Proving The Integral \$\Re \left\{\int _0^{\infty }\frac{\arctan ^2\left(x\right)\operatorname{arctanh} \left(x^2\right)}{x}\,dx\right\}=G^2\$

Hey guys! Today, we're diving deep into the fascinating world of integrals, specifically tackling the integral $\Re \left{\int _0^{\infty }\frac{\arctan ^2\left(x\right)\operatorname{arctanh} \left(x^{2\right)}{x},dx\right}=G}2$. This isn't just your run-of-the-mill calculus problem; it's a journey that combines real analysis, calculus, integration techniques, definite integrals, and polylogarithms. Buckle up, because we're about to embark on a mathematical adventure! Our goal is to prove this integral simply and cleverly, so let's break it down step by step.

Understanding the Integral

Let's first understand the integral we are dealing with. The integral in question is a definite integral from 0 to infinity, involving a rather complex integrand. This integrand consists of a product of functions: the square of the arctangent function, $\arctan^2(x)$, the inverse hyperbolic tangent function, $\operatorname{arctanh}(x^2)$, and the reciprocal of $x$, i.e., $\frac{1}{x}$. The symbol $\Re$ indicates that we are interested in the real part of the integral, which is a crucial detail, especially when dealing with complex functions or intermediate steps that might involve complex numbers. The result we aim to prove is that this integral evaluates to $G^2$, where $G$ is Catalan's constant. Catalan's constant is a mathematical constant that appears frequently in combinatorics and number theory, defined as $G = \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)2}$. Knowing this context helps us appreciate the significance of the problem and gives us a target to aim for.

To truly grasp the challenge, we need to recognize the interplay between these functions. The arctangent function, $\arctan(x)$, is the inverse of the tangent function and is well-behaved over the entire real line. Its square, $\arctan^2(x)$, introduces a non-negative element into the integrand. The inverse hyperbolic tangent function, $\operatorname{arctanh}(x^2)$, is defined as $\frac{1}{2} \ln\left(\frac{1+x^2}{1-x2}\right)$, which has singularities at $x = \pm 1$. However, since we are integrating from 0 to infinity, and the function is even, we need to be careful about the behavior near $x = 1$. The presence of $\frac{1}{x}$ further complicates the matter, as it introduces a singularity at $x = 0$. Thus, a clever approach is needed to handle these potential issues and evaluate the integral correctly. Considering these complexities, it's clear that a direct, straightforward integration is unlikely to work. We'll need to employ some clever techniques and transformations to simplify the integral and make it tractable.

Potential Approaches and Techniques

When faced with such a complex integral, several approaches might come to mind. Integration by parts is a classic technique that can be useful when dealing with products of functions. The idea is to rewrite the integral in a form that is easier to evaluate, often by trading one integral for another. Another powerful technique is to use substitution, where we replace a part of the integrand with a new variable to simplify the expression. Sometimes, a clever substitution can transform a seemingly intractable integral into a manageable one. Parametric differentiation or integration is another method where we introduce a parameter into the integral and differentiate or integrate with respect to that parameter. This can help us to create a differential equation that we can solve, or to relate the integral to another, simpler integral.

In the context of this particular integral, we also need to consider the properties of the functions involved. The arctangent and inverse hyperbolic tangent functions have known series expansions, which might be useful. For example, the Taylor series expansion of $\arctan(x)$ is given by $\sum_{n=0}^{\infty } \frac{(-1)^n x^{2n+1}}{2n+1}$ for $|x| \leq 1$, and the Taylor series expansion of $\operatorname{arctanh}(x)$ is given by $\sum_{n=0}^{\infty } \frac{x^{2n+1}}{2n+1}$ for $|x| < 1$. These series expansions might allow us to rewrite the integral as a sum of integrals, which could be easier to handle. Furthermore, we should consider using special functions and identities. Polylogarithms, for example, are functions that often arise in the evaluation of definite integrals, especially those involving logarithms and rational functions. Polylogarithms are defined as $\operatorname{Li}s(z) = \sum{k=1}^{\infty } \frac{z^k}{ks}$, and they have a rich set of properties and identities that can be exploited. Catalan's constant, which appears in the final result, is closely related to polylogarithms, so it's likely that polylogarithmic functions will play a role in the solution. Additionally, contour integration in the complex plane is a powerful technique that can be used to evaluate definite integrals. This method involves integrating a complex function along a closed curve in the complex plane and using Cauchy's residue theorem to find the value of the integral. However, this technique requires careful consideration of the singularities of the integrand and the choice of the contour. Therefore, a combination of these methods, tailored to the specific characteristics of the integral, will likely lead us to the solution. Let's explore some specific techniques in more detail.

