Evaluating Complex Integrals Using Contour Integration and the Beta Function

In this article, we explore the evaluation of a complex integral with a detailed step-by-step approach, leveraging concepts such as contour integration and the Beta function. The integral in question is given by:

( I int_0^{infty} frac{x^2 - 1}{x^5 - 1} , dx )

1. Simplifying the Integral

To simplify the integral, we start by rewriting the denominator using the finite geometric series. Let's express the denominator in a more convenient form:

( I int_0^{infty} frac{x^2 - 1}{x^5 - 1} , dx int_0^{infty} frac{x^3 - x^2 - x 1}{x^5 - 1} , dx )

1.1 Interval Inversion Substitution

Using the interval inversion substitution, i.e., ( x to x^{-1} ), we can further simplify the integral:

( int_0^{infty} frac{x^n}{x^5 - 1} , dx - int_0^{infty} frac{x^{3-n}}{x^5 - 1} , dx )

Applying this substitution, we obtain:

( I 2 int_0^{infty} frac{x - 1}{x^5 - 1} , dx )

2. Contour Integration

To evaluate the remaining integral, we use contour integration. Consider the contour integral:

( oint_C frac{z - 1}{z^5 - 1} , dz )

where ( C ) is a wedge-shaped contour consisting of the real axis from 0 to R, the circular arc ( Gamma ) with radius R, and the line segment from ( Re^{pi i /5} ) back to the origin.

2.1 Cauchy-Goursat Theorem Application

The contour integral can be split as follows:

( oint_C frac{z - 1}{z^5 - 1} , dz int_0^R frac{x - 1}{x^5 - 1} , dx - int_{Gamma} frac{z - 1}{z^5 - 1} , dz - int_0^R frac{xe^{pi i /5} - 1}{xe^{pi i /5}^5 - 1} cdot e^{pi i /5} , dx )

By the ML-inequality, the integral over ( Gamma ) tends to zero as ( R to infty ). Applying the Cauchy-Goursat Theorem, we find:

( 0 int_0^{infty} frac{x - 1}{x^5 - 1} , dx - e^{2pi i /5} int_0^{infty} frac{x}{x^5 - 1} , dx - e^{pi i /5} int_0^{infty} frac{1}{x^5 - 1} , dx )

3. Evaluating the Integral Using the Beta Function

Next, we evaluate each of the remaining integrals using a contour integral approach. For ( k 0, 1 ), we use the substitution ( t x^5 ):

( int_0^{infty} frac{x^k}{x^5 - 1} , dx frac{1}{5} int_0^{infty} frac{t^{k/5 - 1}}{t - 1} , dt frac{1}{5} Bleft(frac{k 1}{5}, 1 - frac{k 1}{5}right) frac{pi}{5 sinleft(frac{pi (k 1)}{5}right)} )

After substituting ( k 1 ) and ( k 0 ), we get:

( int_0^{infty} frac{x - 1}{x^5 - 1} , dx -e^{2pi i /5} frac{pi}{5 sin(2pi/5)} - e^{pi i /5} frac{pi}{5 sin(pi/5)} )

Simplifying the above expression, we find:

( int_0^{infty} frac{x - 1}{x^5 - 1} , dx -frac{pi}{5} cscleft(frac{2pi}{5}right) )

4. Final Integration

Returning to the original integral:

( I 2 int_0^{infty} frac{x - 1}{x^5 - 1} , dx frac{2pi}{5} cscleft(frac{2pi}{5}right) )

To simplify further, we note:

( cscleft(frac{2pi}{5}right) frac{1}{sinleft(frac{2pi}{5}right)} left(frac{1}{2 sqrt{2}} sqrt{5 sqrt{5}}right)^{-1} sqrt{2 - frac{2}{sqrt{5}}} )

Thus, the final result is:

( boxed{int_0^{infty} frac{1 - x^2}{1 - x x^2 - x^3 x^4} , dx frac{2pi}{5} sqrt{2 - frac{2}{sqrt{5}}} approx 1.32131} )