How do you tell when an improper integral converges or diverges? (specific question)

Question

Hello! I'm having trouble understanding why the improper integral I'm answering is labeled as divergent when it looks like it converges. The improper integral is 1/x³, in-between the interval [-1, 2]. I split the integral so 0 could equal t-> -inf and t->inf. When I did the lim as t->inf, I found the antiderivative to be (-1/2)(1/x²), and plugging in the a and b, leading to this:

a = -1, b = t

(-1/2) limt->inf (1/t² - 1/(-1)²)

(-1/2) limt->inf (1/t² - 1)

Given that 1/x^2 converges to 0, I thought that this side of the integral would converge to -1/2. The other side of the integral is similar:

a = t, b = 2

(-1/2) limt->-inf (1/2² - 1/t²)

(-1/2) limt->-inf (1/4 - 1/t²)

What am I doing wrong? In general, I can solve an integral- I just can't tell when it diverges or converges. Are there any methods to determine this? Sorry if it's a bit messy, I don't know how else to ask this question.

Answer 1

How do you tell when an improper integral converges or diverges? (specific question): Hello! I'm having trouble understanding why the improper integral I'm answering is labeled as divergent when it looks like it converges. The improper integral is 1/x³, in-between the interval [-1, 2]. I split the integral so 0 could equal t-> -inf and t->inf. When I did the lim as t->inf, I found the antiderivative to be (-1/2)(1/x2), and plugging in the a and b.

Detailed Solution:

Question to you: Why t-> ± ∞? Where did you get this information? (0 ≠ ± ∞). The question is clear: in-between the interval [-1, 2]. Thus, you should not go below -1 or above 2.

∫ (1/x³) dx = ∫ x^-3 dx = (x^-2)/(-2) = -1/(2x²)

Since there is Discontinuity/Asymptote at x = 0, this is improper integral that needs to be split, the [-1, 2] into [-1, t] and [s, 2] and see what happen at the Discontinuity/Asymptote at x = 0 by taken Limit as t --> 0 and Limit as s --> 0

Limit as t --> 0

-1/(2x²) Evaluated at interval [-1, t] = -1/2(t)² - (-1/2(-1)²) = -1/2t² + 1/2

Lim t --> 0 (-1/2t² + 1/2) = -1/2(0)² + 1/2 = -1/0 + 1/2 = -∞ + 1/2 = -∞ (Diverges)

Limit as s --> 0

-1/(2x²) Evaluated at interval [s, 2] = -1/2(2)² - (-1/2(s)²) = -1/8 + 1/2s² = -1/8 + 1/2s²

Lim s --> 0 (-1/8 + 1/2s²) = -1/8 + 1/2(0)² = -1/8 + 1/0 = -1/8 + ∞ = ∞ (Diverges)

∫ (1/x³) dx in interval [-1, 2] = Lim t --> 0 (-1/(2x²) Evaluated at interval [-1, t]) + Lim s --> 0 (-1/(2x²) Evaluated at interval [s, 2])

= Lim t --> 0 ( -1/2t² + 1/2) + Lim s --> 0 ( -1/8 + 1/2s²)

= (-1/2(0)² + 1/2) + (-1/8 + 1/2(0)²)

= (-1/0 + 1/2) + (-1/8 + 1/0 )

= (-∞ + 1/2) + (-1/8 + ∞)

= -∞ (Diverges) + ∞ (Diverges)

= Diverges

Therefore: The improper integral: ∫ (1/x³) dx in interval [-1, 2], diverges as x --> 0.

Note: Without calculus, if you graph 1/x³, you will clearly see it increases/decrease without bound as x ==> 0, therefore it Diverges as x ==> 0