Problem 391 - Advanced Calculus

Question

Show that \(1 \int_{0}[\\{\log (1-t)\\} / t] \mathrm{dt}=-\left[\left(1 / 1^{2}\right)+\left(1 / 2^{2}\right)+\left(1 / 3^{2}\right)+\ldots\right.\) \(\left.+\left(1 / n^{2}\right)+\ldots\right]\)

Step-by-Step Solution

Answer

To show that the given integral is equal to the infinite sum, we first integrate by parts and then perform a variable substitution, which allows us to rewrite the integral in terms of an infinite series. We then interchange summation and integration and evaluate the resulting integral. In the end, we find that: \[ \int_0^1 \frac{\log(1-t)}{t} dt = -\left[\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} + \dots \right] \]

Step 1: Identify the given integral

In this exercise, we are asked to show that the following integral: \[ \int_0^1 \frac{\log(1-t)}{t} dt \] is equal to: \[ -\left[\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} + \dots \right] \]

Step 2: Integrate by parts

To evaluate the given integral, we will use integration by parts. Integration by parts is given by the formula: \[ \int udv = uv -\int vdu \] In this case, we will choose our functions u and dv as follows: \(u = \log(1-t)\) which results in \(du =-\frac{dt}{1-t}\) and \(dv =\frac{dt}{t} \) which results in \(v =\log t\) Now we apply the formula for integration by parts: \[ \int_{0}^{1} \frac{\log(1-t)}{t}dt = \left[\log t \cdot \log(1-t)\right]_0^1 -\int_{0}^{1} -\frac{\log t}{1-t}dt \]

Step 3: Evaluate the boundary terms

Let's evaluate the boundary term: \[ \left[\log t \cdot \log(1-t)\right]_0^1 \] As the at the limits (t=0 and t=1) either log t or log(1-t) becomes undefined, this expression goes to zero. So we are left with: \[ \int_{0}^{1} -\frac{\log t}{1-t}dt \]

Step 4: Change the variable

Now, let's make a substitution. Let \(x=t/(1-t)\), then \(t=x/(1+x)\) and \(1-t=1/(1+x)\). Additionally, we need to find the derivative of t in terms of x: \(dt/dx=1/(1+x)^2\), so \(dx= dt(1+x)^2\) Now we can change the variable in the integral: \[ \int_{0}^{1} -\frac{\log t}{1-t}dt=-\int_{0}^{\infty}\log\left(\frac{x}{1+x}\right)\frac{dx}{x} \]

Step 5: Recognize the series expansion

Now, we recognize that the natural logarithm of a fraction can be expressed as an infinite series: \[ \log\left(\frac{x}{1+x}\right) = \left(\log x -\log (1+x)\right) = \sum_{n=1}^{\infty} \frac{(-x)^n}{n} \] This means we can rewrite the integral as: \[ -\int_{0}^{\infty}\log\left(\frac{x}{1+x}\right)\frac{dx}{x}=-\int_{0}^{\infty}\left(\sum_{n=1}^{\infty} \frac{(-x)^n}{n}\right)\frac{dx}{x} \]

Step 6: Interchange summation and integration

Now we interchange the summation and the integration (this is allowed under certain conditions which are satisfied here): \[ -\int_{0}^{\infty}\left(\sum_{n=1}^{\infty} \frac{(-x)^n}{n}\right)\frac{dx}{x}=-\sum_{n=1}^{\infty}\int_{0}^{\infty} \frac{(-x)^{n-1}}{n}dx \]

Step 7: Evaluate the resulting integral

Now we need to evaluate the integral in the sum: \[ \int_{0}^{\infty} \frac{(-x)^{n-1}}{n} dx \] This is a simple power law integral which can be evaluated: \[ \int_{0}^{\infty} \frac{(-x)^{n-1}}{n} dx = \left[\frac{(-x)^n}{n^2}\right]_0^{\infty}= -\frac{1}{n^2} \] By substituting this result in the overall expression, we get: \[ -\sum_{n=1}^\infty \int_{0}^{\infty} \frac{(-x)^{n-1}}{n} dx = -\sum_{n=1}^{\infty} -\frac{1}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^2} \] Now we have shown that the given integral is indeed equal to the infinite sum: \[ \int_0^1 \frac{\log(1-t)}{t} dt = -\left[\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} + \dots \right] \]

Key Concepts

Series Expansion

When dealing with complex integrals, such as \( \int_0^1 \frac{\log(1-t)}{t} dt \), we often use series expansion to simplify the calculation. Series expansion involves expressing a complex function as an infinite sum of simpler terms.

The essence of this method is leveraging the power of known series, like the geometric or binomial series, to represent functions that are otherwise challenging to integrate directly.

• In our exercise, we observed the expression \( \log\left(\frac{x}{1+x}\right) \), which was expanded using the logarithmic series: \( \log x - \log (1+x) = \sum_{n=1}^{\infty} \frac{(-x)^n}{n} \).
• This transformation allows us to work with each term individually, making the integration process far more manageable.

Series expansions are particularly useful because they turn a problem of dealing with a complicated function into a much simpler problem of adding up a series of terms.

Variable Substitution

Variable substitution is a technique that simplifies an integral by changing the variable of integration, often making the evaluation process more straightforward.

In this example, to tackle the integral \( \int_{0}^{1} \frac{-\log t}{1-t}dt \), we used substitution to better fit the integration idea:

• Letting \( x = \frac{t}{1-t} \) , simplifies the expression by transforming both \( t \) and \( 1-t \).
• The derivative relation \( dt = dx/(1+x)^2 \) was subsequently utilized to rewrite the integral in terms of \( x \).
• This substitution provides an easier path for evaluating the integral and investigating its properties.

Substitution plays a key role in transforming limits and bounds of integrals in a way that creates simpler, or more analytically approachable forms to work with.

Convergence of Sums

In the context of infinite series, convergence is a critical concept. We must determine whether an infinite sum approaches a finite limit.

The convergence of sums means that as we add more terms to our series, the total approaches a specific value:

• In our solution, when we replaced the integral with the series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), a clear structure emerged.
• Note that to interchange summation and integration (from step 6), the integral of the series terms must converge.

The ability to determine convergence dictates whether such mathematical manipulations are valid and lead to meaningful results.

Power Law Integration

Power law integration involves integrating functions of the form \( x^n \), which are simple yet ubiquitous in calculus. Participants in our solution can notice that a direct application appeared during the simplification process.

When we expanded the series, power law integrals emerged as indispensable parts of reaching the final answer:

• The integral \( \int_{0}^{\infty} \frac{(-x)^{n-1}}{n} dx \) required the power function to be evaluated on its limits, typically from 0 to infinity.
• Recognizing this allows the integral to simplify into a straightforward computation \( -\frac{1}{n^2} \).

Understanding power law integration aids significantly in situations where multiple series or polynomial representations are involved.