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Evaluating Complex Integrals Involving Logarithms and Inverse Tangents

April 19, 2025Technology4276
Evaluating Complex Integrals Involving Logarithms and Inverse Tangents

Evaluating Complex Integrals Involving Logarithms and Inverse Tangents

Evaluating integrals that involve complex functions such as logarithms and inverse tangents can be a challenging task. In this article, we will explore how to tackle an integral that combines these functions using advanced techniques such as power series expansion and substitution. We will delve into the details and provide a step-by-step explanation.

Introduction to the Problem

The primary integral we are considering is given by:

$$I int_0^1 frac{ln(x^2) arctan(x)}{(1 - x^2)^2} , dx$$

This integral is complex and involves a combination of logarithmic and inverse tangent functions, making it non-standard. To evaluate this integral, we split it into two simpler integrals:

$$I int_0^1 frac{2ln(x)}{(1 - x^2)^2} , dx int_0^1 frac{arctan(x)}{(1 - x^2)^2} , dx$$

Evaluating the First Integral

Let's first look at the first integral:

$$J int_0^1 frac{2ln(x)}{(1 - x^2)^2} , dx$$

To evaluate this integral, we use the power series expansion for the function $frac{1}{1 - x}$:

$$frac{1}{1 - x} sum_{n0}^{infty} (-1)^n x^n$$

By differentiating this series with respect to $x$, we get:

$$frac{1}{(1 - x)^2} -sum_{n1}^{infty} (-1)^n n x^{n-1}$$

Substituting $x^2$ for $x$, we have:

$$frac{1}{(1 - x^2)^2} sum_{n1}^{infty} (-1)^{n-1} n x^{2n-2}$$

Now, substituting this into the integral for $J$, we obtain:

begin{aligned} J sum_{n1}^{infty} (-1)^{n-1} n int_0^1 x^{2n-2} ln(x) , dx sum_{n1}^{infty} frac{(-1)^{n-1} n}{2n-1^2} frac{1}{2} sum_{n1}^{infty} frac{(-1)^n (2n-1)}{2n-1^2} -frac{pi}{8} - frac{1}{2} G end{aligned}

where $G$ is the Catalan’s constant.

Evaluating the Second Integral

Next, we look at the second integral:

$$K int_0^1 frac{arctan(x)}{(1 - x^2)^2} , dx$$

By substituting $x tan(theta)$, we transform the integral to:

$$K int_0^{frac{pi}{4}} frac{theta}{sec^4(theta)} sec^2(theta) , dtheta int_0^{frac{pi}{4}} theta cos^2(theta) , dtheta$$

Using the half-angle identity $cos^2(theta) frac{1 cos(2theta)}{2}$, the integral becomes:

$$K frac{1}{2} int_0^{frac{pi}{4}} theta left(1 frac{cos(2theta)}{2}right) , dtheta$$

Expanding and integrating term by term, we get:

$$K frac{1}{2} left[ theta cdot frac{theta}{2} cdot frac{sin(2theta)}{2} - frac{cos(2theta)}{4} right]_0^{frac{pi}{4}} frac{pi^2}{64} - frac{1}{8}$$

Combining the Results

Combining the results of the two integrals, we obtain:

$$I -frac{pi}{4} - G frac{pi^2}{64} - frac{1}{8} frac{pi^2}{64} - frac{3pi}{16} - frac{1}{8} - G $$

This gives us the final value of the integral, which is approximately $-1.4758$.

Conclusion

In this article, we have explored a detailed step-by-step process for evaluating an integral involving logarithms and inverse tangents. The techniques used include power series expansion, substitution, and integration by parts. These methods are crucial in solving complex integrals and provide a comprehensive approach to tackling similar problems.