Question

\(\int_{0}^{1} \frac{x-1}{\log x} d x=\ldots . . .\)

Short answer

The integral evaluates to the Euler-Mascheroni constant \( \gamma \).

Step-by-step solution

Recognize the Integral Form

The given integral, \( \int_{0}^{1} \frac{x-1}{\log x} \, dx \), is in a form where direct integration might not be straightforward. This form suggests that the integral might be solvable using a specific known formula or transformation technique.

Variable Substitution for Simplification

To solve this integral, we can use the substitution \( x = e^{-t} \), which means \( dx = -e^{-t} \, dt \). This substitution changes the limits of integration: when \( x = 1 \), \( t = 0 \) and when \( x = 0 \), \( t = \infty \). Substituting these into the integral gives: \[ - \int_{\infty}^{0} \frac{(e^{-t}-1)}{-t} e^{-t} \, dt \] This simplifies to: \[ \int_{0}^{\infty} \frac{1-e^{-t}}{t} \, dt \]

Recognize the Result as a Known Integral

The integral \( \int_{0}^{\infty} \frac{1-e^{-t}}{t} \, dt \) is a well-known integral that results in the Euler-Mascheroni constant \( \gamma \). This calculation is based on advanced calculus concepts related to computing limits and the properties of the exponential function, as well as the gamma and digamma functions.

Key concepts

Variable Substitution

Variable substitution is a powerful technique in integral calculus. It involves changing the variable of integration to make the integral easier to solve. This method simplifies complex integrals by transforming them into a more familiar, manageable form. In our problem, we performed variable substitution by setting \( x = e^{-t} \). This substitution modifies the integral's limits and simplifies the integrand.

• Start by identifying a substitution that simplifies the integrand or the limits.
• Change the differential accordingly, here \( dx = -e^{-t} \, dt \).
• Adjust the limits of integration. Since \( x = e^{-t} \), when \( x = 1 \), \( t = 0 \); and when \( x = 0 \), \( t = \infty \).

This process results in a new integral \( \int_{0}^{\infty} \frac{1-e^{-t}}{t} \, dt \), which is easier to evaluate. Variable substitution often offers insight into the behavior of the integral and is an essential tool for solving integrals that are otherwise intractable.

Limits of Integration

Limits of integration are the values that define the start and end of integration on a given integral. They tell us over what interval we are integrating the function. When we use variable substitution, we also need to adjust these limits according to the new variable.

• For the substitution \( x = e^{-t} \), as \( x \) goes from 1 to 0, \( t \) transforms from 0 to \( \infty \).
• Reversing the order of limits, the negative sign is eliminated transforming \( -\int_{\infty}^{0} \) into \( \int_{0}^{\infty} \).

Understanding how to adjust the limits during substitution keeps the mathematical integrity of the problem intact. This ensures that the transformed integral accurately reflects the intended calculation over the correct domain.

Euler-Mascheroni Constant

The Euler-Mascheroni constant, denoted as \( \gamma \), is an important constant in number theory and analysis. Its value is approximately 0.57721. It's often involved in calculations involving limits, series, and integrals.In our problem, the integral \( \int_{0}^{\infty} \frac{1-e^{-t}}{t} \, dt \) is known to equal the Euler-Mascheroni constant. This integral emerges from computations in contexts such as harmonic numbers and the digamma function.

• \( \gamma \) often appears in the study of the growth of the harmonic series \( H_n \), as \( n \to \infty \), where \( H_n - \log n \to \gamma \).
• It’s frequently encountered in the context of special functions like the gamma and beta functions.

Recognizing the value of this integral as an important constant helps in understanding its applications and importance in higher mathematics and natural phenomena.