請問這題微積分怎麼算? find the values of n so that the improper integral converges?

1. ∫(0,1) 1/xⁿ dx

For n ≠ 1,
∫(0,1) 1/xⁿ dx
= lim(t→0⁺) ∫(t,1) 1/xⁿ dt
= 1/(n - 1) lim(t→0⁺) ∫(t,1) d(-1/x⁽ⁱⁿ⁻¹⁾)
= -1/(n - 1) lim(t→0⁺) [ 1/1⁽ⁱⁿ⁻¹⁾ - 1/t⁽ⁱⁿ⁻¹⁾ ]
= -1/(n - 1) + 1/(n - 1) lim(t→0⁺) 1/t⁽ⁱⁿ⁻¹⁾

When n > 1, lim(t→0⁺) 1/t⁽ⁱⁿ⁻¹⁾ →∞
When n < 1, lim(t→0⁺) 1/t⁽ⁱⁿ⁻¹⁾ = 0

For n = 1,
∫(0,1) 1/x dx
= lim(t→0⁺) ∫(t,1) 1/x dt
= lim(t→0⁺) ∫(t,1) d( ln x )
= ln 1 - lim(t→0⁺) ln t
= -∞

∴ n < 1

????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

2. ∫(1,∞) 1/xⁿ dx

For n ≠ 1,
∫(1,∞) 1/xⁿ dx
= lim(t→∞) ∫(1,t) 1/xⁿ dx
= 1/(n - 1) lim(t→∞) ∫(1,t) d(-1/x⁽ⁱⁿ⁻¹⁾)
= -1/(n - 1) lim(t→∞) [ 1/1⁽ⁱⁿ⁻¹⁾ - 1/t⁽ⁱⁿ⁻¹⁾ ]
= -1/(n - 1) + 1/(n - 1) lim(t→∞) 1/t⁽ⁱⁿ⁻¹⁾

When n > 1, lim(t→∞) 1/t⁽ⁱⁿ⁻¹⁾ = 0
When n < 1, lim(t→∞) 1/t⁽ⁱⁿ⁻¹⁾ →∞

For n = 1,
∫(0,1) 1/x dx
= lim(t→∞) ∫(1,t) 1/x dt
= lim(t→∞) ∫(1,t) d( ln x )
= lim(t→∞) ln t - ln 1
= ∞

∴ n > 1

????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

3. ∫(1,∞) 1/xⁿ dx

For n ≠ 1,
∫(0,∞) 1/xⁿ dx
= lim(t→∞) ∫(1/t,t) 1/xⁿ dx
= 1/(n - 1) lim(t→∞) ∫(1/t,t) d(-1/x⁽ⁱⁿ⁻¹⁾)
= -1/(n - 1) lim(t→∞) [ 1/(1/t)⁽ⁱⁿ⁻¹⁾ - 1/t⁽ⁱⁿ⁻¹⁾ ]
= -1/(n - 1) lim(t→∞) [ t⁽ⁱⁿ⁻¹⁾ - 1/t⁽ⁱⁿ⁻¹⁾ ]

When n > 1, lim(t→∞) t⁽ⁱⁿ⁻¹⁾ →∞
When n < 1, lim(t→∞) 1/t⁽ⁱⁿ⁻¹⁾ →∞

For n = 1,
∫(0,∞) 1/xⁿ dx
= lim(t→∞) ∫(1/t,t) 1/x dt
= lim(t→∞) ∫(1/t,t) d( ln x )
= lim(t→∞) ln t - ln (1/t)
= 2 lim(t→∞) ln t
= ∞

∴ There are no values of n so that the improper integral converges.

????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

相片：
我們發生某些問題，請再試一次。

1)
定積分(0->1)1/x^n dx
=[x^(n+1)]/(n+1) |(0->1)
=0+1/(n+1)
=1/(n+1)

2)
定積分(1->無限)1/x^n dx
=[x^(n+1)]/(n+1) |(1->無限)
=無限

3)
定積分(0->無限)1/x^n dx
=[x^(n+1)]/(n+1) |(0->無限)
=無限