Solve: Integral Of (arctan^2(x) * Arctanh(x^2)) / X = G^2

Hey everyone! Today, we're diving deep into a fascinating integral problem that pops up in the realms of real analysis and calculus. It's one of those problems that looks intimidating at first glance, but with a bit of clever maneuvering, we can unravel its secrets. Our mission? To prove that:

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

Where $G$ represents Catalan's constant, a mathematical constant that appears frequently in combinatorics and number theory. So, buckle up, grab your thinking caps, and let's embark on this mathematical adventure together!

Delving into the Depths of Integration

Definite integrals can often appear as formidable challenges, especially when dealing with complex combinations of functions like arctan and arctanh. In this case, the presence of both the arctangent squared and the inverse hyperbolic tangent functions, coupled with the variable in the denominator, creates an intriguing puzzle. To crack this nut, we'll need to employ a blend of techniques, including clever substitutions, integration by parts, and perhaps even a dash of series expansions. Don't worry if these terms sound scary – we'll break them down step-by-step!

Understanding the Players: arctan and arctanh

Before we dive into the nitty-gritty of the integral, let's familiarize ourselves with the key players: $\arctan(x)$ and $\operatorname{arctanh}(x)$.

• arctan(x), also known as the inverse tangent function, gives you the angle whose tangent is $x$. It's a fundamental trigonometric inverse function that pops up in various areas of mathematics and physics.
• arctanh(x), the inverse hyperbolic tangent function, is the inverse of the hyperbolic tangent function, $\tanh(x)$. It's defined as $\operatorname{arctanh}(x) = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $|x| < 1$. This function often appears in problems involving hyperbolic geometry and complex analysis.

The Challenge: A Complex Integral Landscape

Now, let's revisit the integral we're tackling:

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

Notice the following key features:

• The integral is a definite integral, meaning we're evaluating it over a specific interval, from 0 to infinity.
• The integrand (the function inside the integral) is a product of $\arctan^2(x)$, $\operatorname{arctanh}(x^2)$, and $\frac{1}{x}$.
• We're interested in the real part of the integral, denoted by $\operatorname{\mathfrak{R}}$. This might hint at the possibility of complex numbers creeping into our calculations along the way.

Strategies for Taming the Integral Beast

So, how do we even begin to approach such a complex integral? Here are a few strategies that often prove useful in these situations:

1. Substitution: This technique involves replacing a part of the integrand with a new variable to simplify the expression. The goal is to transform the integral into a more manageable form. We should always be on the lookout for substitutions when trying to solve integrals guys.
2. Integration by Parts: This is the integral version of the product rule for differentiation. It's particularly helpful when dealing with products of functions, like the $\arctan^2(x)$ and $\operatorname{arctanh}(x^2)$ we have here. It hinges on the formula $\int u \, dv = uv - \int v \, du$, where we cleverly choose parts of our integrand to be $u$ and $dv$.
3. Series Expansions: Sometimes, we can express functions as infinite sums (series). This can be particularly useful for functions like $\arctan(x)$ and $\operatorname{arctanh}(x)$, which have well-known series representations. Integrating the series term by term can sometimes lead to a solution.
4. Complex Analysis Techniques: Since we're dealing with the real part of the integral, it's possible that techniques from complex analysis, such as contour integration, might be applicable. These methods involve extending the integral into the complex plane and using powerful theorems like Cauchy's integral theorem.

A Potential Path to the Solution: A Step-by-Step Approach

Let's try a combination of these techniques to tackle our integral. A reasonable first step might be to try a substitution to simplify the $\operatorname{arctanh}(x^2)$ term. Let's try substituting $u = x^2$, so $du = 2x \, dx$ and $dx = \frac{du}{2x}$. This gives us:

$$\int _0^{\infty }\frac{\arctan ^2\left(x\right)\operatorname{arctanh} \left(x^2\right)}{x}\,dx = \frac{1}{2}\int _0^{\infty }\frac{\arctan ^2\left(\sqrt{u}\right)\operatorname{arctanh} \left(u\right)}{u}\,du$$

This substitution simplifies the $\operatorname{arctanh}$ argument, but we still have the square root inside the arctangent. Maybe, to make headway, we could then try an integration by parts. To do this, we’ll need to carefully choose our 'u' and 'dv' terms. Remember, the goal is to make the resulting integral simpler than the original. A good starting point might be to let $u = \arctan^2(\sqrt{u})$ and $dv = \frac{\operatorname{arctanh}(u)}{u} \, du$. We'll need to find $du$ and $v$ to proceed. The derivative of $u$ is:

$$du = 2 \arctan(\sqrt{u}) \cdot \frac{1}{1 + u} \cdot \frac{1}{2\sqrt{u}} du = \frac{\arctan(\sqrt{u})}{\sqrt{u}(1 + u)} du$$

Finding $v$ requires integrating $\frac{\operatorname{arctanh}(u)}{u}$. This integral itself might not have a simple closed form, but we can express $\operatorname{arctanh}(u)$ as a series:

$$\operatorname{arctanh}(u) = \sum_{n=0}^{\infty} \frac{u^{2n+1}}{2n+1}$$

Dividing by $u$ gives:

$$\frac{\operatorname{arctanh}(u)}{u} = \sum_{n=0}^{\infty} \frac{u^{2n}}{2n+1}$$

Integrating this term by term, we find:

$$v = \int \frac{\operatorname{arctanh}(u)}{u} du = \int \sum_{n=0}^{\infty} \frac{u^{2n}}{2n+1} du = \sum_{n=0}^{\infty} \frac{u^{2n+1}}{(2n+1)^2}$$

This infinite sum representation for $v$ involves the dilogarithm function, which we'll encounter later. So, applying integration by parts, we get:

$$\frac{1}{2}\int _0^{\infty }\frac{\arctan ^2\left(\sqrt{u}\right)\operatorname{arctanh} \left(u\right)}{u}\,du = \frac{1}{2}\left[ \arctan^2(\sqrt{u}) \sum_{n=0}^{\infty} \frac{u^{2n+1}}{(2n+1)^2} \right]_0^{\infty} - \frac{1}{2} \int_0^{\infty} \sum_{n=0}^{\infty} \frac{u^{2n+1}}{(2n+1)^2} \cdot \frac{\arctan(\sqrt{u})}{\sqrt{u}(1 + u)} du$$

The first term, when evaluated at the limits, requires careful analysis. As $u$ approaches infinity, $\arctan^2(\sqrt{u})$ approaches $(\pi/2)^2$, and the series term also goes to infinity. However, we need to examine the growth rates more closely to determine if this term vanishes or contributes to the integral. The evaluation at 0 is straightforward, giving us 0. This part requires careful treatment, potentially using L'Hôpital's rule after expressing the series in closed form using dilogarithm functions.

The second term looks even more complicated, but it gives us a glimmer of hope. The product of the series and the arctangent term might be simplified further using additional substitutions or series manipulations. Perhaps introducing a new variable $y = \sqrt{u}$ can help us unpack the arctangent term.

The Catalan Constant Connection

At this point, you might be wondering,