Evaluate I = ∫ 0 1 X 2 − X ( X + 1 ) Ln ⁡ X D X I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx I = ∫ 0 1 ​ ( X + 1 ) L N X X 2 − X ​ D X

Evaluating definite integrals can often present a formidable challenge, especially when elementary methods fall short. This article delves into the evaluation of a specific integral, exploring a combination of techniques to arrive at a solution. The integral in question is:

$$I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$$

This integral, at first glance, might seem daunting. The integrand's complexity, featuring a rational function multiplied by the reciprocal of the natural logarithm, hints that standard integration techniques like substitution or integration by parts may not readily yield a closed-form solution. The quest for an antiderivative proves elusive, prompting a shift in strategy towards more advanced methods. Before diving into those methods, it's crucial to understand why directly finding an antiderivative is difficult. The presence of $\ln{x}$ in the denominator, combined with the rational function, creates a complex interplay that standard integration rules struggle to address. The integrand's behavior near the limits of integration, particularly as $x$ approaches 0, also requires careful consideration, as the logarithm function approaches negative infinity, potentially leading to singularities.

In such scenarios, a powerful technique known as parameterized integration often comes to the rescue. This method involves introducing a parameter into the integral, effectively transforming a definite integral into a family of integrals. By carefully choosing the parameter, we can manipulate the integral into a form that is more amenable to differentiation under the integral sign, a technique also known as Feynman's trick. This approach leverages the interplay between integration and differentiation, allowing us to solve integrals that would otherwise be intractable. The beauty of parameterized integration lies in its ability to convert a difficult integral into a differential equation, which can then be solved using standard techniques. This method is particularly effective when the integral involves transcendental functions or combinations of functions that do not have elementary antiderivatives. The success of this method hinges on the judicious choice of the parameter and the ability to differentiate under the integral sign, which requires careful consideration of the conditions for its validity. The resulting differential equation often provides a pathway to the solution that would be otherwise obscured. Understanding the nuances of differentiation under the integral sign is crucial for the successful application of this technique.

A Strategic Transformation: Introducing a Parameter

The key to unlocking this integral lies in introducing a parameter. Let's define a function $I(a)$ as follows:

$$I(a)=\int_{0}^{1}\frac{x^a(x-1)}{x+1}dx$$

Notice that our original integral $I$ corresponds to the derivative of $I(a)$ evaluated at $a=2$. Specifically, if we differentiate $I(a)$ with respect to $a$, we get:

$$\frac{dI}{da} = \int_{0}^{1} \frac{\partial}{\partial a} \left(\frac{x^a(x-1)}{x+1}\right) dx = \int_{0}^{1} \frac{x^a \ln{x} (x-1)}{x+1} dx$$

Thus, our target integral $I$ can be expressed as:

$$I = \left. \frac{dI}{da} \right|_{a=1}$$

This strategic move transforms the original problem into finding $I(a)$, differentiating it, and then evaluating the derivative at $a=1$. This approach leverages the power of differential calculus to tackle the integral. The introduction of the parameter 'a' provides a handle to manipulate the integrand, making it more amenable to integration. The crucial step is to recognize that the original integral is related to the derivative of the parameterized integral. This connection allows us to leverage the techniques of differential calculus to solve the integral. The process of differentiating under the integral sign, however, requires careful attention to ensure its validity. The conditions for differentiating under the integral sign, such as the continuity of the integrand and its partial derivative with respect to the parameter, must be verified. The transformation of the integral into a parameterized form is a powerful technique that can be applied to a wide range of integrals. The success of this method often depends on the judicious choice of the parameter and the ability to evaluate the resulting integral. The parameterized integral $I(a)$ now becomes the focal point of our investigation. We will explore methods to evaluate this integral, which will ultimately lead us to the solution of the original problem.

