 'u' and 'dv' terms to simplify the integral. For example, if we let $\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)$ be one of the terms, we might be able to simplify the integral.
4. Series Expansion: If we can express part of the integrand as a convergent series, we can sometimes integrate term by term. This approach could be useful for the logarithmic term, as the natural logarithm has a well-known series expansion.
5. Complex Analysis and Contour Integration: For certain integrals, especially those with singularities, complex analysis techniques such as contour integration can be extremely effective. This involves integrating a complex function along a closed path in the complex plane and using the residue theorem to evaluate the integral.

Each of these integration methods offers a unique pathway to tackle the integral. The choice of method often depends on the specific characteristics of the integrand and the integration interval. In our case, the presence of the square root, logarithm, and singularities suggests that a combination of these techniques might be necessary. We'll start by exploring substitutions and symmetry arguments to see if we can simplify the integral, and then consider other techniques as needed. Let's dive in and see what we can uncover!

Leveraging Substitution for Simplification

One of the most promising approaches for tackling this integral is to employ a clever substitution. The square root term, $\sqrt{\frac{1+x}{1-x}}$, is a clear indicator that a trigonometric substitution might be beneficial. Let's consider the substitution:

$$x = \cos(2\theta)$$

Why this substitution? Well, it allows us to use trigonometric identities to simplify the square root. When $x = \cos(2\theta)$, we have:

$$\sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{1+\cos(2\theta)}{1-\cos(2\theta)}}$$

Using the trigonometric identities:

$$1 + \cos(2\theta) = 2\cos^2(\theta)$$

$$1 - \cos(2\theta) = 2\sin^2(\theta)$$

Our square root term simplifies beautifully:

$$\sqrt{\frac{1+\cos(2\theta)}{1-\cos(2\theta)}} = \sqrt{\frac{2\cos^2(\theta)}{2\sin^2(\theta)}} = \sqrt{\cot^2(\theta)} = |\cot(\theta)|$$

Now, we need to consider the interval of integration. When $x$ ranges from -1 to 1, $2\theta$ ranges from $\pi$ to 0 (or equivalently, $0$ to $\pi$), so $\theta$ ranges from $\frac{\pi}{2}$ to 0. In this interval, $\cot(\theta)$ is positive, so $|\cot(\theta)| = \cot(\theta)$.

We also need to find $dx$ in terms of $d\theta$. Differentiating $x = \cos(2\theta)$, we get:

$$dx = -2\sin(2\theta) \, d\theta$$

Finally, let's rewrite the logarithmic term in terms of $\theta$. We have:

$$\frac{2x^2 + 2x + 1}{2x^2 - 2x + 1} = \frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}$$

While this expression doesn't simplify immediately, it's now in terms of trigonometric functions, which might make it easier to manipulate later.

By applying this trigonometric substitution, we've managed to significantly simplify the square root term. This is a crucial step forward in tackling the integral. However, we still need to deal with the transformed logarithmic term and the other parts of the integrand. In the next section, we'll explore how to further simplify the integral using the new substitution and potentially other techniques like integration by parts or symmetry arguments.

Reimagining the Integral with Trigonometric Insight

Let's recap where we are. We've made the substitution $x = \cos(2\theta)$, which transformed the integral. We found that $\sqrt{\frac{1+x}{1-x}}$ became $\cot(\theta)$, and $dx$ became $-2\sin(2\theta) \, d\theta$. Our integral now looks like this:

$$I = \int_{\frac{\pi}{2}}^0 \frac{1}{\cos(2\theta)} \cot(\theta) \ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) (-2\sin(2\theta)) \, d\theta$$

We can simplify this a bit further. Recall that $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. Plugging these in, we get:

$$I = \int_{\frac{\pi}{2}}^0 \frac{1}{\cos(2\theta)} \frac{\cos(\theta)}{\sin(\theta)} \ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) (-4\sin(\theta)\cos(\theta)) \, d\theta$$

Notice that the $\sin(\theta)$ and $\cos(\theta)$ terms cancel out nicely:

$$I = \int_{\frac{\pi}{2}}^0 \frac{-4}{\cos(2\theta)} \ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) \, d\theta$$

To make the limits of integration look more conventional, let's flip them and change the sign:

$$I = 4 \int_0^{\frac{\pi}{2}} \frac{1}{\cos(2\theta)} \ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) \, d\theta$$

Now, the integral looks somewhat cleaner, but the logarithmic term is still a bit unwieldy. At this point, we might consider a few different paths:

1. Further Trigonometric Simplification: We could try to simplify the argument of the logarithm using more trigonometric identities. However, this might lead to a more complex expression.
2. Integration by Parts: We could try integration by parts, but it's not immediately clear what a good choice for 'u' and 'dv' would be.
3. Series Expansion: We could attempt to expand the logarithm as a series, but this might also get messy.
4. Symmetry Arguments: Perhaps there's some symmetry hidden within the integral that we can exploit.

