Calculus II·Convergence of Improper Integrals

Improper Integral Convergence by p-Test

Given the improper integral ∫_a^∞ c/x^p dx, decide whether it converges and, if so, compute its value. Convergence is determined by the p-test: the integral converges iff p > 1.

✓ All 6 test cases pass.

Inputs

c (numerator constant)

The constant multiplier in the integrand c/x^p.

p (exponent)

Convergence threshold for the integral ∫_a^∞ c/x^p dx is p > 1.

a (lower bound)

Lower limit. Must be > 0.

Test cases

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#	Inputs	Expected	Got	Status
1	{"c":1,"p":2,"a":1}	1	1 $\int_{1}^{\infty} \frac{1}{x^{2}} \, dx = \frac{1 \cdot 1^{1-2}}{2 - 1} = 1.00000$	✓
2	{"c":1,"p":3,"a":1}	0.5	0.5 $\int_{1}^{\infty} \frac{1}{x^{3}} \, dx = \frac{1 \cdot 1^{1-3}}{3 - 1} = 0.500000$	✓
3	{"c":2,"p":2,"a":1}	2	2 $\int_{1}^{\infty} \frac{2}{x^{2}} \, dx = \frac{2 \cdot 1^{1-2}}{2 - 1} = 2.00000$	✓
4	{"c":1,"p":2,"a":2}	0.5	0.5 $\int_{2}^{\infty} \frac{1}{x^{2}} \, dx = \frac{1 \cdot 2^{1-2}}{2 - 1} = 0.500000$	✓
5	{"c":1,"p":1,"a":1}	diverges (p ≤ 1)	diverges (p ≤ 1) $\int_{1}^{\infty} \frac{1}{x^{1}} \, dx \text{ diverges since } p = 1 \le 1$	✓
6	{"c":1,"p":0.5,"a":1}	diverges (p ≤ 1)	diverges (p ≤ 1) $\int_{1}^{\infty} \frac{1}{x^{0.5}} \, dx \text{ diverges since } p = 0.5 \le 1$	✓

Source: Pattern: p-integral test. Reproduces the standard Cal II result, e.g. Stewart 'Calculus' 8e §7.8.

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