Evaluating Integral X*arctan(x)/(1-x^2) * Log^2((1 + X^2)/2)

In this article, we delve into the intricate process of evaluating a fascinating definite integral. Our focus is on integrals that involve a blend of trigonometric and logarithmic functions. Specifically, we aim to tackle the integral:

$$\int_0^1 \frac{x\arctan(x)}{1-x^2} \log^2 \left( \frac{1 + x^2 }{2} \right) \textrm{d}x$$

Integrals of this nature, especially those combining the arctangent function and logarithms, often present a significant challenge. They require a delicate application of various integration techniques and a deep understanding of special functions. Our journey to solve this integral will involve employing substitution methods, leveraging series representations, and possibly, exploring connections to known mathematical constants or special functions. We will navigate through the complexities, providing a detailed and clear exposition of each step involved.

The integral under consideration belongs to a broader class of integrals that frequently appear in advanced calculus and mathematical analysis. These types of problems are not only academically intriguing but also have applications in diverse fields such as physics and engineering. The arctangent function, which is the inverse tangent, often emerges in problems related to angles and geometric representations. Logarithmic functions, on the other hand, are fundamental in describing growth and decay processes. The interplay between these two functions, combined with rational functions like the one in our integrand, leads to rich mathematical structures and challenging problems.

The presence of the squared logarithm adds another layer of complexity. This term suggests that we might need to consider techniques such as integration by parts or differentiating under the integral sign (also known as Feynman's technique) to reduce the integral to a manageable form. We might also explore the possibility of using series expansions for both the arctangent and logarithmic functions to transform the integral into an infinite sum, which can then be evaluated term by term or by recognizing a known series. Before diving into the detailed solution, let's appreciate the elegance and sophistication of such integrals, which serve as excellent exercises in mastering advanced calculus techniques. The goal is not just to find the final answer, but to understand the journey and the mathematical tools used along the way. This understanding will be invaluable in tackling similar problems in the future.

When approaching integrals involving products of arctan and logarithms, it's crucial to have a strategic toolkit of methods. These methods often hinge on exploiting the unique properties of arctangent and logarithmic functions. One common technique is u-substitution, where a suitable substitution can simplify the integrand, making it more amenable to integration. For instance, in our case, letting $u = x^2$ might be a promising starting point, as it simplifies both the rational function and the logarithmic term. However, the success of this method largely depends on the specific form of the integral and requires careful consideration.

Another powerful approach is to leverage the series representations of arctangent and logarithmic functions. The arctangent function, for example, has a well-known Maclaurin series expansion:

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

Similarly, the logarithm can be expressed as a series, although the specific form depends on the argument of the logarithm. In our integral, we have $\log((1 + x^2)/2)$, which can be manipulated using logarithmic identities and then potentially expanded into a series. By substituting these series into the integral, we can transform the problem into evaluating an infinite sum of integrals, which, if convergent, can be easier to handle. This method requires careful attention to the convergence of the series and the interchange of summation and integration, but it often provides a pathway to a solution.

Integration by parts is another classic technique that can be particularly useful when dealing with products of functions. The method involves choosing parts of the integrand as 'u' and 'dv', and applying the formula:

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

The key to success here is to strategically choose 'u' and 'dv' such that the new integral $\int v \, du$ is simpler than the original one. In our case, we might consider choosing the logarithmic term as 'u' and the rest of the integrand as 'dv', or vice versa, depending on which choice leads to a simplification. This method might need to be applied iteratively, especially when dealing with higher powers of logarithms, like the squared logarithm in our integral. Furthermore, another advanced technique that can be quite effective is differentiating under the integral sign, also known as Feynman's technique. This method involves introducing a parameter into the integral, differentiating with respect to that parameter, evaluating the resulting simpler integral, and then integrating back to obtain the solution to the original problem. This technique is particularly powerful when the integral is difficult to evaluate directly, but its derivative with respect to a parameter is more tractable. The choice of the parameter is crucial and often requires a bit of intuition and experience.

