Question

Find conditions on \(p\) and \(q\) such that the integral converges. (a) \(\int_{-1}^{1} \frac{(\cos \pi x / 2)^{q}}{\left(1-x^{2}\right)^{p}} d x\) (b) \(\int_{-1}^{1}(1-x)^{p}(1+x)^{q} d x\) (c) \(\int_{0}^{\infty} \frac{x^{p} d x}{\left(1+x^{2}\right)^{q}}\) (d) \(\int_{1}^{\infty} \frac{[\log (1+x)]^{p}(\log x)^{q}}{x^{p+q}} d x\) (e) \(\int_{1}^{\infty} \frac{(\log (1+x)-\log x)^{q}}{x^{p}} d x\) (f) \(\int_{0}^{\infty} \frac{(x-\sin x)^{q}}{x^{p}} d x\)

Short answer

(a) \(\int_{-1}^{1} \frac{(\cos \pi x / 2)^{q}}{\left(1-x^{2}\right)^{p}} dx\) (b) \(\int_{-1}^{1}(1-x)^{p}(1+x)^{q} dx\) (c) \(\int_{0}^{\infty} \frac{x^{p} dx}{\left(1+x^{2}\right)^{q}}\) (d) \(\int_{1}^{\infty} \frac{[\log (1+x)]^{p}(\log x)^{q}}{x^{p+q}} dx\) (e) \(\int_{1}^{\infty} \frac{(\log (1+x)-\log x)^{q}}{x^{p}} dx\) (f) \(\int_{0}^{\infty} \frac{(x-\sin x)^{q}}{x^{p}} dx\) Answer: (a) \(p < \frac{1}{2}\) (b) \(p > -1\) and \(q > -1\) (c) \(p > 2q-1\) (d) \(q > 0\) (e) \(p > 0\) (f) \(q > p-1\)

Step-by-step solution

Write the integral using comparison test and determine converging conditions

We have: \(\int_{-1}^{1} \frac{(\cos \pi x / 2)^{q}}{\left(1-x^{2}\right)^{p}} dx\) Since \((\cos \pi x/2)^q \le 1\), we can compare this integral with: \(\int_{-1}^{1} \frac{1}{\left(1-x^{2}\right)^{p}} dx\) For this integral to converge, we need the exponents to sum up to greater than 1: \(-2p > -1\) So, we have the condition \(p < \frac{1}{2}\) for this integral to converge. (b) Integral B:

Write the integral using comparison test and determine converging conditions

We have: \(\int_{-1}^{1}(1-x)^{p}(1+x)^{q} dx\) Observe that \(0\le (1-x)^{p}(1+x)^{q} \le (1-x)^{p} + (1+x)^{q}\) Compare with the sum of two integrals: \(\int_{-1}^{1} (1-x)^{p} dx + \int_{-1}^{1} (1+x)^{q} dx\) Both integrals are convergent when their respective exponents are greater than -1, Hence, the converging conditions are \(p > -1\) and \(q > -1\). (c) Integral C:

Write the integral using comparison test and determine converging conditions

We have: \(\int_{0}^{\infty} \frac{x^{p} dx}{\left(1+x^{2}\right)^{q}}\) Compare this integral with: \(\int_{0}^{\infty} \frac{x^{p} dx}{x^{2q}}\) We need the sum of the exponents to be greater than -1, so: \(p - 2q > -1\) The converging condition is \(p > 2q-1\). (d) Integral D:

Analyze the integral and find converging conditions for each factor

We have: \(\int_{1}^{\infty} \frac{[\log (1+x)]^{p}(\log x)^{q}}{x^{p+q}} dx\) We can analyze the two factors of the logarithm separately: i) For the factor \((\log x)^{q}\), compare with the integral: \(\int_{1}^\infty \frac{dx}{x^{1+q}}\) For this integral to converge, we need \(1+q > 1\), i.e. \(q > 0\). The converging condition is \(q > 0\). (e) Integral E:

Write the integral using comparison test and determine converging conditions

We have: \(\int_{1}^{\infty} \frac{(\log (1+x)-\log x)^{q}}{x^{p}} dx\) Notice that since \(0 < \log(1+x) - \log(x) < \frac{1}{x}\), we can compare with the following integral: \(\int_{1}^{\infty} \frac{1}{x^{p+1}} dx\) For this integral to converge, we need \(p+1 > 1\), so: \(p > 0\) The converging condition is \(p > 0\). (f) Integral F:

Write the integral using comparison test and determine converging conditions

We have: \(\int_{0}^{\infty} \frac{(x-\sin x)^{q}}{x^{p}} dx\) Since \((x-\sin x)^{q} \le x^{q}\), compare with: \(\int_{0}^{\infty} \frac{x^{q}}{x^{p}} dx\) For this integral to converge, we need the sum of the exponents to be greater than -1, so: \(q - p > -1\) The converging condition is \(q > p-1\).

Key concepts

Comparison Test

The comparison test is a valuable method in determining the convergence properties of integrals, especially when dealing with complex functions. Essentially, the idea behind this test is to compare the original complex integral with a simpler integral that is easier to evaluate. This comparison helps us understand whether the original integral is convergent or divergent.

To use the comparison test effectively, follow these steps:

• Identify a function that bounds the given integral from above or below and is simpler to integrate.
• Check that the simpler function converges (or diverges) over the interval of integration.
• Compare the simpler function's behavior to the original function. If the simpler function converges and bounds the original function from above, then the original function converges as well.
• Conversely, if the simpler function diverges, then the original function also diverges if it is bounded by the simpler function from below.

This approach was employed in all six parts of the example problem, helping to systematically break down each integral to the fundamental question of convergence by exploiting known behaviors of simpler, comparable functions.

Integral Conditions

Integral convergence depends on specific conditions involving the exponent values in the integrand. When evaluating improper integrals, it's crucial to understand the role that the exponents play in determining convergence or divergence.

Let's consider the first example given, \[ \int_{-1}^{1} \frac{(\cos \pi x / 2)^{q}}{\left(1-x^{2}\right)^{p}} dx, \]and the condition \( p < 1/2 \) for convergence. This condition arises because the function becomes challenging to integrate near the endpoints \( -1 \) and \( 1 \), where \( 1-x^2 \) approaches zero, potentially causing the integral to diverge. The exponent \( p \) needs to be less than \( 1/2 \) to balance this behavior and ensure convergence.

Similarly, other integrals have specific conditions for convergence, generally derived from analyzing the behavior as the variable approaches critical points in the domain (such as 0, \( \infty \), -1, or 1). Understanding these conditions is key to determining the convergence of more complex integrals.

Integration Techniques

When approaching integrals, especially those with complex expressions or limits leading to complications, using specific integration techniques can simplify the process. Often, combining these techniques with the comparison test and integral conditions leads to a clearer understanding of whether an integral converges.

There are several common techniques:

• Substitution: This involves changing variables to simplify an integrand, making the integration process more manageable. It's particularly useful when dealing with symmetry or simplifying irrational expressions.
• Partial Fraction Decomposition: This breaks down rational functions into simpler fractions that are easier to integrate.
• Integration by Parts: This technique applies especially to products of functions, enabling us to express the integral in forms that are easier to solve.

While these aren't explicitly required in the exercise solutions provided, knowing these techniques can offer alternative routes to solving integrals and understanding their behavior. They can also complement the comparison test by providing alternative approaches to verify the convergence conditions identified.