Let's explore the possibility of symmetry arguments. This often involves looking for functions that are even or odd with respect to some point. If we can identify such symmetry, we might be able to simplify the integral significantly.

Unveiling Symmetry: A Key to Simplification

In the quest to solve this integral, symmetry can be a powerful ally. Let's examine our current form of the integral:

$$I = 4 \int_0^{\frac{\pi}{2}} \frac{1}{\cos(2\theta)} \ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) \, d\theta$$

To investigate symmetry, a common technique is to make a substitution that reflects the integration interval around its midpoint. In this case, the midpoint of the interval $[0, \frac{\pi}{2}]$ is $\frac{\pi}{4}$. So, let's try the substitution:

$$\phi = \frac{\pi}{2} - \theta$$

This substitution effectively reflects the interval around $\frac{\pi}{4}$. Now, we need to express everything in terms of $\phi$:

• $d\theta = -d\phi$
• When $\theta = 0$, $\phi = \frac{\pi}{2}$
• When $\theta = \frac{\pi}{2}$, $\phi = 0$
• $\cos(2\theta) = \cos(\pi - 2\phi) = -\cos(2\phi)$

Now, let's tackle the logarithmic term. This is where things get interesting:

$$\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1} = \frac{2\cos^2(\pi - 2\phi) + 2\cos(\pi - 2\phi) + 1}{2\cos^2(\pi - 2\phi) - 2\cos(\pi - 2\phi) + 1}$$

Since $\cos(\pi - x) = -\cos(x)$, we have:

$$\frac{2(-\cos(2\phi))^2 + 2(-\cos(2\phi)) + 1}{2(-\cos(2\phi))^2 - 2(-\cos(2\phi)) + 1} = \frac{2\cos^2(2\phi) - 2\cos(2\phi) + 1}{2\cos^2(2\phi) + 2\cos(2\phi) + 1}$$

Notice something remarkable? This is the reciprocal of the original argument of the logarithm! So,

$$\ln\left(\frac{2\cos^2(2\theta) + 2\cos(2\theta) + 1}{2\cos^2(2\theta) - 2\cos(2\theta) + 1}\right) = \ln\left(\left(\frac{2\cos^2(2\phi) - 2\cos(2\phi) + 1}{2\cos^2(2\phi) + 2\cos(2\phi) + 1}\right)\right) = -\ln\left(\frac{2\cos^2(2\phi) + 2\cos(2\phi) + 1}{2\cos^2(2\phi) - 2\cos(2\phi) + 1}\right)$$

Now, let's rewrite the integral in terms of $\phi$:

$$I = 4 \int_{\frac{\pi}{2}}^0 \frac{1}{-\cos(2\phi)} \left(-\ln\left(\frac{2\cos^2(2\phi) + 2\cos(2\phi) + 1}{2\cos^2(2\phi) - 2\cos(2\phi) + 1}\right)\right) (-d\phi)$$

Simplifying, we get:

$$I = -4 \int_0^{\frac{\pi}{2}} \frac{1}{\cos(2\phi)} \ln\left(\frac{2\cos^2(2\phi) + 2\cos(2\phi) + 1}{2\cos^2(2\phi) - 2\cos(2\phi) + 1}\right) \, d\phi$$

But wait! This is just -I! So, we have:

$$I = -I$$

This implies that:

$$2I = 0$$

Therefore:

$$I = 0$$

We've done it! By cleverly exploiting the symmetry of the integral, we've shown that the integral evaluates to 0. This is a beautiful result that highlights the power of symmetry arguments in simplifying complex integrals.

Conclusion: The Elegance of Mathematical Solutions

In conclusion, we've successfully navigated the intricate landscape of the integral:

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx$$

By employing a combination of techniques, including trigonometric substitution and, most importantly, a keen observation of symmetry, we've arrived at the elegant solution:

$$I = 0$$

This journey underscores the importance of strategic problem-solving in mathematics. The initial form of the integral appeared quite daunting, but by breaking it down into smaller parts and applying the right tools, we were able to reveal its underlying structure. The trigonometric substitution was crucial in simplifying the square root term, but it was the exploitation of symmetry that ultimately led us to the final answer.

This problem serves as a reminder that mathematical solutions are not always about brute-force calculation. Often, the most elegant solutions come from a deep understanding of the problem's properties and the creative application of mathematical principles. So, the next time you encounter a challenging integral, remember to look for hidden symmetries, consider clever substitutions, and, most importantly, enjoy the process of discovery! Keep exploring, keep questioning, and keep solving, guys! You've got this!