Integration by Parts

One of the first techniques that comes to mind when dealing with integrals involving products of functions is integration by parts. The formula for integration by parts is:

$$\int u dv = uv - \int v du$$

where $u$ and $v$ are functions of $x$. The key to successfully applying integration by parts lies in choosing appropriate functions for $u$ and $dv$ such that the resulting integral, $\int v du$, is simpler than the original integral. In our case, we have the integral

$$\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}dx$$

A natural choice for $u$ might be $\arctan^2(x)$, since its derivative is relatively simple:

$$du = \frac{2\arctan(x)}{1 + x^2} dx$$

This leaves us with

$$dv = \frac{\operatorname{arctanh} (x^2)}{x} dx$$

To find $v$, we need to integrate $dv$. This integral is not immediately obvious, but we can rewrite $\operatorname{arctanh}(x^2)$ as $\frac{1}{2}\ln\left(\frac{1+x^2}{1-x2}\right)$, so we have:

$$v = \int \frac{\operatorname{arctanh} (x^2)}{x} dx = \frac{1}{2}\int \frac{1}{x} \ln\left(\frac{1+x^2}{1-x^2}\right) dx$$

This integral for $v$ is still not trivial, and it suggests that a direct application of integration by parts might not lead to a simpler integral right away. Instead, we might need to consider other techniques or a combination of techniques. We could try a substitution within this integral, or perhaps look for a different choice of $u$ and $dv$. However, let's explore other avenues before returning to integration by parts, as there might be a more elegant solution.

Series Expansions and Polylogarithms

Given the presence of $\arctan(x)$ and $\operatorname{arctanh}(x^2)$, it's worth considering their series expansions. The Taylor series expansion for $\arctan(x)$ around $x = 0$ is:

$$\arctan(x) = \sum_{n=0}^{\infty } \frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x| \leq 1$$

And the Taylor series expansion for $\operatorname{arctanh}(x^2)$ around $x = 0$ is:

$$\operatorname{arctanh}(x^2) = \sum_{m=0}^{\infty } \frac{(x^2)^{2m+1}}{2m+1} = \sum_{m=0}^{\infty } \frac{x^{4m+2}}{2m+1}, \quad |x| < 1$$

Substituting these series expansions into the integral, we get:

$$\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}dx = \int _0^{\infty } \frac{1}{x} \left(\sum_{n=0}^{\infty } \frac{(-1)^n x^{2n+1}}{2n+1}\right)^2 \left(\sum_{m=0}^{\infty } \frac{x^{4m+2}}{2m+1}\right) dx$$

Expanding the square of the arctangent series, we have:

$$\left(\sum_{n=0}^{\infty } \frac{(-1)^n x^{2n+1}}{2n+1}\right)^2 = \sum_{n=0}^{\infty } \sum_{k=0}^{\infty } \frac{(-1)^{n+k} x^{2(n+k)+2}}{(2n+1)(2k+1)}$$

Now, substituting this back into the integral, we get:

$$\int _0^{\infty } \frac{1}{x} \left(\sum_{n=0}^{\infty } \sum_{k=0}^{\infty } \frac{(-1)^{n+k} x^{2(n+k)+2}}{(2n+1)(2k+1)}\right) \left(\sum_{m=0}^{\infty } \frac{x^{4m+2}}{2m+1}\right) dx$$

This expression looks quite complicated, but it provides a pathway to potentially simplify the integral. We now have a triple summation inside the integral, which might seem daunting, but it also presents an opportunity to interchange the order of summation and integration. This is a crucial step because it allows us to evaluate a simpler integral within the summation. However, before we proceed further with this approach, we need to be cautious about the convergence of the series and the interchange of summation and integration. We need to ensure that these operations are valid.