To solve the definite integral involving x*arctan(x)/(1-x^2) * log^2((1 + x^2)/2), we will embark on a step-by-step journey, employing a combination of techniques tailored to the problem's unique structure. First, let's restate the integral for clarity:

$$\int_0^1 \frac{x\arctan(x)}{1-x^2} \log^2 \left( \frac{1 + x^2 }{2} \right) \textrm{d}x$$

The initial step involves a strategic substitution to simplify the integral. A natural choice is to let $u = x^2$. This substitution transforms the integral as follows:

$$x^2 = u \implies 2x \, dx = du \implies dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}}$$

The limits of integration also change: when $x = 0$, $u = 0$, and when $x = 1$, $u = 1$. Substituting these into the integral, we get:

$$\frac{1}{2} \int_0^1 \frac{\arctan(\sqrt{u})}{1-u} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

This substitution has simplified the rational part of the integrand, but the presence of $\arctan(\sqrt{u})$ and the squared logarithm still poses a challenge. To further tackle this, we can consider using the series representation of $\arctan(\sqrt{u})$. Recall the Maclaurin series for $\arctan(x)$:

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

Replacing $x$ with $\sqrt{u}$, we obtain:

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

Substituting this series into our integral, we have:

$$\frac{1}{2} \int_0^1 \frac{1}{1-u} \left( \sum_{n=0}^{\infty} \frac{(-1)^n u^{n+\frac{1}{2}}}{2n+1} \right) \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

Now, we face the challenge of interchanging the summation and integration. This step requires careful justification, as it is valid only under certain conditions, such as uniform convergence. Assuming we can interchange the summation and integration (which we will verify later), we get:

$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_0^1 \frac{u^{n+\frac{1}{2}}}{1-u} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

This transformation has shifted our problem to evaluating a family of integrals of the form:

$$I_n = \int_0^1 \frac{u^{n+\frac{1}{2}}}{1-u} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

Evaluating these integrals $I_n$ requires further techniques. We might consider using integration by parts or exploring other series representations. The term $\frac{1}{1-u}$ can be expanded as a geometric series:

$$\frac{1}{1-u} = \sum_{k=0}^{\infty} u^k, \quad |u| < 1$$

Substituting this into $I_n$, we get:

$$I_n = \int_0^1 u^{n+\frac{1}{2}} \left( \sum_{k=0}^{\infty} u^k \right) \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

Again, we need to interchange the summation and integration, leading to:

$$I_n = \sum_{k=0}^{\infty} \int_0^1 u^{n+k+\frac{1}{2}} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

Now, we are faced with integrals of the form:

$$J_{n,k} = \int_0^1 u^{n+k+\frac{1}{2}} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

These integrals can be tackled using integration by parts. Let's choose $v = \log^2((1 + u)/2)$ and $dw = u^{n+k+\frac{1}{2}} du$. Then, we need to compute $dv$ and $w$. This process will generate a series of terms, and we will need to carefully manage these terms and potentially recognize patterns or known mathematical constants. The steps outlined here provide a roadmap for solving the integral. The actual computation involves meticulous algebraic manipulation and a deep understanding of series and integration techniques.

Continuing our exploration of advanced techniques and considerations for evaluating the integral, we've arrived at a point where we need to tackle the integral $J_{n,k}$:

$$J_{n,k} = \int_0^1 u^{n+k+\frac{1}{2}} \log^2 \left( \frac{1 + u}{2} \right) \textrm{d}u$$

As previously mentioned, integration by parts is a viable strategy here. Let's set:

$$v = \log^2 \left( \frac{1 + u}{2} \right) \quad \text{and} \quad dw = u^{n+k+\frac{1}{2}} du$$

Then, we compute $dv$ and $w$:

$$dv = 2 \log \left( \frac{1 + u}{2} \right) \cdot \frac{1}{\frac{1 + u}{2}} \cdot \frac{1}{2} du = \frac{\log \left( \frac{1 + u}{2} \right)}{1 + u} du$$

$$w = \int u^{n+k+\frac{1}{2}} du = \frac{u^{n+k+\frac{3}{2}}}{n + k + \frac{3}{2}}$$

