Solving the Integral of arctan(x)/(x(1 x^2)) Using Advanced Techniques

Integration and differentiation under the integral sign, also known as Feynman's trick, are powerful tools for evaluating integrals, especially when dealing with functions that are not straightforward to integrate directly. In this article, we provide a detailed step-by-step solution to the integral (I int_0^{infty} frac{arctan(x)}{x(1 x^2)} dx). We use a combination of advanced techniques, including parametrization, complex analysis, and the residue theorem to solve this problem.

Introduction to the Integral

Our goal is to evaluate the following integral:

[I int_0^{infty} frac{arctan(x)}{x(1 x^2)} dx]

This is a challenging integral that requires some advanced techniques to solve. We will begin by using a more general form of the integral, which will allow us to use the technique of differentiation under the integral sign.

Step 1: Generalizing the Integral

Let us consider the more general integral:

[I_a int_0^{infty} frac{arctan(ax)}{x(1 x^2)} dx, quad a > 0]

where the parameter (a) is introduced to help us differentiate the integral with respect to this parameter. We will compute (I_1) at the end.

Step 2: Differentiating Under the Integral Sign

Differentiate (I_a) with respect to (a):

[frac{d}{da} I_a int_0^{infty} frac{partial}{partial a} left( frac{arctan(ax)}{x(1 x^2)} right) dx]

Applying the chain rule, we get:

[frac{d}{da} I_a int_0^{infty} frac{x}{1 ax^2} cdot frac{1}{x(1 x^2)} dx int_0^{infty} frac{1}{(1 ax^2)(1 x^2)} dx]

Step 3: Simplifying the Integral

To simplify the integral, we can use the residue theorem or known results from integral tables. The residue theorem is particularly useful for such integrals. Let's use this approach to evaluate:

[I_a int_0^{infty} frac{1}{(1 ax^2)(1 x^2)} dx]

The poles of the integrand are at (x pm i) and (x pm frac{i}{sqrt{a}}). We will consider the upper-half plane for our contour, and the residues at the poles inside this region will contribute to the integral.

Step 4: Computing the Integral Using Residues

The residue at (x frac{i}{sqrt{a}}) is:

[text{Residue} lim_{x to frac{i}{sqrt{a}}} left( x - frac{i}{sqrt{a}} right) frac{1}{(1 ax^2)(1 x^2)} frac{1}{2sqrt{a}(1 aleft(frac{i}{sqrt{a}}right)^2)} frac{1}{2sqrt{a}(2)} frac{1}{4sqrt{a}}]

The integral over the semicircular contour in the upper half-plane is then:

[I_a 2pi i cdot frac{1}{4sqrt{a}} frac{pi}{2sqrt{a}}]

Step 5: Integrating with Respect to a

We integrate the result with respect to (a) to find (I_a):

[I_a int frac{pi}{2sqrt{a}} da frac{pi}{2} ln|a| C]

To determine the constant (C), we evaluate the integral at (a 0):

[I_0 0 frac{pi}{2} ln|0| C implies C 0]

Thus, we have:

[I_a frac{pi}{2} ln|a|]

Step 6: Evaluating the Original Integral

Finally, we substitute (a 1) and (r 1) into the expression for (I_1):

[I_1 frac{pi}{2} ln|1| frac{pi}{2} ln 2]

Therefore, the value of the original integral is:

[I int_0^{infty} frac{arctan(x)}{x(1 x^2)} dx frac{pi}{2} ln 2]