Improper Integrals – Rigorous Definitions, Tests & Applications

To classify an integral as improper, we must check the limits of integration and the behavior of the function within those limits. In this case, the limits 0 and 5 are finite numbers. However, the function f(x) = 1/(x-3) is undefined at x = 3 because the denominator becomes zero, creating a vertical asymptote. Since x = 3 is inside the interval [0, 5], the function is unbounded at that point. Therefore, the integral is improper because the integrand has an infinite discontinuity within the interval of integration.

We start by rewriting the improper integral as a limit. We define the integral as the limit as b approaches infinity of the integral of e⁻²ˣ from 0 to b. Next, we find the antiderivative of e⁻²ˣ, which is (-½)e⁻²ˣ. We then evaluate this antiderivative from 0 to b. Plugging in the upper limit gives (-½)e^(-2b), and plugging in the lower limit gives (-½)e^(0), which simplifies to -1/2. Subtracting the lower limit value from the upper limit value gives (-½)e^(-2b) - (-½), or 1/2 - 1/(2e^(2b)). Finally, we take the limit as b approaches infinity. The term 1/(2e^(2b)) approaches 0, leaving us with the final answer of 1/2.

We begin by setting up the integral as the limit as b approaches infinity of the integral of 1/x from 1 to b. The antiderivative of 1/x is ln|x|. We evaluate ln(x) from 1 to b, which results in ln(b) - ln(1). Since ln(1) is 0, the expression simplifies to ln(b). We then evaluate the limit of ln(b) as b approaches infinity. Because the natural logarithm function grows without bound as its input increases, the limit is infinity. Therefore, the integral does not result in a finite number, meaning it diverges.

We first identify why this integral is improper. The function 1/√(x) is undefined at x = 0, which is the lower limit of integration. Because the interval of integration is from 0 to 4, we are approaching x = 0 from values larger than 0 (the positive side). Therefore, we must replace the lower limit 0 with a variable, say a, and take the limit as a approaches 0 from the right (denoted as 0+). This yields the correct setup: the limit as a approaches 0 from the right of the integral from a to 4.

The denominator is sqrt(4 - x²). If x = 2, the denominator is 0. If x = -2, the denominator is 0. An improper integral can only handle one singularity at a limit at a time. Therefore, we must pick an arbitrary point in the middle (like x = 0) and split the integral into two parts: one from -2 to 0 (handling the singularity at -2) and one from 0 to 2 (handling the singularity at 2).

The formula for surface area involves 2πysqrt(1+(y')²). For y=1/x, this simplifies to an expression that is greater than 2π*(1/x). By the Direct Comparison Test, since the integral of 1/x from 1 to infinity diverges (harmonic series behavior), the integral for the surface area must also diverge. This leads to the paradox that the horn has finite volume but infinite surface area.

The Limit Comparison Test states that if the limit as x approaches infinity of f(x)/g(x) equals a finite, positive constant (L > 0), then both integrals behave the same way. We chose g(x) = 1/x. We know from the p-test that the integral of 1/x from 1 to infinity diverges. Since the limit is 3 (a finite positive number), the integral of f(x) must also diverge, just like g(x).

Direct comparison is difficult here because the denominator (2x³ - 1) makes the fraction larger than (x+5)/2x³, and finding a ceiling function is messy. Instead, we use the Limit Comparison Test. We compare the integrand to the leading terms: x / 2x³ = 1/(2x²). Let g(x) = 1/x². We take the limit as x approaches infinity of [(x + 5) / (2x³ - 1)] divided by [1/x²]. This simplifies to the limit of (x³ + 5x²) / (2x³ - 1). As x goes to infinity, this limit is 1/2. Since 1/2 is a finite positive number, our integral behaves exactly like the integral of 1/x². Since the integral of 1/x² converges (p = 2 > 1), the original integral also converges.

This is a tricky integral because there is an infinite discontinuity at x = 0, which is inside the interval [-1, 1]. We must split the integral at the discontinuity: integral from -1 to 0 plus integral from 0 to 1. Let's look at the integral from 0 to 1. We set this up as the limit as a approaches 0 from the right of the integral from a to 1 of x^(-2). The antiderivative is -1/x. Evaluating from a to 1 gives (-1/1) - (-1/a) = -1 + 1/a. As a approaches 0 from the right, 1/a approaches infinity. Since this part of the integral diverges, the entire integral diverges. We do not need to calculate the other side.

We set up the limit as b approaches infinity of the integral from 1 to b. We use integration by parts with u = ln(x) and dv = x^(-2) dx. Then du = (1/x) dx and v = -1/x. The formula gives uv - integral of v du, which becomes -ln(x)/x - integral of (-1/x²) dx. This simplifies to -ln(x)/x + integral of x^(-2) dx. The antiderivative of x^(-2) is -1/x. So the full antiderivative is -ln(x)/x - 1/x. We evaluate from 1 to b: [-ln(b)/b - 1/b] - [-ln(1)/1 - 1/1]. Since ln(1) is 0, the lower limit evaluation is -(-1) = 1. For the upper limit, as b approaches infinity, 1/b goes to 0. The term ln(b)/b is infinity/infinity, so we apply L'Hôpital's Rule to get (1/b)/1, which approaches 0. The result is 0 - 0 + 1 = 1.

We set up the integral as the limit as b approaches infinity of the integral of sin(x) from 0 to b. The antiderivative of sin(x) is -cos(x). Evaluating from 0 to b gives -cos(b) - (-cos(0)), which simplifies to 1 - cos(b). We then take the limit of (1 - cos(b)) as b approaches infinity. The cosine function oscillates continuously between -1 and 1 and never settles on a single number. Because the limit does not exist due to oscillation, the improper integral diverges.

The integrand 1/(x-2) has a vertical asymptote at x = 2, which is in the middle of the interval [0, 4]. We must split the integral into two parts: the integral from 0 to 2 and the integral from 2 to 4. Let's examine the first part: the integral from 0 to 2. We write this as the limit as b approaches 2 from the left of the integral from 0 to b. The antiderivative is ln|x-2|. Evaluating at limits gives ln|b-2| - ln|0-2| = ln|b-2| - ln(2). As b approaches 2, |b-2| approaches 0, and ln(0) approaches negative infinity. Since the first part diverges, the entire integral diverges.

We use u-substitution. Let u = eˣ, which implies du = eˣ dx. We must also change the limits of integration. When x = 0, u = e⁰= 1. As x approaches infinity, u approaches infinity. The integral becomes the integral of 1/(1 + u²) du from 1 to infinity. This is a standard improper integral. The antiderivative of 1/(1 + u²) is arctan(u). We evaluate the limit as b approaches infinity of [arctan(b) - arctan(1)]. As b approaches infinity, arctan(b) approaches π/2. We know that arctan(1) is π/4. Therefore, the result is π/2 - π/4 = π/4.

Set up: limit as b→∞ of ∫ from 0 to b of x e^(-x²) dx. Use substitution u = x², du = 2x dx, so x dx = du/2. When x=0, u=0; when x=b, u=b². The integral becomes ∫ from u=0 to u=b² of e^(-u) * (du/2) = (½) ∫ from 0 to b² of e^(-u) du = (½)[-e^(-u)] from 0 to b² = (½)(-e^(-b²) + e^0) = (½)(1 - e^(-b²)). Now take limit as b→∞: e^(-b²) → 0, so limit = (½)(1 - 0) = 1/2. Thus converges to 1/2.

The integrand is unbounded at x=1. For integrals of the form ∫ from a to b of (1/(x-c)^p) dx with singularity at c inside, convergence occurs if p