Problem 384 - Calculus Volume 1

Question

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition \(\ln (x)=\int_{1}^{x} \frac{d t}{t}, \quad\) using properties of the definite integral and making no further assumptions. Use a change of variable in the integral \(\int_{1}^{x y} \frac{1}{t} d t\) to show that \(\ln x y=\ln x+\ln y\) for \(x, y>0.\)

Step-by-Step Solution

Answer

Using integration and substitution, we proved that \(\ln(xy) = \ln(x) + \ln(y)\).

Step 1: Recognize the Given Integral

We are given the integral expression for the natural logarithm: \(\ln(xy) = \int_{1}^{xy} \frac{1}{t} \, dt\). Our goal is to use a change of variables to prove \(\ln(xy) = \ln(x) + \ln(y)\).

Step 2: Set Up the Change of Variables

We'll use a substitution to change the variable inside the integral. Let \(u = \frac{t}{y}\), which implies that \(t = uy\) and \(dt = y \cdot du\).

Step 3: Transform the Limits of Integration

For the lower limit when \(t = 1\), \(u = \frac{1}{y}\). For the upper limit when \(t = xy\), \(u = x\). Now the integral becomes \(\int_{1}^{xy} \frac{1}{t} \, dt = \int_{1/y}^{x} \frac{1}{uy} y \, du\).

Step 4: Simplify the Integral

Simplify the integral by substituting the expression for \(dt\) and factor out constants. We have:\[ \int_{1/y}^{x} \frac{1}{uy} \cdot y \, du = \int_{1/y}^{x} \frac{1}{u} \, du.\]

Step 5: Separate the Integral into Two Parts

This new integral can be split into two parts from 1 to \(x\) and from \(1/y\) to 1:\[ \int_{1/y}^{x} \frac{1}{u} \, du = \int_{1/y}^{1} \frac{1}{u} \, du + \int_{1}^{x} \frac{1}{u} \, du.\]

Step 6: Evaluate Each Integral

Now evaluate each integral based on the definition of natural logarithms:- \(\int_{1/y}^{1} \frac{1}{u} \, du = -\ln(y)\) (as \(u\) moves from \(1/y\) to 1).- \(\int_{1}^{x} \frac{1}{u} \, du = \ln(x)\).Therefore,\[ \ln(xy) = -\ln(y) + \ln(x) + \ln(y).\]

Step 7: Combine and Simplify Results

Combine the evaluated integrals: the negative and positive \(\ln(y)\) terms cancel each other out. Thus, we are left with:\[ \ln(xy) = \ln(x) + \ln(y).\]

Step 8: Conclusion

We have shown using integration and substitution that \(\ln(xy) = \ln(x) + \ln(y)\) for \(x, y > 0\), confirming the logarithmic identity.

Key Concepts

Definite Integral

The concept of a definite integral is fundamental in calculus. It tells us the accumulation of quantities, akin to finding the area under a curve.
Interestingly, a definite integral has an upper and lower limit, providing a net value over a specific interval.

• Symbolically written as \int_{a}^{b} f(t) \, dt\.
• It gives a precise value rather than a general formula.

This conceptual framework is employed in defining functions like the natural logarithm. In the exercise, the definite integral from 1 to \( xy \) of \( \frac{1}{t} \) now represents \( \ln(xy) \).
The boundaries 1 and \( xy \) specify the range for computation. It forms the starting point for proving logarithmic identities using integration.

Integration by Substitution

Integration by substitution is a method used to solve integrals more easily by transforming variables.
It simplifies the integration process by aligning our problem with a basic integral form.

• Identify part of the integral as \( u \) (a new variable).
• Express the differential, \( dt \), in terms of \( du \).
• Change the bounds of the integral according to this substitution.

In the given exercise, we set \( u = \frac{t}{y} \), which translates \( t \) into \( uy \), allowing natural decomposition of the integral.
This transforms the integral limits and variable, effectively simplifying the calculation.

Logarithmic Identities

Logarithmic identities are essential properties that simplify calculations with logarithms.
These identities allow us to break down complex logarithmic expressions into manageable parts.

• The exercise proves a key identity: \( \ln(xy) = \ln(x) + \ln(y) \).
• It is supported by properties of integrals and substitution.

Such identities help in multiplication and division involving logs, turning these operations into a sum or a difference.
In this exercise, by proving \( \ln(xy) = \ln(x) + \ln(y) \), we affirm how integral properties give us core logarithmic truths.