Prove Integral: $\Re \{\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x}dx\}=G^2$

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Hey guys! Today, we're diving deep into a fascinating integral problem that pops up in the realms of real analysis and calculus. Specifically, we're tackling the definite integral:

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

Where $G$ represents Catalan's constant. Sounds intimidating? Don't worry, we'll break it down step-by-step, making it super clear and even fun! This journey involves some clever techniques and a bit of mathematical wizardry, so buckle up and let's get started!

Understanding the Integral and Its Components

To really nail this integral, we first need to understand the key players involved. Let’s break down each component:

• Arctangent Function (arctan(x)): The arctangent function, also known as the inverse tangent, gives us the angle whose tangent is x. It’s a staple in calculus and pops up in various integration problems. Understanding its properties and behavior is crucial. Remember, it's bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
• Inverse Hyperbolic Tangent Function (arctanh(x²)): The inverse hyperbolic tangent function, denoted as $\operatorname{arctanh}(x^2)$, is the inverse of the hyperbolic tangent function. It's defined as $\operatorname{arctanh}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$ for $|x| < 1$. In our case, we have $\operatorname{arctanh}(x^2)$, which means we are dealing with the inverse hyperbolic tangent of $x^2$.
• The Integrand's Structure: Our integrand is a product of $\arctan^2(x)$ and $\operatorname{arctanh}(x^2)$, all divided by x. This structure suggests that we might need to use techniques like integration by parts, series expansions, or clever substitutions to simplify the integral.
• Limits of Integration: We're integrating from 0 to $\infty$. This infinite upper limit often hints at the need for special considerations, such as improper integrals and limits.

So, with these components in mind, let’s map out a strategy for solving this integral. We'll likely need to combine multiple techniques to get to our desired result of $G^2$, where $G$ is Catalan's constant.

Strategies for Tackling the Integral

Okay, let's strategize! When faced with such a complex integral, there are a few key techniques we can consider. Here's a rundown of potential approaches:

• Integration by Parts: This technique is a classic for integrals involving products of functions. The formula is $\int u dv = uv - \int v du$. The trick here is to choose u and dv wisely. We might consider setting $u = \arctan^2(x)$ and $dv = \frac{\operatorname{arctanh}(x^2)}{x} dx$ or vice versa. However, the derivatives and integrals of these functions can get messy, so we need to be cautious.
• Series Expansion: Both $\arctan(x)$ and $\operatorname{arctanh}(x^2)$ have well-known series expansions. We could express these functions as infinite series and then multiply and integrate term by term. This method can sometimes transform a difficult integral into a manageable series summation. The series expansions are:
  • $\arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$ for $|x| \leq 1$
  • $\operatorname{arctanh}(x^2) = \sum_{n=0}^{\infty} \frac{x^{4n+2}}{2n+1}$ for $|x| < 1$
• Substitution: A clever substitution can sometimes simplify the integral by transforming the integrand into a more manageable form. For example, we might try $u = x^2$ or $u = \arctan(x)$. The key is to look for parts of the integrand that might simplify with a change of variables.
• Complex Analysis Techniques: Since we are dealing with a real part of an integral, it might be beneficial to explore complex analysis techniques, such as contour integration. This method involves integrating a complex function along a path in the complex plane and using Cauchy's integral theorem or residue theorem to evaluate the integral.

Given the structure of our integral, using series expansions seems like a promising approach. It allows us to convert the integral into a sum, which we can then try to evaluate. Let's dive deeper into this method.

Applying Series Expansions

Alright, let's roll up our sleeves and apply the series expansions! As we discussed earlier, we have the series representations for $\arctan(x)$ and $\operatorname{arctanh}(x^2)$:

• $\arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$ for $|x| \leq 1$
• $\operatorname{arctanh}(x^2) = \sum_{n=0}^{\infty} \frac{x^{4n+2}}{2n+1}$ for $|x| < 1$

Since we have $\arctan^2(x)$, we need to square the series expansion of $\arctan(x)$. This can be a bit tricky, but let's proceed carefully.

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

This squaring operation involves multiplying the series by itself and collecting terms. It results in a double summation, which can be written as:

$\arctan^2(x) = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^{n+m} x^{2(n+m+1)}}{(2n+1)(2m+1)}$

Now, we multiply this by the series expansion of $\operatorname{arctanh}(x^2)$ and divide by x:

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

This results in a triple summation:

$\frac{\arctan^2(x) \operatorname{arctanh}(x^2)}{x} = \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^{n+m} x^{2n+2m+4k+3}}{(2n+1)(2m+1)(2k+1)}$

Now, we integrate term by term with respect to x from 0 to $\infty$. This is where things get interesting, and we'll see how this series representation can lead us to Catalan's constant.