Applying integration by parts, we have:

$$J_{n,k} = vw - \int w \, dv$$

$$J_{n,k} = \left[ \frac{u^{n+k+\frac{3}{2}}}{n + k + \frac{3}{2}} \log^2 \left( \frac{1 + u}{2} \right) \right]_0^1 - \int_0^1 \frac{u^{n+k+\frac{3}{2}}}{n + k + \frac{3}{2}} \cdot \frac{\log \left( \frac{1 + u}{2} \right)}{1 + u} du$$

Evaluating the first term at the limits of integration:

$$\left[ \frac{u^{n+k+\frac{3}{2}}}{n + k + \frac{3}{2}} \log^2 \left( \frac{1 + u}{2} \right) \right]_0^1 = \frac{1}{n + k + \frac{3}{2}} \log^2(1) - \lim_{u \to 0} \frac{u^{n+k+\frac{3}{2}}}{n + k + \frac{3}{2}} \log^2 \left( \frac{1 + u}{2} \right) = 0$$

The limit goes to zero because the polynomial term dominates the logarithmic term as $u$ approaches 0. So, we are left with:

$$J_{n,k} = - \frac{1}{n + k + \frac{3}{2}} \int_0^1 \frac{u^{n+k+\frac{3}{2}} \log \left( \frac{1 + u}{2} \right)}{1 + u} du$$

Let's denote the new integral as $K_{n,k}$:

$$K_{n,k} = \int_0^1 \frac{u^{n+k+\frac{3}{2}} \log \left( \frac{1 + u}{2} \right)}{1 + u} du$$

To tackle $K_{n,k}$, we can again use the series expansion. We can express $\frac{1}{1 + u}$ as a geometric series:

$$\frac{1}{1 + u} = \sum_{m=0}^{\infty} (-1)^m u^m, \quad |u| < 1$$

Substituting this into $K_{n,k}$, we get:

$$K_{n,k} = \int_0^1 u^{n+k+\frac{3}{2}} \left( \sum_{m=0}^{\infty} (-1)^m u^m \right) \log \left( \frac{1 + u}{2} \right) du$$

Interchanging summation and integration (again requiring careful justification):

$$K_{n,k} = \sum_{m=0}^{\infty} (-1)^m \int_0^1 u^{n+k+m+\frac{3}{2}} \log \left( \frac{1 + u}{2} \right) du$$

Let's denote the new integral as $L_{n,k,m}$:

$$L_{n,k,m} = \int_0^1 u^{n+k+m+\frac{3}{2}} \log \left( \frac{1 + u}{2} \right) du$$

To evaluate $L_{n,k,m}$, we can use integration by parts one more time. Let's set:

$$v = \log \left( \frac{1 + u}{2} \right) \quad \text{and} \quad dw = u^{n+k+m+\frac{3}{2}} du$$

Then,

$$dv = \frac{1}{1 + u} du$$

$$w = \frac{u^{n+k+m+\frac{5}{2}}}{n + k + m + \frac{5}{2}}$$

Applying integration by parts:

$$L_{n,k,m} = \left[ \frac{u^{n+k+m+\frac{5}{2}}}{n + k + m + \frac{5}{2}} \log \left( \frac{1 + u}{2} \right) \right]_0^1 - \int_0^1 \frac{u^{n+k+m+\frac{5}{2}}}{n + k + m + \frac{5}{2}} \cdot \frac{1}{1 + u} du$$

The first term evaluates to:

$$\left[ \frac{u^{n+k+m+\frac{5}{2}}}{n + k + m + \frac{5}{2}} \log \left( \frac{1 + u}{2} \right) \right]_0^1 = \frac{\log(1)}{n + k + m + \frac{5}{2}} - \lim_{u \to 0} \frac{u^{n+k+m+\frac{5}{2}} \log \left( \frac{1 + u}{2} \right)}{n + k + m + \frac{5}{2}} = 0$$

So, we have:

$$L_{n,k,m} = - \frac{1}{n + k + m + \frac{5}{2}} \int_0^1 \frac{u^{n+k+m+\frac{5}{2}}}{1 + u} du$$

Let's denote the new integral as $M_{n,k,m}$:

$$M_{n,k,m} = \int_0^1 \frac{u^{n+k+m+\frac{5}{2}}}{1 + u} du$$

To evaluate $M_{n,k,m}$, we can again use the series expansion for $\frac{1}{1 + u}$:

$$M_{n,k,m} = \int_0^1 u^{n+k+m+\frac{5}{2}} \left( \sum_{p=0}^{\infty} (-1)^p u^p \right) du$$

