Will it converge?

Calculus Level 3

Which of the following integrals converge?

I. ∫ 0 ∞ ln ⁡ ( x ) x 2 + 1 d x \displaystyle{\int_0^\infty \frac{\ln(x)}{x^2 + 1} \, dx}

II. ∫ 1 ∞ 1 ( x − 1 ) 2 d x \displaystyle{\int_1^\infty \frac{1}{(x-1)^2} \, dx}

III. ∫ 0 π / 2 tan ⁡ ( x ) d x \displaystyle{\int_0^{\pi/2} \sqrt{\tan(x)} \, dx}

I I and II I and III I, II, and III II

Henry Maltby by Henry Maltby , 26, USA

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1 solution

Henry Maltby
May 4, 2016

Relevant wiki: Improper Integrals

I. This integral converges. Note that the integrand is defined at neither endpoint, but with the u u -substitution u = 1 x u = \tfrac{1}{x} , ∫ 0 1 ln ⁡ ( x ) x 2 + 1 d x = ∫ 1 ∞ − ln ⁡ ( u ) u 2 + 1 d x . \int_0^1 \frac{\ln(x)}{x^2 + 1} \, dx = \int_1^\infty - \frac{\ln(u)}{u^2 + 1} \, dx.

Furthermore, note that ∫ 1 ∞ ln ⁡ ( x ) x 2 + 1 d x < ∫ 1 ∞ x x 2 + 1 d x < ∫ 1 ∞ 1 x 3 / 2 d x \int_1^\infty \frac{\ln(x)}{x^2 + 1} \, dx < \int_1^\infty \frac{\sqrt{x}}{x^2 + 1} \, dx < \int_1^\infty \frac{1}{x^{3/2}} \, dx converges by the p p -series test. So, not only does the integral in question converge, but it converges to ∫ 0 1 ln ⁡ ( x ) x 2 + 1 d x + ∫ 1 ∞ ln ⁡ ( x ) x 2 + 1 d x = 0. \int_0^1 \frac{\ln(x)}{x^2 + 1} \, dx + \int_1^\infty \frac{\ln(x)}{x^2 + 1} \, dx = 0.

II. This integral does not converge. Note that the integrand is defined at neither endpoint, and with the u u -substitution u = 1 x − 1 u = \frac{1}{x - 1} ∫ 1 2 1 ( x − 1 ) 2 d x = ∫ 1 ∞ 1 d x \int_1^2 \frac{1}{(x-1)^2} \, dx = \int_1^\infty 1 \, dx does not converge.

III. This integral converges. Note that the integrand is undefined at its right endpoint, yet with the u u -substitution u = tan ⁡ ( x ) u = \tan(x) , ∫ 0 π / 2 tan ⁡ ( x ) d x = ∫ 0 ∞ u u 2 + 1 d u < ∫ 0 ∞ 1 u 3 / 2 d u \int_0^{\pi/2} \sqrt{\tan(x)} \, dx = \int_0^\infty \frac{\sqrt{u}}{u^2 + 1} \, du < \int_0^\infty \frac{1}{u^{3/2}} \, du which converges by the p p -series test.

Therefore, the answer is I and III \boxed{\text{I and III}} .

11 Helpful 1 Interesting 0 Brilliant 0 Confused

Instead of doing the p-series test and finding a bound that always works, I find that it's easier to consider the power-series expansion about the undefined point, and see that it doesn't contain any x a x^a terms with − 1 ≤ a < 0 -1 \leq a < 0 .

Calvin Lin Staff - 5 years ago

Last argument is wrong. The last integral written does not converge, because the integration starts from zero. Argument is wrong but solution is correct.

Milan Stojanovic - 1 year, 1 month ago

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