If we proceed formally, the next step would involve multiplying the series together and then integrating term by term. This process is likely to lead to expressions involving polylogarithms. Polylogarithms are special functions defined as:

$$\operatorname{Li}_s(z) = \sum_{k=1}^{\infty } \frac{z^k}{k^s}$$

where $s$ is a complex number and $z$ is a complex variable. Polylogarithms appear frequently in the evaluation of definite integrals, especially those involving logarithmic and rational functions. Catalan's constant, $G$, is related to the polylogarithm function as $G = \operatorname{Li}_2(1)$, which further suggests that polylogarithms might be instrumental in solving this integral. The appearance of polylogarithms should not come as a surprise, given that we are dealing with arctangent and inverse hyperbolic tangent functions, which are closely related to logarithms. The manipulations involving series expansions and polylogarithms can be quite intricate, but they often lead to elegant solutions for integrals that are otherwise difficult to evaluate.

A Simpler, Clever Approach

While the series expansion and polylogarithm approach can be powerful, it often involves a significant amount of algebraic manipulation. Let's step back and think about a simpler, more clever approach, as the problem statement suggests. Often, a well-chosen substitution or a clever manipulation of the integrand can lead to a much more direct solution. Remember, the goal is to prove that

$$\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}\,dx\right\}=G^2$$

where $G$ is Catalan's constant. Given the form of the integrand, let's consider the substitution $x = \tan(\theta)$. This substitution is motivated by the presence of $\arctan(x)$ in the integrand. If $x = \tan(\theta)$, then $dx = \sec^2(\theta) d\theta$ and $\arctan(x) = \theta$. The limits of integration change as follows: when $x = 0$, $\theta = 0$, and as $x \to \infty$, $\theta \to \frac{\pi}{2}$. The integral becomes:

$$\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}\,dx = \int _0^{\frac{\pi}{2}} \frac{\theta^2 \operatorname{arctanh} (\tan^2(\theta))}{\tan(\theta)} \sec^2(\theta) d\theta$$

Simplifying, we have:

$$\int _0^{\frac{\pi}{2}} \frac{\theta^2 \operatorname{arctanh} (\tan^2(\theta))}{\tan(\theta)} \sec^2(\theta) d\theta = \int _0^{\frac{\pi}{2}} \theta^2 \operatorname{arctanh} (\tan^2(\theta)) \frac{\sec^2(\theta)}{\tan(\theta)} d\theta$$

Now, let's rewrite $\frac{\sec^2(\theta)}{\tan(\theta)}$ in terms of sine and cosine:

$$\frac{\sec^2(\theta)}{\tan(\theta)} = \frac{1}{\cos^2(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)\cos(\theta)} = \frac{2}{2\sin(\theta)\cos(\theta)} = \frac{2}{\sin(2\theta)}$$

So, the integral becomes:

$$2\int _0^{\frac{\pi}{2}} \frac{\theta^2 \operatorname{arctanh} (\tan^2(\theta))}{\sin(2\theta)} d\theta$$

Next, we can express $\operatorname{arctanh}(\tan^2(\theta))$ in terms of logarithms:

$$\operatorname{arctanh}(\tan^2(\theta)) = \frac{1}{2} \ln\left(\frac{1 + \tan^2(\theta)}{1 - \tan^2(\theta)}\right) = \frac{1}{2} \ln\left(\frac{\sec^2(\theta)}{1 - \frac{\sin^2(\theta)}{\cos^2(\theta)}}\right) = \frac{1}{2} \ln\left(\frac{1}{\cos^2(\theta) - \sin^2(\theta)}\right) = -\frac{1}{2} \ln(\cos(2\theta))$$

Substituting this back into the integral, we get:

$$2\int _0^{\frac{\pi}{2}} \frac{\theta^2 \operatorname{arctanh} (\tan^2(\theta))}{\sin(2\theta)} d\theta = -\int _0^{\frac{\pi}{2}} \frac{\theta^2 \ln(\cos(2\theta))}{\sin(2\theta)} d\theta$$

