Evaluating the Integral ∫01 ln(1 - x5) dx Using Power Series and Digamma Function

In this article, we will explore the evaluation of the indefinite integral (int_{0}^{1} ln(1 - x^5) , dx). We will employ two methods: power series expansion and the use of the digamma function. Both approaches offer valuable insights into solving such integrals.

Using a Power Series Expansion

The natural logarithm (ln(1 - x^5)) can be expanded using the Taylor series:

[ ln(1 - x^5) -sum_{n1}^{infty} frac{x^{5n}}{n} quad text{for } |x| This expansion allows us to rewrite the integral as:

[ int_{0}^{1} ln(1 - x^5) , dx -int_{0}^{1} sum_{n1}^{infty} frac{x^{5n}}{n} , dx. ]

By interchanging the sum and the integral (justified by uniform convergence), we obtain:

[ int_{0}^{1} ln(1 - x^5) , dx -sum_{n1}^{infty} frac{1}{n} int_{0}^{1} x^{5n} , dx. ]

The integral (int_{0}^{1} x^{5n} , dx) can be computed as:

[ int_{0}^{1} x^{5n} , dx frac{1}{5n 1}. ]

Substituting this back into the series, we get:

[ int_{0}^{1} ln(1 - x^5) , dx -sum_{n1}^{infty} frac{1}{n(5n 1)}. ]

This series representation, although theoretically exact, converges very slowly and is not practical for numerical evaluation.

Numerical Evaluation

For practical computation, numerical methods like Gaussian quadrature or adaptive quadrature are often employed. In Python, for instance, we can use the following code snippet:

```pythonfrom  import quadimport numpy as npresult, error  quad(lambda x: np.log(1 - x**5), 0, 1)print(result)```

Running this code will yield a numerical approximation of the integral.

Approximate Result

The numerical evaluation of the integral gives:

[ int_{0}^{1} ln(1 - x^5) , dx approx -0.323366. ]

Using the Digamma Function

Another method involves representing the integral in terms of the digamma function (psi). We start by considering the Maclaurin series for the logarithm:

[ int_{0}^{1} -sum_{k1}^{infty} frac{x^{5k}}{k} , dx -sum_{k1}^{infty} frac{1}{k} int_{0}^{1} x^{5k} , dx -sum_{k1}^{infty} frac{1}{k(5k 1)}. ]

Using partial fractions, we can write:

[ int_{0}^{1} ln(1 - x^5) , dx -sum_{k1}^{infty} left( frac{1}{k} - frac{5}{5k 1} right) -sum_{k1}^{infty} left( frac{1}{k} - frac{1}{k frac{1}{5}} right). ]

To simplify this, we use the digamma recurrence relation:

[ psi(x 1) frac{1}{x} psi(x). ]

Substituting (x frac{1}{5}), we get:

[ -sum_{k1}^{infty} left( psileft(1 - frac{1}{5}right) - psileft(frac{1}{5}right) right). ]

Using Gauss' Digamma Theorem, we can derive the closed form:

[ I -5 ln(10) - frac{pi}{2} cotleft(frac{pi}{5}right) - 2 cosleft(frac{2pi}{5}right) lnleft(sinleft(frac{pi}{5}right)right) - 2 cosleft(frac{4pi}{5}right) lnleft(sinleft(frac{2pi}{5}right)right). ]

Simplifying the trigonometric functions, we obtain:

[ I approx -0.28818. ]