Integrating the Series and Catalan's Constant

Okay, let's integrate this beast! We have the triple summation:

$\int_0^{\infty} \frac{\arctan^2(x) \operatorname{arctanh}(x^2)}{x} dx = \int_0^{\infty} \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^{n+m} x^{2n+2m+4k+3}}{(2n+1)(2m+1)(2k+1)} dx$

Integrating term by term, we get:

$\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \sum_{k=0}^{\infty} \frac{(-1)^{n+m}}{(2n+1)(2m+1)(2k+1)} \int_0^{\infty} x^{2n+2m+4k+3} dx$

Here's where we encounter a slight issue: the integral $\int_0^{\infty} x^{2n+2m+4k+3} dx$ diverges! This means we need to be extra careful with how we handle the integration. A common trick is to introduce a convergence factor. Let's consider the integral:

$\int_0^{\infty} x^{2n+2m+4k+3} e^{-\epsilon x} dx$

where $\epsilon$ is a small positive number. This will make the integral converge, and we can take the limit as $\epsilon$ approaches 0 later. Using the substitution $u = \epsilon x$, we get:

$\int_0^{\infty} x^{2n+2m+4k+3} e^{-\epsilon x} dx = \frac{1}{\epsilon^{2n+2m+4k+4}} \int_0^{\infty} u^{2n+2m+4k+3} e^{-u} du$

The integral on the right is a Gamma function:

$\int_0^{\infty} u^{2n+2m+4k+3} e^{-u} du = \[(2n+2m+4k+3) = (2n+2m+4k+3)!$

However, we are still facing divergence issues as $\epsilon$ approaches 0. This indicates that we need a different approach or a more sophisticated technique to evaluate the original integral.

Let’s reconsider our strategy. Series expansions, while powerful, can sometimes lead to complicated expressions. Maybe there’s another way to skin this cat! How about we revisit integration by parts or explore some other substitutions?

Revisiting Integration by Parts

Okay, let's give integration by parts another shot. Sometimes, a fresh perspective can make all the difference! The formula for integration by parts is:

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

The key here is choosing the right u and dv. In our case, we have:

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

Let's try setting:

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

Then we need to find du and v.

First, let's find du:

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

Now, let's find v. This involves integrating $\frac{\operatorname{arctanh}(x^2)}{x}$:

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

To solve this integral, let's use the substitution $t = x^2$, so $dt = 2x dx$ and $dx = \frac{dt}{2x}$:

$v = \int \frac{\operatorname{arctanh}(t)}{x} \frac{dt}{2x} = \frac{1}{2} \int \frac{\operatorname{arctanh}(t)}{t} dt$

Now, we can use the series expansion for $\operatorname{arctanh}(t)$:

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

So, our integral becomes:

$v = \frac{1}{2} \int \frac{1}{t} \sum_{n=0}^{\infty} \frac{t^{2n+1}}{2n+1} dt = \frac{1}{2} \int \sum_{n=0}^{\infty} \frac{t^{2n}}{2n+1} dt$

Integrating term by term, we get:

$v = \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2n+1} \int t^{2n} dt = \frac{1}{2} \sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)^2} + C$

Substituting back $t = x^2$, we have:

$v = \frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2} + C$

This series looks promising! It involves the squares of the terms $(2n+1)$ in the denominator, which is a hallmark of expressions related to Catalan's constant.

Now, let's plug u, dv, du, and v back into the integration by parts formula and see where it leads us.

Applying Integration by Parts Formula

Alright, let's plug everything into the integration by parts formula: $\int u dv = uv - \int v du$. We have:

• $u = \arctan^2(x)$
• $dv = \frac{\operatorname{arctanh}(x^2)}{x} dx$
• $du = \frac{2 \arctan(x)}{1+x^2} dx$
• $v = \frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2}$

So, our integral becomes:

$\int _0^{\infty }\frac{\arctan ^2(x)\operatorname{arctanh} (x^2)}{x} dx = \left[ \arctan^2(x) \cdot \frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2} \right]_0^{\infty} - \int_0^{\infty} \frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2} \cdot \frac{2 \arctan(x)}{1+x^2} dx$

Let's analyze the first term. As $x$ approaches infinity, $\arctan^2(x)$ approaches $(\frac{\pi}{2})^2 = \frac{\pi^2}{4}$. However, the series $\sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2}$ also tends to infinity, so we have an indeterminate form. Evaluating the limit of this term will require careful consideration.

Let's simplify the second integral:

$\int_0^{\infty} \frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{4n+2}}{(2n+1)^2} \cdot \frac{2 \arctan(x)}{1+x^2} dx = \int_0^{\infty} \sum_{n=0}^{\infty} \frac{x^{4n+2} \arctan(x)}{(2n+1)^2 (1+x^2)} dx$

This integral looks quite challenging as well. We have a series inside an integral, and the integrand involves a rational function multiplied by $\arctan(x)$.

At this point, it's clear that while integration by parts has given us some interesting expressions, it hasn't directly led us to the solution. The indeterminate form in the first term and the complexity of the second integral suggest that we might need to explore other avenues.

Perhaps a different substitution or a more specialized technique is required. Or maybe, just maybe, we need a little bit of magic! Let's take a step back and see if we can find a clever trick or a known result that can help us crack this integral.

The Clever Trick: A Differential Identity

Alright, guys, sometimes the best way to solve a tough problem is to pull a rabbit out of a hat – or, in this case, a clever differential identity! This is where the