This integral does not converge as x → ∞ x \to \infty

Calculus Level 5

For x > 0 x > 0 , define f ( x ) f(x) as follows: f ( x ) = ∫ 0 x ( t 2 + 1 − t ) d t . f(x) = \int_{0}^{x} \left( \sqrt{t^2+1} - t \right) \, dt. Let L = lim ⁡ x → ∞ ( f ( x ) − ln ⁡ x 2 ) . L = \displaystyle\lim_{x\to\infty} \left(f(x) - \dfrac{\ln x}{2}\right).

If L = a + ln ⁡ b c L = \dfrac{a + \ln{b}}{c} for positive integers a , b , a, b, and c c with gcd ⁡ ( a , c ) = 1 \gcd(a,c) = 1 , then find the value of a + b + c a + b + c .


The answer is 9.

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Ariel Gershon by Ariel Gershon , 26, Canada

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1 solution

First Last
Nov 28, 2016

Integrating ∫ 0 x 1 + t 2 d t \displaystyle\int_{0}^{x}\sqrt{1+t^2}dt by parts results in 1 2 ( x 1 + x 2 − ln ⁡ ( 1 + x 2 − x ) ) \displaystyle \frac{1}{2}\big(x \sqrt{1+x^2}-\ln{(\sqrt{1+x^2}-x) \big)} . Thus,

f ( x ) − ln ⁡ x 2 = 1 2 ( x 1 + x 2 − ln ⁡ ( 1 + x 2 − x ) − x 2 − ln ⁡ x ) = 1 2 ( x 1 + x 2 − x 2 − ln ⁡ ( x 1 + x 2 − x 2 ) ) f(x) - \frac{ \ln x } { 2 } = \displaystyle\frac{1}{2}\big( x \sqrt{1+x^2}- \ln{( \sqrt{1+x^2}-x)} -x^2- \ln{x} \big) \\ = \displaystyle\frac{1}{2}\big( x \sqrt{1+x^2}- x^2 - \ln{(x \sqrt{1+x^2}-x^2)} \big)

Let g ( x ) = x 1 + x 2 − x 2 g(x) = x \sqrt{ 1 + x^2} - x^2 , then f ( x ) − ln ⁡ x 2 = g ( x ) − ln ⁡ g ( x ) 2 f(x) - \frac{ \ln x } { 2 } = \frac{ g(x) - \ln g(x) } { 2 } .

We seek to find the limit of g ( x ) g(x) by rewriting g(x) as lim ⁡ x → ∞ g ( x ) = lim ⁡ x → ∞ 1 + 1 x 2 − 1 1 x 2 \displaystyle\lim_{x\to\infty}g(x) = \lim_{x\to\infty} \frac{\sqrt{1+\frac{1}{x^2}}-1}{\frac{1}{x^2}} . By applying L'Hopital's Rule, we obtain

lim ⁡ x → ∞ − 1 x 3 1 + 1 x 2 − 2 x 3 = 1 2 \lim_{x\to\infty}\frac{\frac{-1}{x^3\sqrt{1+\frac{1}{x^2}}}}{\frac{-2}{x^3}} = \frac{1}{2}

Hence, lim ⁡ x → ∞ f ( x ) − ln ⁡ x 2 = 1 2 × ( 1 2 − ln ⁡ 1 2 ) = 1 + ln ⁡ 4 4 \lim_{x\rightarrow \infty } f(x) - \frac{ \ln x } { 2 } = \displaystyle\frac{1}{2}\times\big( \frac{1}{2}-\ln{\frac{1}{2}}\big) = \frac{1+\ln{4}}{4} .

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Nice solution! However you don't need L'Hopital's rule:

lim ⁡ x → ∞ ( x x 2 + 1 − x 2 ) = lim ⁡ x → ∞ x x 2 + 1 + x = lim ⁡ x → ∞ 1 1 + 1 x 2 + 1 = 1 2 \begin{aligned} \lim_{x\to\infty} \left(x\sqrt{x^2+1} - x^2\right) & = \lim_{x\to\infty}\dfrac{x}{\sqrt{x^2+1}+x} \\ & = \lim_{x\to\infty} \dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1 } \\ & = \dfrac{1}{2} \end{aligned}

Ariel Gershon - 4 years, 6 months ago

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I like this trick with multiplying by the conjugate!

E.g. Showing by first principle that for f ( x ) = x f(x) = \sqrt{x} ,
f ′ ( x ) = lim ⁡ h → 0 x + h − x h = lim ⁡ h → 0 1 x + h + x = 1 2 x . f'(x) = \lim_{h\rightarrow 0 } \frac{ \sqrt{ x+h } - \sqrt{x} } { h } =\lim_{h\rightarrow 0 } \frac{ 1 } { \sqrt{ x+h } + \sqrt{x} } = \frac{1}{ 2 \sqrt{x} } .

Calvin Lin Staff - 4 years, 6 months ago

just the same approach , i have used .

Rudraksh Sisodia - 4 years, 6 months ago

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