Evaluating the Definite Integral \(\displaystyle\int_0^1 \ln \left(\frac{1}{x}−1\right)\) dx

The question asks us to find the value of the definite integral \(\displaystyle\int_0^1 \ln \left(\frac{1}{x}−1\right)\) dx. This integral is taken over the interval from 0 to 1. The function inside the integral is \(\ln \left(\frac{1}{x}−1\right)\).

First, let's simplify the expression inside the logarithm: \(\frac{1}{x} − 1 = \frac{1}{x} − \frac{x}{x} = \frac{1-x}{x}\) So the integral becomes: \(\int_0^1 \ln \left(\frac{1-x}{x}\right) dx\)

Using the property of logarithms \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\), we can rewrite the integrand: \(\ln \left(\frac{1-x}{x}\right) = \ln(1-x) - \ln(x)\) The integral is now: \(\int_0^1 (\ln(1-x) - \ln(x)) dx = \int_0^1 \ln(1-x) dx - \int_0^1 \ln(x) dx\)

This is an improper integral because \(\ln(x)\) is undefined at \(x=0\) and \(\ln(1-x)\) is undefined at \(x=1\) (specifically, it approaches \(-\infty\)). We need to evaluate each part separately using limits.

Step 1: Evaluate \(\int_0^1 \ln(x) dx\)

The antiderivative of \(\ln(x)\) is \(x \ln x - x\). We evaluate this from 0 to 1 using limits: \(\int_0^1 \ln(x) dx = \lim_{a \to 0^+} [x \ln x - x]_a^1\) \(= (1 \ln 1 - 1) - \lim_{a \to 0^+} (a \ln a - a)\) \(= (0 - 1) - \lim_{a \to 0^+} (a \ln a - a)\) For the limit \(\lim_{a \to 0^+} a \ln a\), we can rewrite it as \(\lim_{a \to 0^+} \frac{\ln a}{1/a}\) and use L'Hopital's rule: \(\lim_{a \to 0^+} \frac{\frac{1}{a}}{-\frac{1}{a^2}} = \lim_{a \to 0^+} (-a) = 0\) So, \(\lim_{a \to 0^+} (a \ln a - a) = 0 - 0 = 0\). Therefore, \(\int_0^1 \ln(x) dx = -1 - 0 = -1\)

Step 2: Evaluate \(\int_0^1 \ln(1-x) dx\)

Let's use a substitution. Let \(u = 1-x\). Then \(du = -dx\). When \(x=0\), \(u = 1-0 = 1\). When \(x=1\), \(u = 1-1 = 0\). The integral becomes: \(\int_{x=0}^{x=1} \ln(1-x) dx = \int_{u=1}^{u=0} \ln(u) (-du)\) \(= - \int_1^0 \ln(u) du = \int_0^1 \ln(u) du\) This is the same form as the integral evaluated in Step 1. The variable name doesn't change the value of the definite integral. Therefore, \(\int_0^1 \ln(1-x) dx = -1\)

Step 3: Combine the results

The original integral is the difference between the two evaluated integrals: \(\int_0^1 \ln \left(\frac{1}{x}−1\right) dx = \int_0^1 \ln(1-x) dx - \int_0^1 \ln(x) dx\) $$ = (-1) - (-1) $$ $$ = -1 + 1 = 0 $$

The value of the integral is 0.

Revision Table: Integral Evaluation Summary

Additional Information: Improper Integrals and Logarithms

An integral \(\int_a^b f(x) dx\) is called an improper integral if the function \(f(x)\) is not continuous on the interval \([a, b]\) or if one or both limits of integration are infinite. In this problem, the integrand \(\ln\left(\frac{1-x}{x}\right)\) is not defined at \(x=0\) and \(x=1\).

• At \(x \to 0^+\), \(\frac{1-x}{x} \to \frac{1}{0^+} \to +\infty\), so \(\ln\left(\frac{1-x}{x}\right) \to +\infty\).
• At \(x \to 1^-\), \(\frac{1-x}{x} \to \frac{0^+}{1} \to 0^+\), so \(\ln\left(\frac{1-x}{x}\right) \to -\infty\).

Because of these discontinuities where the function approaches infinity or negative infinity, the integral must be evaluated using limits as demonstrated in the step-by-step solution.

The standard integral \(\int \ln(x) dx = x \ln x - x + C\) is a useful result in calculus. When evaluating definite integrals of \(\ln(x)\) over intervals including 0, the limit \(\lim_{x \to 0^+} x \ln x = 0\) is crucial.