Improper Integrals – Advanced Techniques, Comparison & Gamma Links

To identify the singularity, we look at the behavior of the function tan(x) = sin(x)/cos(x) at the limits of integration. At x = 0, tan(0) = 0, which is defined. However, at x = π/2, cos(π/2) is 0, making the tangent function undefined. As x approaches π/2 from the left, tan(x) approaches infinity. Therefore, there is a vertical asymptote at the upper limit x = π/2, making this an improper integral.

The integrand is undefined at the upper limit x = 4 because the denominator becomes √(0). We set this up as the limit as b approaches 4 from the left of the integral from 0 to b of (4-x)^(-½). Using substitution where u = 4-x and du = -dx (so dx = -du), the antiderivative is -2(4-x)^(½). We evaluate this from 0 to b: [-2(4-b)^(½)] - [-2(4-0)^(½)]. This simplifies to -2√(4-b) + 2√(4). As b approaches 4, the term -2√(4-b) approaches 0. The remaining term is 2√(4) = 2 * 2 = 4.

To test for absolute convergence, we look at the integral of |sin(x)/x²|. We know that |sin(x)| is always less than or equal to 1. Therefore, |sin(x)/x²| ≤ 1/x². We can compare the integral of |sin(x)/x²| to the integral of 1/x². Since the integral of 1/x² from 1 to infinity converges (p-test with p=2), the integral of the absolute value converges by the Direct Comparison Test. Since the absolute value converges, the original integral converges absolutely.

The integrand f(x) = xe^(-x²) is an odd function because f(-x) = (-x)e^(-(-x)²) = -xe^(-x²) = -f(x). However, for improper integrals, we must verify convergence on both sides separately. We evaluate the integral from 0 to infinity first. Using substitution u = -x², du = -2x dx, the integral becomes -1/2 e^u. Evaluating from 0 to infinity gives a finite value of 1/2. Similarly, the integral from negative infinity to 0 gives -1/2. Since both halves converge, we can add them: 1/2 + (-½) = 0.

We first use partial fraction decomposition on 1/(x² - 1) = 1/((x-1)(x+1)). We write this as A/(x-1) + B/(x+1). Solving for the constants, we get A = 1/2 and B = -1/2. The integral becomes (½) * integral of [1/(x-1) - 1/(x+1)]. The antiderivative is (½)[ln|x-1| - ln|x+1|], which can be combined using log rules to (½)ln|(x-1)/(x+1)|. We evaluate the limit as b approaches infinity of this expression from 2 to b. At the upper limit, we have (½)ln((b-1)/(b+1)). As b goes to infinity, the argument (b-1)/(b+1) approaches 1, and ln(1) is 0. At the lower limit x=2, we have (½)ln((2-1)/(2+1)) = (½)ln(⅓). The definite integral is 0 - (½)ln(⅓). Since -ln(⅓) = ln(3), the final answer is (½)ln(3).

The function ln(x)/x is undefined at x = 0 (vertical asymptote). We set up the limit as a approaches 0 from the right of the integral from a to 1. We use u-substitution with u = ln(x), so du = (1/x) dx. The integral becomes the integral of u du, which has an antiderivative of u²/2. Substituting back, we get (ln(x))² / 2. We evaluate from a to 1: [(ln(1))² / 2] - [(ln(a))² / 2]. Since ln(1) = 0, the first term is 0. The second term is -(ln(a))² / 2. As a approaches 0 from the right, ln(a) approaches negative infinity. Squaring it yields positive infinity. Thus, the limit is negative infinity, and the integral diverges.

We apply the p-test for integrals of the form 1/x^p on the interval (0, 1]. Note that this is different from the interval [1, infinity). On (0, 1], the integral converges if p = 1.

A: p=1, diverges.

B: p=2, diverges.

D: p=3, diverges.

C: Here, the function is x^(-½), so p = 1/2. Since 1/2

The integrand is undefined at x = 2 because the denominator becomes zero. We set up the limit as b approaches 2 from the left of the integral from 0 to b. We recognize the integrand 1/√(a² - x²) as the derivative of arcsin(x/a). Here a = 2, so the antiderivative is arcsin(x/2). We evaluate this from 0 to b: arcsin(b/2) - arcsin(0/2). As b approaches 2, b/2 approaches 1. The limit becomes arcsin(1) - arcsin(0). Since arcsin(1) = π/2 and arcsin(0) = 0, the result is π/2.