Interchanging summation and integration:

$$M_{n,k,m} = \sum_{p=0}^{\infty} (-1)^p \int_0^1 u^{n+k+m+p+\frac{5}{2}} du$$

The integral is now straightforward:

$$\int_0^1 u^{n+k+m+p+\frac{5}{2}} du = \frac{1}{n + k + m + p + \frac{7}{2}}$$

Thus, we have:

$$M_{n,k,m} = \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

Now, we can trace back through our substitutions and integrations by parts to express the original integral in terms of this final series. This involves carefully substituting the results back into the previous equations:

$$L_{n,k,m} = - \frac{1}{n + k + m + \frac{5}{2}} M_{n,k,m} = - \frac{1}{n + k + m + \frac{5}{2}} \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

$$K_{n,k} = \sum_{m=0}^{\infty} (-1)^m L_{n,k,m} = - \sum_{m=0}^{\infty} (-1)^m \frac{1}{n + k + m + \frac{5}{2}} \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

$$J_{n,k} = - \frac{1}{n + k + \frac{3}{2}} K_{n,k} = \frac{1}{n + k + \frac{3}{2}} \sum_{m=0}^{\infty} (-1)^m \frac{1}{n + k + m + \frac{5}{2}} \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

$$I_n = \sum_{k=0}^{\infty} J_{n,k} = \sum_{k=0}^{\infty} \frac{1}{n + k + \frac{3}{2}} \sum_{m=0}^{\infty} (-1)^m \frac{1}{n + k + m + \frac{5}{2}} \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

Finally, the original integral is:

$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} I_n = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \sum_{k=0}^{\infty} \frac{1}{n + k + \frac{3}{2}} \sum_{m=0}^{\infty} (-1)^m \frac{1}{n + k + m + \frac{5}{2}} \sum_{p=0}^{\infty} \frac{(-1)^p}{n + k + m + p + \frac{7}{2}}$$

This expression represents the solution to the integral as a quadruple infinite series. Evaluating this series analytically is a formidable task, and it might require advanced techniques from complex analysis or special functions. Alternatively, one could resort to numerical methods to approximate the value of the series and, thus, the integral. This detailed step-by-step solution demonstrates the complexity involved in evaluating such integrals and the array of techniques that might be necessary to reach a solution.

In conclusion, the evaluation of the definite integral involving products of arctan and logarithms:

$$\int_0^1 \frac{x\arctan(x)}{1-x^2} \log^2 \left( \frac{1 + x^2 }{2} \right) \textrm{d}x$$

is a complex task that showcases the interplay of various advanced calculus techniques. We began by employing a substitution to simplify the integral, followed by the application of series representations for the arctangent function and the rational term. This led us to a quadruple infinite series, which, while representing the solution, is challenging to evaluate analytically.

Throughout the process, we encountered the need to interchange summation and integration multiple times, a step that requires careful justification using concepts from real analysis, such as uniform convergence. Integration by parts was another key technique, applied iteratively to reduce the complexity of the integrals. Each step required meticulous algebraic manipulation and a deep understanding of the properties of logarithmic and trigonometric functions.

The final expression for the integral is a testament to the intricate nature of such problems. While we derived a formal solution in terms of an infinite series, obtaining a closed-form expression might be elusive. In such cases, numerical methods can provide accurate approximations, offering a practical way to estimate the value of the integral.

The journey to solve this integral underscores the importance of having a diverse toolkit of mathematical techniques and the ability to apply them strategically. It also highlights the beauty and challenge of mathematical analysis, where seemingly simple integrals can lead to profound and complex expressions. The process of tackling such problems not only enhances our problem-solving skills but also deepens our appreciation for the rich structure of mathematics. The lessons learned here can be applied to a wide range of similar problems, making this exercise a valuable contribution to our mathematical understanding. The integral serves as a reminder that in mathematics, the journey is just as important as the destination, and the techniques we develop along the way are invaluable tools in our mathematical explorations.