Question

For what value of \(c\) does the integral \(\int_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+1}}-\frac{c}{x+1}\right) d x\) converge? Evaluate the integral for this value of \(c\).

Short answer

The integral converges for the value \(c = 1\) and the value of the integral for this value of \(c\) is \(0\).

Step-by-step solution

Set the two parts of the integrand equal

Assume that for the integral to converge, the two parts of the integrand need to be equal for all \(x \), giving the condition \(\frac{1}{\sqrt{x^{2}+1}} = \frac{c}{x+1}\). Solving this for \(c\) gives \(c = \frac{x+1}{\sqrt{x^{2}+1}}\).

Confirming our assumption

Substitute \(x = 0\) into the expression we obtained for \(c\) to find that \(c = 1\). This should hold true for all \(x\), and indeed it does. If we substitute \(x = 0\) into the original integrand, the resulting integral \(\int_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+1}}-\frac{1}{x+1}\right) dx\) does indeed converge. Hence, \(c = 1\) is the value that makes the integral converge.

Evaluate the integral for c=1

Now, we substitute \(c = 1\) into the integral and evaluate it. The integrand simplifies to \(0\) for all values of \(x\). Accordingly, the integral evaluates to \(0\).

Key concepts

Calculus

Calculus, a fundamental branch of mathematics, encompasses the study of change and motion using two core concepts: differentiation and integration. Differentiation focuses on rates of change and slopes of curves, while integration, its inverse, concerns the accumulation of quantities, such as areas under curves.

To solve the given exercise, integration plays a crucial role. Specifically, we look at an improper integral, which stretches to infinity. Improper integrals often require specific techniques and attention to convergence to ensure that the infinite area being calculated results in a finite number.

Improper Integrals

Improper integrals arise when the interval of integration is infinite or the function has an infinite discontinuity. The integral in the problem, \( \int_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+1}}-\frac{c}{x+1}\right) dx \), is an example of an improper integral because it extends to infinity.

To deal with this, we consider the limit of a proper integral as one limit of integration approaches infinity. This concept stretches beyond the graph's finite bounds, presenting challenges in determining whether the area under the curve is finite (converges) or infinite (diverges).

In our solution, we have tackled this by finding a value for \(c\) that ensures the two functions in the integrand offset each other adequately to produce a convergent result.

Convergence Tests for Integrals

Determining the convergence of an integral is pivotal in calculus, especially for improper integrals. To this end, convergence tests provide the necessary techniques. One common method is the comparison test, wherein we compare the improper integral to a known, simpler integral's convergence.

In the exercise, we compare the two terms in the integrand. We then set them equal to find a singular value of \(c\) that allows the integral value to remain finite as \(x \to \infty\). We confirmed through substitution that \(c = 1\) satisfies this condition. When \(c = 1\), the integrand simplifies to \(0\), hence, the integral also simplifies to \(0\), clearly demonstrating convergence as there are no areas to sum to infinity. This finding was essential in confirming that the improper integral converges for \(c = 1\).