Improper Integrals – Basics, Definitions & First Examples

An improper integral is defined by two specific conditions. The first condition is when the interval of integration is unbounded, meaning at least one of the limits of integration is positive or negative infinity. The second condition is when the integrand function has a discontinuity or becomes unbounded (goes to infinity) at some point within the limits of integration or at an endpoint. Therefore, the correct choice is the one that states the integral has one or both limits infinite or the integrand is unbounded.

We evaluate this improper integral by first rewriting it as a limit. The upper limit is infinity, so we replace it with a variable, say 'b', and take the limit as b approaches infinity. The integral becomes: limit as b→∞ of ∫ from 1 to b of (1/x²) dx. Next, we find the antiderivative of 1/x², which is -1/x. We then evaluate this antiderivative from 1 to b: [-1/x] from 1 to b = (-1/b) - (-1/1) = (-1/b) + 1. Finally, we take the limit as b approaches infinity: limit as b→∞ of (1 - 1/b). As b grows very large, 1/b approaches 0. Therefore, the limit is 1 - 0 = 1. The integral converges to 1.

We check each integral for the conditions that make an integral improper. An integral is improper if it has an infinite limit or if the integrand is unbounded within the integration interval. For option B, the integrand is 1/(x-3). This function becomes undefined and goes to infinity when x = 3. Since x = 3 is within the interval from 1 to 5, the integrand is unbounded at that point. This makes the integral improper. The other integrals have finite limits and their integrands (x², sin(x), x³) are defined and continuous on their respective intervals, so they are proper integrals.

This integral has an infinite upper limit, so it is improper. We set it up as a limit: limit as b→∞ of ∫ from 0 to b of e⁻ˣ dx. First, find the antiderivative of e⁻ˣ, which is -e⁻ˣ. We evaluate this from 0 to b: [-e⁻ˣ] from 0 to b = [-e^(-b)] - [-e^(0)] = -e^(-b) + 1. Now we take the limit as b→∞: limit as b→∞ of (1 - e^(-b)). Since e^(-b) = 1/(e^b), as b gets very large, e^(-b) approaches 0. Therefore, the limit is 1 - 0 = 1. The integral converges to 1.

This integral is improper because the integrand 1/√x is unbounded (goes to infinity) as x approaches 0 from the right. The lower limit is the point of discontinuity. We rewrite it as a limit: limit as a→0⁺ of ∫ from a to 4 of (1/√x) dx. First, rewrite 1/√x as x^(-½). Its antiderivative is (x^(½)) / (½) = 2√x. Now evaluate from a to 4: [2√x] from a to 4 = 2√4 - 2√a = 2*2 - 2√a = 4 - 2√a. Take the limit as a approaches 0 from the positive side: lima→0⁺ (4 - 2√a). As a gets very close to 0, √a also approaches 0. So the limit is 4 - 0 = 4. The integral converges to 4.

We evaluate the integral using a limit. Set up: limit as b→∞ of ∫ from 1 to b of (1/x) dx. The antiderivative of 1/x is ln|x|. Evaluate from 1 to b: [ln|x|] from 1 to b = ln(b) - ln(1) = ln(b) - 0 = ln(b). Now take the limit as b→∞: limit as b→∞ of ln(b). The natural logarithm function increases without bound as its argument goes to infinity. Therefore, ln(b) approaches infinity. This means the limit does not exist as a finite number; the integral diverges (specifically, to infinity).

The integral is improper because its lower limit is negative infinity. The standard procedure for evaluating an improper integral with an infinite limit is to replace the infinite limit with a variable and then take the limit as that variable approaches the infinity. Specifically, for an integral from -∞ to 0, we replace -∞ with a variable, say 'a', and then take the limit as 'a' approaches negative infinity. This gives us: limit as a→-∞ of ∫ from a to 0 of e²ˣ dx. This is the correct first step.

This is an improper integral due to the infinite upper limit. We set it up as a limit: limit as b→∞ of ∫ from 2 to b of (1/(x ln x)) dx. We use substitution to integrate. Let u = ln x, then du = (1/x) dx. The integral becomes ∫ (1/u) du = ln|u| = ln|ln x|. Now evaluate from 2 to b: [ln|ln x|] from 2 to b = ln(ln b) - ln(ln 2). Now take the limit as b→∞: limit as b→∞ of [ln(ln b) - ln(ln 2)]. As b increases, ln b increases, and ln(ln b) also increases without bound. Therefore, the limit is infinity, meaning the integral diverges.

An integral is improper if the integrand has a discontinuity or becomes infinite within the interval of integration. For the integral ∫₀¹ (1/x) dx, at x = 0, the function 1/x is undefined and its limit is infinity. This means the integrand is unbounded at the lower limit of integration, x = 0. Therefore, the integral is improper due to the unbounded integrand at the endpoint x=0.