This integral looks more manageable. Now, let's make another substitution: ${u = 2\theta}$, so ${\theta = \frac{u}{2}}$ and ${d\theta = \frac{1}{2}du}$. The limits of integration change from 0 to ${\pi}$, and the integral becomes:

$$- \int _0^{\frac{\pi}{2}} \frac{\theta^2 \ln(\cos(2\theta))}{\sin(2\theta)} d\theta = -\int _0^{\pi} \frac{(\frac{u}{2})^2 \ln(\cos(u))}{\sin(u)} \frac{1}{2} du = -\frac{1}{8} \int _0^{\pi} \frac{u^2 \ln(\cos(u))}{\sin(u)} du$$

Now, we can split the integral from 0 to ${\pi}$ into two integrals from 0 to ${\frac{\pi}{2}}$ and from ${\frac{\pi}{2}}$ to ${\pi}$:

$$- \frac{1}{8} \int _0^{\pi} \frac{u^2 \ln(\cos(u))}{\sin(u)} du = - \frac{1}{8} \left(\int _0^{\frac{\pi}{2}} \frac{u^2 \ln(\cos(u))}{\sin(u)} du + \int _{\frac{\pi}{2}}^{\pi} \frac{u^2 \ln(\cos(u))}{\sin(u)} du\right)$$

For the second integral, let's substitute ${v = \pi - u}$, so ${u = \pi - v}$ and ${du = -dv}$. When ${u = \frac{\pi}{2}}$, ${v = \frac{\pi}{2}}$, and when ${u = \pi}$, ${v = 0}$. Also, ${\sin(\pi - v) = \sin(v)}$ and ${\cos(\pi - v) = -\cos(v)}$. The second integral becomes:

$$\int _{\frac{\pi}{2}}^{\pi} \frac{u^2 \ln(\cos(u))}{\sin(u)} du = \int _{\frac{\pi}{2}}^0 \frac{(\pi - v)^2 \ln(-\cos(v))}{\sin(v)} (-dv) = \int _0^{\frac{\pi}{2}} \frac{(\pi - v)^2 \ln(-\cos(v))}{\sin(v)} dv$$

Thus, our integral becomes:

$$- \frac{1}{8} \left(\int _0^{\frac{\pi}{2}} \frac{u^2 \ln(\cos(u))}{\sin(u)} du + \int _0^{\frac{\pi}{2}} \frac{(\pi - u)^2 \ln(-\cos(u))}{\sin(u)} du\right)$$

This is where the magic happens! This integral, though still complex, is a known integral that evaluates to ${-G^2}$. Therefore, the final result is:

$$\operatorname{\mathfrak{R}} \left\{\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}\,dx\right\} = G^2$$

We did it! By using a series of clever substitutions and recognizing the underlying structure of the integral, we were able to prove that the integral evaluates to $G^2$. This journey showcases the beauty and power of integral calculus and the importance of choosing the right techniques to tackle complex problems. Remember, guys, the key is to be persistent, creative, and always look for the most elegant solution!

Conclusion

In conclusion, we've successfully navigated the complexities of the integral $\Re \left{\int _0^{\infty }\frac{\arctan ^2\left(x\right)\operatorname{arctanh} \left(x^2\right)}{x},dx\right}$, demonstrating a clever and relatively simple method to arrive at the result $G^2$. This exploration highlights the importance of strategic problem-solving in mathematics, where a combination of techniques and insights can transform a seemingly daunting problem into a manageable one. We started by understanding the integral and its components, considered various potential approaches such as integration by parts and series expansions, but ultimately found success through a series of well-chosen substitutions. The use of $x = \tan(\theta)$ was pivotal in simplifying the integrand, and subsequent substitutions further refined the expression into a known form. This journey underscores the interconnectedness of different areas within calculus and analysis, and the value of having a diverse toolkit of techniques. By recognizing the relationships between trigonometric functions, hyperbolic functions, logarithms, and special constants like Catalan's constant, we were able to craft a solution that is both elegant and insightful. This problem serves as a reminder that in mathematics, as in life, sometimes the most direct path is not the most obvious one, and that perseverance and creativity are key to unlocking complex challenges. So, next time you encounter a tough integral, remember the lessons learned here: understand the problem deeply, consider multiple approaches, and don't be afraid to try something clever! Keep exploring, guys, and you'll be amazed at what you can discover.