Question

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition ln(x)=?x1dtt, using properties of the definite integral and making no further assumptions. Use the identity \(\ln (x)=\int_{1}^{x} \frac{d t}{t}\) to derive the identity \(\ln \left(\frac{1}{x}\right)=-\ln x\).

Short answer

\(\ln \left(\frac{1}{x}\right) = -\ln x\).

Step-by-step solution

Understand the Problem

We want to show that \( \ln \left(\frac{1}{x}\right) = -\ln x \) using the definition of the natural logarithm as a definite integral: \( \ln(x) = \int_{1}^{x} \frac{dt}{t} \). This involves evaluating \( \ln \left(\frac{1}{x}\right) \) using this integral representation.

Express \( \ln \left(\frac{1}{x}\right) \) as an Integral

According to the definition of the natural logarithm, we have:\[\ln \left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{dt}{t}.\] We substitute \( x \) with \( \frac{1}{x} \) in the integral.

Change of Variable in the Integral

To simplify the integral, we'll use a substitution. Let \( u = \frac{1}{t} \), then \( du = -\frac{1}{t^2} dt \), which implies \( dt = -t^2 du \). Rewriting \( t \) as \( \frac{1}{u}\), adjust the limits of the integral accordingly:when \( t = 1 \), \( u = 1 \) and when \( t = \frac{1}{x} \), \( u = x \). So, the integral becomes:\[\int_{1}^{1/x} \frac{dt}{t} = \int_{1}^{x} \frac{-1}{u} du.\]

Evaluate the Integral

The integral \( \int_{1}^{x} \frac{-1}{u} du \) can be solved straightforwardly:\[\int_{1}^{x} -\frac{1}{u} du = \left[-\ln(u)\right]_{1}^{x} = -\ln(x) - (-\ln(1)) = -\ln(x) - 0 = -\ln(x).\] Substituting back, \( \ln \left(\frac{1}{x}\right) = -\ln(x)\), which is the desired identity.

Key concepts

Natural Logarithm

In mathematics, the natural logarithm is the logarithm to the base of the number "e", where "e" is an irrational constant approximately equal to 2.71828. It is denoted as \( \ln(x) \). The natural logarithm has a characteristic property that can be derived from its definition, \( \ln(x) = \int_{1}^{x} \frac{dt}{t} \). This represents the area under the curve \( y = \frac{1}{t} \) from 1 to \( x \). It is an important function in calculus because it simplifies the differentiation and integration of exponential functions. The natural logarithm is a continuous and increasing function, with \( \ln(1) = 0 \). It is primarily used in problems involving growth processes, like populations or compound interest, because of its natural link to the exponential function.

Definite Integral

A definite integral, represented as \( \int_{a}^{b} f(x) \, dx \), calculates the net area under a curve \( f(x) \) from the point \( x = a \) to \( x = b \). Unlike indefinite integrals, which represent a family of functions, definite integrals yield a specific numerical value. The process of finding this area involves evaluating the antiderivative at the upper limit and subtracting the value of the antiderivative at the lower limit. In our context, the natural logarithm is expressed as a definite integral: \( \ln(x) = \int_{1}^{x} \frac{dt}{t} \). This form is fundamental as it illustrates the link between calculus and the natural logarithm function by interpreting \( \ln(x) \) as the area under a simple hyperbolic curve \( \frac{1}{t} \).

Change of Variables

Changing variables, also known as substitution, is a powerful technique in calculus that simplifies the process of integration. It involves replacing a variable with another expression, thereby transforming the integral into a more manageable form. In the exercise, to evaluate \( \ln \left(\frac{1}{x}\right) \), we make a substitution: let \( u = \frac{1}{t} \). This changes the variable and requires us to adjust the integral’s limits of integration. When \( t = 1 \), \( u = 1 \); and when \( t = \frac{1}{x} \), \( u = x \). These adjustments convert \( \int_{1}^{1/x} \) to \( \int_{1}^{x} \), changing the function to \( -\frac{1}{u} \) which simplifies the computation. This technique is a versatile tool in both definite and indefinite integration.

Integration Techniques

Various techniques simplify solving integrals, each suitable for different types of functions. One fundamental method is the change of variables, as seen in this exercise. This involves finding a suitable substitution to transform the integral into an easier one. Another basic technique is integration by parts, which is useful for products of functions. There are also specific strategies such as partial fraction decomposition, tailored for rational functions, and trigonometric substitution, which address integrals with square roots. Mastering these techniques is crucial for efficiently tackling a wide range of integral calculus problems. In the discussed problem, changing the variables reduced a complex integral into one that's straightforward to solve, thereby illustrating the power of these methods.