This integral is improper at both ends: x=0 (infinite discontinuity) and x=infinity (infinite interval). However, we can handle it with a single substitution. Let u = √(x). Then du = 1/(2√(x)) dx, which implies 2 du = (1/√(x)) dx. The limits change: if x=0, u=0; if x approaches infinity, u approaches infinity. The integral becomes the integral of 2e^(-u) du from 0 to infinity. The antiderivative of 2e^(-u) is -2e^(-u). Evaluating the limit as b approaches infinity: [-2e^(-b)] - [-2e^0]. As b goes to infinity, e^(-b) goes to 0. The result is 0 - (-2) = 2.

We evaluate the integral of e^(kx) from 0 to b. The antiderivative is (1/k)e^(kx). Evaluating from 0 to b gives (1/k)e^(kb) - (1/k)e⁰= (1/k)(e^(kb) - 1). We take the limit as b approaches infinity.

If k > 0, e^(kb) approaches infinity, so the integral diverges.

If k = 0, the integrand is 1, and the integral of 1 from 0 to infinity clearly diverges.

If k

The integrand f(x) = x / (1 + x²)² is an odd function because f(-x) = -x / (1 + (-x)²)² = -f(x). If the integral converges, the symmetry over the interval (-infinity, infinity) implies the answer is 0. To be rigorous, we check the integral from 0 to infinity. Let u = 1 + x², so du = 2x dx, or (½)du = x dx. The integral becomes (½) integral of u^(-2) du. The antiderivative is (½)(-1/u) = -1/(2(1+x²)). Evaluating from 0 to infinity: limit as b goes to infinity of [-1/(2(1+b²))] - [-1/2]. The first term goes to 0, leaving 1/2. Since the positive half converges to 1/2, and the function is odd, the negative half converges to -1/2. The sum is 0.

First, we factor the denominator: x² - 6x + 9 = (x-3)². The integrand is 1/(x-3)². There is a vertical asymptote at x = 3, which lies inside the interval [0, 4]. We must split the integral at x = 3. Consider the part from 3 to 4. We set up the limit as a approaches 3 from the right of the integral from a to 4 of (x-3)^(-2). The antiderivative is -1/(x-3). Evaluating from a to 4 gives [-1/(1)] - [-1/(a-3)] = -1 + 1/(a-3). As a approaches 3 from the right, the denominator (a-3) is a tiny positive number, so 1/(a-3) approaches positive infinity. Since this part diverges, the whole integral diverges.

The integrand has vertical asymptotes at both -1 and 1. We split the integral at x = 0.

Part 1: Integral from -1 to 0. Limit as a -> -1+ of [arcsin(0) - arcsin(a)] = 0 - (-pi/2) = π/2.

Part 2: Integral from 0 to 1. Limit as b -> 1- of [arcsin(b) - arcsin(0)] = pi/2 - 0 = π/2.

Since both parts converge, we add them: pi/2 + pi/2 = pi.

We look for a comparison function. On the interval from 0 to infinity, the denominator (x + 1) is always greater than 1 (for x > 0). Therefore, the fraction e⁻ˣ / (x + 1) is strictly less than e⁻ˣ / 1. We know that the integral of e⁻ˣ from 0 to infinity converges (it equals 1). Since our integrand is positive and bounded above by a convergent integral, it must also converge.

Using the definition provided, Γ(2) is the integral of x²⁻¹e⁻ˣ from 0 to infinity, which simplifies to the integral of x*e⁻ˣ from 0 to infinity. We solve this using integration by parts. Let u = x and dv = e⁻ˣ dx. Then du = dx and v = -e⁻ˣ. The formula gives -xe⁻ˣ - integral of -e⁻ˣ dx. This simplifies to -xe⁻ˣ - e⁻ˣ. Evaluating the limit as b approaches infinity: [-be^(-b) - e^(-b)] - [0 - 1]. Using L'Hôpital's rule on the first term, both terms involving b go to 0. The result is 0 - (-1) = 1.