Evaluating the Integral Using Series Expansion and Alternating Series

Integrals are fundamental in calculus, and certain integrals can be challenging to evaluate directly. In this article, we will demonstrate a method for evaluating the integral I int;?1 (ln(1-x)/x) dx using series expansion and properties of the alternating series. This technique leverages the Taylor series for ln(1-x) and properties of the Riemann zeta function.

Step 1: Express ln(1-x) in Taylor Series

First, we express the natural logarithm function, ln(1-x), in its Taylor series representation. The Taylor series for ln(1-x) around x 0 is given by:

ln(1-x) sum;n1∞ (-1)(n-1) xn/n for |x| 1.

Substituting this series into the integral, we get:

I int;0?1 (1/x) sum;n1∞ (-1)(n-1) xn/n dx

Step 2: Interchange Sum and Integral

Since the series converges uniformly on [0, 1], we can interchange the sum and integral. This yields:

I sum;n1∞ (-1)(n-1) int;0?1 xn-1 dx.

The integral of xn-1 from 0 to 1 is:

int;0?1 xn-1 dx 1/n.

Thus, the integral expression simplifies to:

I sum;n1∞ (-1)(n-1) / n2.

Step 3: Recognize the Series as the Alternating Riemann Zeta Function

The series sum;n1∞ (-1)(n-1) / n2 is recognized as the alternating Basel sum, which is a special case of the Riemann zeta function at s 2:

sum;n1∞ (-1)(n-1) / n2 ζ(2) sin π/2 π2/12.

Therefore, the value of the original integral is:

I π2/12.

Final Result

The final result of the integral evaluation is:

int;0?1 (ln(1-x)/x) dx π2/12.

Proof of the Alternating Basel Sum

For completeness, let's derive the result of the series sum;n1∞ (-1)(n-1) / n2. We can write:

sum;n1∞ (-1)(n-1) / n2 sum;n1∞ 1/(2n-1)2 - sum;n1∞ 1/(2n)2

The second series is:

sum;n1∞ 1/(2n)2 1/4 sum;n1∞ 1/n2 1/4 ζ(2) π2/24.

The first series is the sum of the reciprocals of the squares of the odd numbers, which is known to be:

sum;n1∞ 1/(2n-1)2 1/2 sum;n1∞ 1/n2 1/2 ζ(2) π2/8.

Thus, the series simplifies to:

sum;n1∞ (-1)(n-1) / n2 π2/8 - π2/24 π2/12.

In conclusion, the value of the integral I is: