A floor function integral

Calculus Level pending

Let ⌊ n ⌋ \lfloor n\rfloor be the greatest integer less than or equal to x x . For example, ⌊ 2 ⌋ = 2 \lfloor 2\rfloor=2 and ⌊ π ⌋ = 3 \lfloor \pi\rfloor =3 .

What is ∫ 0 n ⌊ 2 x ⌋ d x \int_0^n\lfloor2^x\rfloor dx ?

Note that k ! = k ( k − 1 ) ( k − 2 ) . . . ( 3 ) ( 2 ) ( 1 ) { k ∈ Z + } k!=k(k-1)(k-2)...(3)(2)(1)\ \left \{k\in\mathbb{Z}^+\right \} .

log ⁡ 2 [ ( 2 n − 1 ) ! ] \log_2\left [(2^n-1)!\right ] n 2 n − log ⁡ 2 [ ( 2 n ) ! ] n2^n-\log_2\left [(2^n)!\right ] n 2 n − log ⁡ 2 [ ( 2 n − 1 ) ! ] n2^n-\log_2\left [(2^n-1)!\right ] n 2 n n2^n log ⁡ 2 ( 2 n ) ! \log_2(2^n)!

Matthew Christopher by Matthew Christopher , 17, United Kingdom

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1 solution

⌊ 2 x ⌋ \lfloor 2^x\rfloor takes on all integer values from 1 1 to 2 n − 1 2^n-1 in the domain interval [ 0 , n ) [0, n) .

[https://www.desmos.com/calculator/ighh5kisms](https://www.desmos.com/calculator/ighh5kisms) https://www.desmos.com/calculator/ighh5kisms In the image, the blue graph is 2 x 2^x and the red graph is ⌊ 2 x ⌋ \lfloor 2^x\rfloor .

The integral is the sum of areas of rectangles that increase in height by 1 1 and decrease in width, as follows:

∫ 0 n ⌊ 2 x ⌋ d x = 1 ( log ⁡ 2 2 − log ⁡ 2 1 ) + 2 ( log ⁡ 2 3 − log ⁡ 2 2 ) + . . . + ( 2 n − 1 ) ( log ⁡ 2 2 n − log ⁡ 2 ( 2 n − 1 ) ) = − log ⁡ 2 1 − log ⁡ 2 3 − . . . − log ⁡ 2 ( 2 n − 1 ) + ( 2 n − 1 ) log ⁡ 2 ( 2 n ) = 2 n log ⁡ 2 ( 2 n ) − ( log ⁡ 2 1 + log ⁡ 2 3 + . . . + log ⁡ 2 ( 2 n ) ) = n 2 n − log ⁡ 2 [ 1 ⋅ 2 ⋅ 3 ⋅ . . . ⋅ ( 2 n − 2 ) ( 2 n − 1 ) ( 2 n ) ] = n 2 n − log ⁡ 2 [ ( 2 n ) ! ] \begin{aligned}\int_0^n\lfloor2^x\rfloor dx&=1(\log_22-\log_21)+2(\log_23-\log_22)+...+(2^n-1)(\log_2{2^n}-\log_2(2^n-1))\\&=-\log_21-\log_23-...-\log_2{(2^n-1)}+(2^n-1)\log_2(2^n)\\&=2^n\log_2(2^n)-\left (\log_21+\log_23+...+\log_2(2^n)\right )\\&=n2^n-\log_2\left [1\cdot2\cdot3\cdot...\cdot(2^n-2)(2^n-1)(2^n)\right ]\\&=\color{#20A900}{\boxed{n2^n-\log_2\left [ (2^n)!\right ]}}\end{aligned}

1 Helpful 1 Interesting 2 Brilliant 0 Confused

Really nice write-up. One small additional point that may be interesting: if you use Stirling's approximation for ( 2 n ) ! (2^n)! , you find n 2 n − log ⁡ 2 [ ( 2 n ) n ] ≈ n 2 n − ( n 2 n − 2 n log ⁡ e 2 ) = 2 n log ⁡ e 2 n 2^n - \log_2 \left[ \left( 2^n \right)^n \right] \approx n 2^n - \left(n 2^n - \frac{2^n}{\log_e 2} \right) = \frac{2^n}{\log_e 2}

...which you might recognise as the result of a very similar integral to the one in the question. This suggests an alternative approach, but I'm not sure it'd be enough to separate the two similar answer options.

Chris Lewis - 10 months, 2 weeks ago

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