This integral is improper because both limits are infinite. We must split it into two integrals: ∫ from -∞ to 0 and ∫ from 0 to ∞. We write: ∫ from -∞ to ∞ f(x) dx = limit as a→-∞ of ∫ from a to 0 f(x) dx + limit as b→∞ of ∫ from 0 to b f(x) dx. First, find the antiderivative of 1/(1+x²), which is arctan(x). For the first part: limit as a→-∞ of [arctan(x)] from a to 0 = arctan(0) - limit as a→-∞ arctan(a) = 0 - (-π/2) = π/2. For the second part: limit as b→∞ of [arctan(x)] from 0 to b = limit as b→∞ arctan(b) - arctan(0) = (π/2) - 0 = π/2. Adding the two results: π/2 + π/2 = π. Therefore, the integral converges to π.

We check each statement. Statement A is true; for example, ∫ from 1 to ∞ of 1/x² dx converges to 1. Statement B is true; it is a known p-test for improper integrals: ∫ from 1 to ∞ of 1/x^p dx converges if p > 1 and diverges if p ≤ 1. Statement C is false; an integral with an unbounded integrand can still converge if the singularity is "integrable." For example, ∫₀¹ 1/√x dx converges to 2. The unboundedness does not automatically mean divergence; it depends on how fast the function goes to infinity. Statement D is true; an integral can have both features, like ∫ from 0 to ∞ of (1/√x) dx, which is improper at 0 (unbounded) and at ∞ (infinite limit). Therefore, the false statement is C.

The integral is improper because ln(x) is undefined (goes to -∞) as x→0⁺. We rewrite it as a limit: limit as a→0⁺ of ∫ from a to 1 of ln(x) dx. To integrate ln(x), use integration by parts. Let u = ln(x), dv = dx. Then du = (1/x) dx, v = x. The integral ∫ ln(x) dx = x ln(x) - ∫ x*(1/x) dx = x ln(x) - ∫ 1 dx = x ln(x) - x. Now evaluate from a to 1: [x ln(x) - x] from a to 1 = (1*ln(1) - 1) - (a ln(a) - a) = (0 - 1) - (a ln(a) - a) = -1 - a ln(a) + a. Now take the limit as a→0⁺. We need limit a→0⁺ of (a ln(a)) and limit a→0⁺ of a. The limit of a is 0. For a ln(a), rewrite as ln(a)/(1/a) and use L'Hôpital's rule or know that the limit is 0. So the expression becomes -1 - 0 + 0 = -1. The integral converges to -1.

This is an improper integral of the form ∫ from 1 to ∞ of 1/x^p dx, where p = 0.5. The p-test states that such an integral converges if p > 1 and diverges if p ≤ 1. Here, p = 0.5, which is less than 1. Therefore, the integral diverges. Statement A is incorrect; positivity of the exponent is not sufficient for convergence (p must be >1). Statement C is incorrect; merely decreasing is not enough (the harmonic series 1/x diverges). Statement D is incorrect; the fact that the function goes to zero is necessary but not sufficient for convergence of the integral.

The integrand 1/(x-1)² is unbounded at x = 1, which is within the interval (0,2). We must split the integral at the point of discontinuity: ∫₀¹ (1/(x-1)²) dx + ∫ from 1 to 2 of (1/(x-1)²) dx. Consider the right-hand piece: ∫ from 1 to 2 of (1/(x-1)²) dx. Rewrite as limit as c→1⁺ of ∫ from c to 2 of (1/(x-1)²) dx. To find the antiderivative of (x-1)⁻², let u = x-1, so du = dx. The integral becomes ∫u⁻² du = -u⁻¹ = -1/u = -1/(x-1). Evaluating this from c to 2 gives (-1/(2-1)) - (-1/(c-1)) = -1 - (-1/(c-1)) = -1 + 1/(c-1). As c→1⁺, (c-1) → 0⁺, so 1/(c-1) → +∞. Therefore, the limit is infinity, so this part diverges to infinity. Since one part diverges, the whole integral diverges.

We set up the integral as limit as b→∞ of ∫ from 0 to b of sin(x) dx. The antiderivative of sin(x) is -cos(x). Evaluate from 0 to b: [-cos(x)] from 0 to b = -cos(b) - (-cos(0)) = -cos(b) + 1. Now consider the limit as b→∞: limit as b→∞ of (1 - cos(b)). The cosine function oscillates between -1 and 1 and does not approach a specific value as b→∞. Therefore, the limit does not exist. This means the improper integral diverges (not to infinity, but by oscillation). The correct choice is that it diverges because the limit does not exist as a finite number.