Euler and Mascheroni would be proud of you!

Calculus Level 2

∫ 1 ∞ ( 1 ⌊ x ⌋ − 1 x ) d x = ? \displaystyle \int_1^\infty \left(\dfrac{1}{\lfloor x\rfloor}-\dfrac{1}{x}\right) dx~ = ~?

Submit your answer to 3 decimal places.


The answer is 0.577.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Chew-Seong Cheong
Sep 16, 2016

Relevant wiki: Euler-Mascheroni Constant

I = ∫ 1 ∞ ( 1 ⌊ x ⌋ − 1 x ) d x = lim ⁡ n → ∞ ∑ k = 1 n ∫ k k + 1 ( 1 k − 1 x ) d x where k is a positive integer. = lim ⁡ n → ∞ ∑ k = 1 n [ x k − ln ⁡ x ] k k + 1 = lim ⁡ n → ∞ ∑ k = 1 n ( k + 1 k − k k − ln ⁡ ( k + 1 ) + ln ⁡ k ) = lim ⁡ n → ∞ ( ∑ k = 1 n 1 k − ln ⁡ n ) = lim ⁡ n → ∞ ( H n − ln ⁡ n ) where H n is a harmonic number. = γ ≈ 0.577 where γ is Euler-Mascheroni constant. \begin{aligned} I & = \int_1^\infty \left(\frac 1{\lfloor x \rfloor} - \frac 1x \right) dx \\ & = \lim_{n \to \infty} \sum_{k=1}^n \int_k^{k+1} \left(\frac 1k - \frac 1x \right) dx & \small \color{#3D99F6}{\text{where }k \text{ is a positive integer.}} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \left[\frac xk - \ln x \right]_k^{k+1} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \left(\frac {k+1}k - \frac kk - \ln (k+1) + \ln k \right) \\ & = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac 1k - \ln n \right) \\ & = \lim_{n \to \infty} \left(H_n - \ln n \right) & \small \color{#3D99F6}{\text{where }H_n \text{ is a harmonic number.}} \\ & = \gamma \approx \boxed{0.577} & \small \color{#3D99F6}{\text{where }\gamma \text{ is Euler-Mascheroni constant.}} \end{aligned}

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Nice solution there! But there's a slight mistake.

It should actually be,

lim ⁡ n → ∞ ( ∑ k = 1 n 1 k − ln ⁡ n ) \displaystyle \lim_{n\to\infty}\left(\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}-\ln n\right)

Instead of,

lim ⁡ n → ∞ ∑ k = 1 n ( 1 k − ln ⁡ n ) \displaystyle \lim_{n\to\infty}\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{k}-\ln n\right)

as you've already taken the sum ∑ k = 1 n − ln ⁡ ( k + 1 k ) = − ln ⁡ n \displaystyle \sum_{k=1}^{n}-\ln\left(\dfrac{k+1}{k}\right) = -\ln n

Tapas Mazumdar - 4 years, 9 months ago

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Thanks. I have done the changes.

Chew-Seong Cheong - 4 years, 9 months ago

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