Question

Prove that $$ \int_{0}^{x} \frac{\ln (1-t)}{t} d t=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \quad \text { for } x \in[-1,1] $$

Short answer

The equality is proven by expanding \(\ln(1-t)\) as a series and integrating term by term.

Step-by-step solution

Consider the Series Expansion

Recall the series expansion for \( \ln(1-t) \): \[ \ln(1-t) = - \sum_{n=1}^{\infty} \frac{t^n}{n}, \quad \text{for } |t| < 1. \]

Change the Order of Integration and Summation

Substitute the series expansion into the integral: \[ \int_{0}^{x} \frac{\ln (1-t)}{t} dt = \int_{0}^{x} \frac{-\sum_{n=1}^{\infty} \frac{t^n}{n}}{t} dt. \] Change the order of summation and integration: \[ = -\sum_{n=1}^{\infty} \int_{0}^{x} \frac{t^n}{nt} dt = -\sum_{n=1}^{\infty} \int_{0}^{x} \frac{t^{n-1}}{n} dt. \]

Solve the Inner Integral

Evaluate the integral: \[ \int_{0}^{x} t^{n-1} dt. \] This integral evaluates to: \[ \left[ \frac{t^n}{n} \right]_{0}^{x} = \frac{x^n}{n}. \]

Substitute Back into the Summation

Substitute the evaluated integral back into the summation:\[ -\sum_{n=1}^{\infty} \frac{x^n}{n^2}. \]Since the negative sign from the series expansion cancels out with the negative sign in the integration, the expression simplifies to:\[ \sum_{n=1}^{\infty} \frac{x^n}{n^2}. \]

Conclusion

Thus, we have shown that \[ \int_{0}^{x} \frac{\ln (1-t)}{t} dt = \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \quad \text{for } x \in[-1,1]. \]

Key concepts

Series Expansion

In mathematics, a series expansion is a representation of a function as a sum of its terms calculated from its series. This can often make complex functions easier to work with and more approachable for integration or differentiation. In the original problem, we consider the function \( \ln(1-t) \), which appears frequently in calculus. For this particular exercise, we use the series expansion:

• \( \ln(1-t) = -\sum_{n=1}^{\infty} \frac{t^n}{n} \) when \(|t| < 1\).

This expansion is known as a Taylor series where the function is written as an infinite sum. In practice, using the series expansion allows us to break down the logarithmic function into simpler polynomial terms derived from the geometry of the unit circle. By transforming \( \ln(1-t) \) this way, we can then proceed with the integration much more conveniently.

Order of Integration and Summation

The order of integration and summation refers to the rearranged sequence of operations, which can simplify solving integrals involving series expansions. In calculus, iterated integrals and summations often allow for such rearrangements if convergence conditions are met. This is crucial for making calculations more manageable. In the original step by step solution, the order of integration and summation is rearranged as follows:

• The series expansion of the logarithmic function is substituted inside the integral.
• Then, the integration is separated for each term in the series.

This rearrangement requires confirming that both operations are interchangeable for the given conditions on \(x\). The procedure helps transform the problem into a more straightforward calculation where each term of the series is integrated individually. By doing so, integration becomes more straightforward and results in a new series expression.

Evaluating Integrals

Evaluating integrals is a fundamental skill in calculus, often involving methods such as substitution, integration by parts, or changing the order of operations as seen previously. In the exercise, the integral:

• \( \int_{0}^{x} t^{n-1} \, dt \)

was solved using basic integration rules, which transforms the power of \(t\) into a division by the same power incremented by one:

• The result is \( \left[ \frac{t^n}{n} \right]_{0}^{x} = \frac{x^n}{n} \).

Successfully evaluating the integral is key to solving the series and proving the result. This method shows the power of elementary integration when applied to a sum of terms coming from a potentially daunting series.

Convergence of Series

The convergence of a series is a crucial concept that tells us whether the infinite sum of terms indeed approaches a finite value. In the given problem, convergence is guaranteed due to specific conditions on \(x\). The series:

• \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \)

converges for \(-1 < x < 1\), due to the nature of geometric and p-series in mathematics. Understanding convergence is essential here:

• As \(x\) approaches 1 or -1, the terms of the series grow smaller rapidly, ensuring the total sum stays finite.
• Each term's denominator grows quadratically, assuring convergence by the p-series test with \(p = 2\). This specific condition is necessary to interchange the summation and integration steps previously discussed.

Exploring convergence helps ensure all operations are valid mathematically, allowing the use of series safely in integrations like the one in this exercise.