ISI 2016 B.Math Entrance (2)

Calculus Level 3

∫ 1 n ⌊ x ⌋ x − ⌊ x ⌋ d x \large \int_{1}^{n} \lfloor x \rfloor^{x - \lfloor x \rfloor} \, dx

Evaluate the integral above in terms of positive integer n > 2 n> 2 .

1 + 1 ln ⁡ 2 + 2 ln ⁡ 3 + ⋯ + n − 2 ln ⁡ ( n − 1 ) 1+\frac{1}{\ln 2}+\frac{2}{\ln 3}+\cdots+\frac{n-2}{\ln (n-1)} 1 + 2 3 ln ⁡ 2 − 2 2 ln ⁡ 2 + ⋯ + ( n − 1 ) n ln ⁡ ( n − 1 ) − ( n − 1 ) n − 1 ln ⁡ ( n − 1 ) 1+\frac{2^3}{\ln2}-\frac{2^2}{\ln2}+\cdots+\frac{(n-1)^n}{\ln(n-1)}-\frac{(n-1)^{n-1}}{\ln(n-1)} 1 2 + 2 2 3 + ⋯ + n n + 1 n + 1 \dfrac{1}{2} + \dfrac{2^2}{3}+\cdots+\dfrac{n^{n+1}}{n+1}

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Md Zuhair by Md Zuhair , 20, India

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2 solutions

Chew-Seong Cheong
Apr 14, 2017

Relevant wiki: Integration of Piecewise Functions

I = ∫ 1 n ⌊ x ⌋ x − ⌊ x ⌋ d x = ∫ 1 n ⌊ x ⌋ { x } d x where { x } is fractional part of x = ∑ k = 1 n − 1 ∫ 0 1 k x d x = 1 + ∑ k = 2 n − 1 k x ln ⁡ k ∣ 0 1 = 1 + ∑ k = 2 n − 1 k − 1 ln ⁡ k = 1 + 1 ln ⁡ 2 + 2 ln ⁡ 3 + ⋯ + n − 2 ln ⁡ ( n − 1 ) \begin{aligned} I & = \int_1^n \lfloor x \rfloor ^{x-\lfloor x \rfloor} dx \\ & = \int_1^n \lfloor x \rfloor ^{\{x\}} dx & \small \color{#3D99F6} \text{where }\{x\} \text{ is fractional part of }x \\ & = \sum_{k=1}^{n-1} \int_0^1 k^x \ dx \\ & = 1 + \sum_{k=2}^{n-1} \frac {k^x}{\ln k} \ \bigg|_0^1 \\ & = 1 + \sum_{k=2}^{n-1} \frac {k - 1}{\ln k} \\ & = \boxed{1+\dfrac 1{\ln 2} + \dfrac 2{\ln 3}+\cdots+\dfrac {n-2}{\ln(n-1)}} \end{aligned}

6 Helpful 1 Interesting 1 Brilliant 0 Confused

Md, you don't need to use \displaystyle and \dfrac within \ [ \ ]. It is automatic. Braces { } are not needed after \displaystyle and \large. You can use \to instead of \rightarrow. You don't need braces if the operand is only one character. For example, \frac 12 ( 1 2 \frac 12 ), \sqrt 2 ( 2 \sqrt 2 ), \frac 1{\ln 2} ( 1 ln ⁡ 2 \frac 1{\ln 2} ), \sum_1^n ( ∑ 1 n \sum_1^n ).

Chew-Seong Cheong - 4 years, 1 month ago
Tapas Mazumdar
Apr 14, 2017

Relevant wiki: Integration of Piecewise Functions

∫ 1 n ⌊ x ⌋ x − ⌊ x ⌋ = ∑ t = 1 n − 1 ∫ t t + 1 t x − t d x = ∫ 1 2 1 x − 1 d x + ∑ t = 2 n − 1 ∫ t t + 1 t x − t d x = 1 + ∑ t = 2 n − 1 { ( t x − t ln ⁡ t ) ∣ x = t x = t + 1 } = 1 + ∑ t = 2 n − 1 t − 1 ln ⁡ t = 1 + 1 ln ⁡ 2 + 2 ln ⁡ 3 + 3 ln ⁡ 4 + ⋯ + n − 2 ln ⁡ ( n − 1 ) \begin{aligned} \displaystyle \int_1^n {\lfloor x \rfloor}^{x - \lfloor x \rfloor} &= \sum_{t=1}^{n-1} \int_t^{t+1} t^{x-t} \, dx \\ &= \int_1^2 1^{x-1} \, dx + \sum_{t=2}^{n-1} \int_t^{t+1} t^{x-t} \, dx \\ &= 1 + \sum_{t=2}^{n-1} \left\{ \left( \dfrac{t^{x-t}}{\ln t} \right) {\huge |}_{x=t}^{x=t+1} \right\} \\ &= 1 + \sum_{t=2}^{n-1} \dfrac{t-1}{\ln t} \\ &= 1 + \dfrac{1}{\ln 2} + \dfrac{2}{\ln 3} + \dfrac{3}{\ln 4} + \cdots + \dfrac{n-2}{\ln (n-1)} \end{aligned}

5 Helpful 1 Interesting 1 Brilliant 0 Confused

The correct option has a typo here

1 + 1 ln ⁡ 2 + 2 ln ⁡ 3 + 4 ln ⁡ 4 + ⋯ + n − 2 ln ⁡ ( n − 1 ) 1 + \dfrac{1}{\ln 2} + \dfrac{2}{\ln 3} + {\color{#D61F06} \dfrac{4}{\ln 4}} + \cdots + \dfrac{n-2}{\ln (n-1)}

Tapas Mazumdar - 4 years, 1 month ago

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Thanks. I see that this problem has been edited. Those who previously answered XXX has been marked correct.

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