Evaluating the Parameterized Integral $I(a)$

Now, let's focus on evaluating $I(a)$. We can rewrite the integrand using long division or by adding and subtracting terms:

$$\frac{x^a(x-1)}{x+1} = x^a \left(1 - \frac{2}{x+1}\right) = x^a - \frac{2x^a}{x+1}$$

Therefore,

$$I(a) = \int_{0}^{1} x^a dx - 2 \int_{0}^{1} \frac{x^a}{x+1} dx$$

The first integral is straightforward:

$$\int_{0}^{1} x^a dx = \frac{x^{a+1}}{a+1} \Big|_{0}^{1} = \frac{1}{a+1}$$

The second integral requires a bit more finesse. We can express $\frac{1}{x+1}$ as a geometric series:

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

Substituting this into the second integral, we get:

$$\int_{0}^{1} \frac{x^a}{x+1} dx = \int_{0}^{1} x^a \sum_{n=0}^{\infty} (-x)^n dx$$

We can interchange the integral and the summation (justification required, but we'll assume it's valid for now):

$$\int_{0}^{1} x^a \sum_{n=0}^{\infty} (-x)^n dx = \sum_{n=0}^{\infty} (-1)^n \int_{0}^{1} x^{a+n} dx$$

Each integral in the summation is now easy to evaluate:

$$\int_{0}^{1} x^{a+n} dx = \frac{x^{a+n+1}}{a+n+1} \Big|_{0}^{1} = \frac{1}{a+n+1}$$

Thus,

$$\int_{0}^{1} \frac{x^a}{x+1} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{a+n+1}$$

Putting it all together, we have:

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

This expression for $I(a)$ is a crucial step forward. It represents the parameterized integral in terms of an infinite series. While this series representation might seem complex, it provides a pathway to differentiate $I(a)$ with respect to $a$, which is the key to finding our original integral. The manipulation of the integrand, by expressing it as a geometric series, is a common technique in integral evaluation. This approach transforms the integral into an infinite sum of simpler integrals, which can often be evaluated more easily. The interchange of the integral and summation, however, requires careful justification. In this case, the uniform convergence of the series within the interval of integration ensures the validity of the interchange. The resulting series representation of the integral is a powerful tool, allowing us to analyze the behavior of the integral and to differentiate it with respect to the parameter. The infinite sum representation of $I(a)$ might appear daunting, but it is a critical step in our journey towards solving the original integral. It is a testament to the power of series representation in tackling complex mathematical problems.

Differentiating and Evaluating at $a=1$

Now we need to differentiate $I(a)$ with respect to $a$:

$$\frac{dI}{da} = \frac{d}{da} \left(\frac{1}{a+1} - 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{a+n+1}\right)$$

Differentiating term by term, we get:

$$\frac{dI}{da} = -\frac{1}{(a+1)^2} + 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(a+n+1)^2}$$

Finally, we evaluate this derivative at $a=1$:

$$\left. \frac{dI}{da} \right|_{a=1} = -\frac{1}{(1+1)^2} + 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(1+n+1)^2} = -\frac{1}{4} + 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(n+2)^2}$$

Let's rewrite the summation by shifting the index:

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

Recall the well-known series for $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$:

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}$$

So,

$$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} - \frac{(-1)^1}{1^2} = -\frac{\pi^2}{12} + 1$$

Substituting this back into our expression for the derivative:

$$\left. \frac{dI}{da} \right|_{a=1} = -\frac{1}{4} + 2 \left(-\frac{\pi^2}{12} + 1\right) = -\frac{1}{4} - \frac{\pi^2}{6} + 2 = \frac{7}{4} - \frac{\pi^2}{6}$$

Therefore, the final result is:

$$I = \frac{7}{4} - \frac{\pi^2}{6}$$

This elegant solution demonstrates the power of parameterized integration. By introducing a parameter, we transformed a seemingly intractable integral into a manageable problem. The steps involved differentiating under the integral sign, expressing the integrand as a series, and leveraging known series representations. Each step required careful consideration and justification, highlighting the importance of rigor in mathematical analysis. The final result, a combination of rational and irrational numbers, is a testament to the intricate connections within mathematics. The journey from the initial integral to the final solution showcases the beauty and power of calculus techniques. The successful evaluation of this integral underscores the importance of strategic problem-solving in mathematics. The ability to recognize the applicability of parameterized integration, to manipulate the integrand, and to leverage known series representations are all crucial skills for a mathematician. The final answer, a concise and elegant expression, is a reward for the persistent pursuit of a solution.

In this article, we successfully evaluated the integral $I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$ using the technique of parameterized integration. This method, combined with series manipulation and careful differentiation, allowed us to arrive at the solution $I = \frac{7}{4} - \frac{\pi^2}{6}$. This example showcases the versatility and power of calculus in tackling challenging problems.