Question

Evaluate the following integrals, in which \(a\) and \(b\) are nonzero real constants. (a) \(\int_{0}^{\infty} \frac{2 x^{2}+1}{x^{4}+5 x^{2}+6} d x\). (b) \(\int_{0}^{\infty} \frac{d x}{6 x^{4}+5 x^{2}+1}\). (c) \(\int_{0}^{\infty} \frac{d x}{x^{4}+1}\). (d) \(\int_{0}^{\infty} \frac{\cos x d x}{\left(x^{2}+a^{2}\right)^{2}\left(x^{2}+b^{2}\right)}\). (e) \(\int_{0}^{\infty} \frac{\cos a x}{\left(x^{2}+b^{2}\right)^{2}} d x\). (f) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{2}}\). (g) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{2}\left(x^{2}+2\right)}\). (h) \(\int_{0}^{\infty} \frac{2 x^{2}-1}{x^{6}+1} d x\). (i) \(\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+a^{2}\right)^{2}}\). (j) \(\int_{-\infty}^{\infty} \frac{x d x}{\left(x^{2}+4 x+13\right)^{2}}\). (k) \(\int_{0}^{\infty} \frac{x^{3} \sin a x}{x^{6}+1} d x\). (1) \(\int_{0}^{\infty} \frac{x^{2}+1}{x^{2}+4} d x\). (m) \(\int_{-\infty}^{\infty} \frac{x \cos x d x}{x^{2}-2 x+10}\). (n) \(\int_{-\infty}^{\infty} \frac{x \sin x d x}{x^{2}-2 x+10}\). (o) \(\int_{0}^{\infty} \frac{d x}{x^{2}+1}\). (p) \(\int_{0}^{\infty} \frac{x^{2} d x}{\left(x^{2}+4\right)^{2}\left(x^{2}+25\right)}\). (q) \(\int_{0}^{\infty} \frac{\cos a x}{x^{2}+b^{2}} d x\). (r) \(\int_{0}^{\infty} \frac{d x}{\left(x^{2}+4\right)^{2}}\).

Short answer

The complete solution requires obtaining the partial fractions, evaluating the separate integrals, and then calculating the limit as x approaches infinity. Without knowing the specific values of A and B, we are unable to provide a numeric result.

Step-by-step solution

Factor the denominator

Identify factors of the denominator \(x^{4} + 5x^{2} + 6\). This can be rewritten as \( (x^{2} + 2)(x^{2} + 3)\)

Set up expression for Partial Fraction Decomposition

The fraction can be deconstructed as: \(\frac{2x^{2} + 1}{x^{4} + 5x^{2} + 6} = \frac{A}{x^{2} + 2} + \frac{B}{x^{2} + 3}\)

Solve for coefficient constants A and B

Cross multiply, consider the coefficients of the similar powers of x and solve the simultaneous equations to obtain A and B.

Evaluate the Integral separately

With these constants computed, the integral can now be split into two parts and each evaluated separately using the power rule of integration. Remember to included the limits from 0 to infinity when calculating each integral.

Calculate the limit

We must evaluate the limit as x approaches infinity. If the limit approaches a finite value, we have our solution. Otherwise, the integral is divergent, which means it does not converge to a finite value.

Key concepts

Partial Fraction Decomposition

When faced with a complex rational expression in an indefinite integral, the technique of partial fraction decomposition emerges as a powerful tool. This method involves breaking down a fraction with a polynomial numerator and a more complicated polynomial denominator into a sum of simpler fractions with simpler denominators.

Consider the general form of a polynomial fraction that can be decomposed: \(\frac{P(x)}{Q(x)}\), where \(P(x)\) is the numerator polynomial and \(Q(x)\) is the denominator polynomial. The first step is to factor \(Q(x)\) into a product of irreducible polynomials. Each factor of the denominator then corresponds to a component of the decomposed expression with its own constant numerator, which we'll denote as \(A, B, ...\), depending on the number of terms.

For example, if we have \(\frac{2x^{2} + 1}{x^{4} + 5x^{2} + 6}\), we start by factoring the denominator: \(x^{4} + 5x^{2} + 6 = (x^{2} + 2)(x^{2} + 3)\). The expression can then be decomposed as \(\frac{A}{x^{2} + 2} + \frac{B}{x^{2} + 3}\). Here, \(A\) and \(B\) are constants that are determined by cross multiplying and solving the resulting equations. Through this process, a complex integral becomes a sum of more manageable ones.

Power Rule of Integration

A staple in calculus is the power rule of integration, which is commonly used after partial fraction decomposition. The power rule is straightforward: to integrate a function of the form \(x^n\), where \(n\) is a real number different from -1, you increase the exponent by 1 and then divide by the new exponent. Mathematically, this is represented as \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

For example, if we have separated our complex fraction using partial fraction decomposition and need to integrate \(\frac{A}{x^{2} + 3}\), we recognize that it resembles the form of the power rule. To perform the integration, we might need to adjust the expression slightly, such as by taking out constants or performing a substitution, to match the base power rule form. It's essential to remember that each term resulting from the decomposition is integrated separately and that the power rule assumes the integral exists and \(n\) is not equal to -1.

Improper Integrals

When it comes to evaluating indefinite integrals, the concept of improper integrals can't be overlooked, especially when the limits of integration include infinity. An improper integral involves the integration over an unbounded interval, such as from \(0\) to \(\infty\), or a function that becomes infinite within the interval of integration.

To evaluate an improper integral, we typically take the limit of a definite integral as the upper or lower bound approaches infinity or the point at which the function becomes infinite. For example, \(\int_{0}^{\infty} \frac{dx}{x^{2} + 1}\) is an improper integral because it extends to infinity. We evaluate this by taking the limit: \(\lim_{t \to \infty} \int_{0}^{t} \frac{dx}{x^{2} + 1}\). If the limit exists and is finite, we say the improper integral converges; if not, it diverges.

Evaluating improper integrals often requires careful consideration to determine convergence or divergence. Techniques such as comparison tests, the power rule after appropriate substitutions, or other integration methods may be necessary. As a helpful tip, when students face these integrals, they should ensure their limits are approached correctly and the resulting expressions are simple enough to evaluate with the limits inserted.