Integral Integral13

Calculus Level 3

0 1 log ( 1 x 5 + 1 x 4 + 1 x 3 + 1 x 2 + 1 x + 1 ) d x = 1 2 ( π a + log ( b ) ) \large \int_0^1 \log \left(\frac{1}{x^5}+\frac{1}{x^4}+\frac{1}{x^3}+\frac{1}{x^2}+\frac{1}{ x}+1\right) \, dx=\frac{1}{2} \left(\pi \sqrt{a}+\log (b)\right)

where a a and b b are positive integers. Submit a + b a+b .


The answer is 435.

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2 solutions

Chew-Seong Cheong
Dec 25, 2017

I = 0 1 log ( 1 x 5 + 1 x 4 + 1 x 3 + 1 x 2 + 1 x + 1 ) d x = 0 1 log ( 1 + x + x 2 + x 3 + x 4 + x 5 x 5 ) d x = 0 1 log ( 1 x 6 x 5 ( 1 x ) ) d x = 0 1 log ( 1 x 6 ) d x 5 0 1 log x d x 0 1 log ( 1 x ) d x By integration by parts = x log ( 1 x 6 ) 0 1 + 0 1 6 x 6 1 x 6 d x 5 ( x log x x ) 0 1 ( 1 x ) log ( 1 x ) x 0 1 = x log ( 1 x 6 ) 0 1 6 0 1 d x + 6 0 1 1 1 x 6 d x + 6 = x log ( 1 x 6 ) 0 1 6 6 0 1 1 ( x 1 ) ( x 2 + x + 1 ) ( 1 + x ) ( x 2 x + 1 ) d x + 6 = x log ( 1 x 6 ) 0 1 6 0 1 1 6 ( 1 x 1 1 x + 1 + x 2 x 2 x + 1 x + 2 x 2 + x + 1 ) d x = x log ( 1 x 6 ) 0 1 log ( x 1 ) 0 1 + log ( x + 1 ) 0 1 1 2 0 1 2 x 1 3 x 2 x + 1 d x + 1 2 0 1 2 x + 1 + 3 x 2 + x + 1 d x = lim x 1 ( x log ( 1 x 6 ) log ( 1 x ) ) + log 2 + 1 2 log ( x 2 + x + 1 x 2 x + 1 ) 0 1 + 3 2 0 1 1 x 2 x + 1 d x + 3 2 0 1 1 x 2 + x + 1 d x = lim x 1 log ( 1 x 6 1 x ) + log 2 + 1 2 log 3 + 2 0 1 1 4 3 ( x 1 2 ) 2 + 1 d x + 2 0 1 1 4 3 ( x + 1 2 ) 2 + 1 d x = lim x 1 log ( n = 0 5 x n ) + log 2 + 1 2 log 3 + 3 1 3 1 3 1 x 2 + 1 d x + 3 1 3 3 1 x 2 + 1 d x = log 6 + log 2 + 1 2 log 3 + 3 1 3 3 1 x 2 + 1 d x = 1 2 ( 2 log 6 + 2 log 2 + log 3 ) + 3 tan 1 x 1 3 3 = 1 2 ( log ( 6 2 2 2 3 ) ) + 3 ( tan 1 3 + tan 1 1 3 ) = 1 2 log 432 + 3 ( tan 1 3 + 1 3 1 3 × 1 3 ) = 1 2 log 432 + 3 × π 2 = 1 2 ( π 3 + log 432 ) \begin{aligned} I & = \int_0^1 \log \left(\frac 1{x^5} + \frac 1{x^4} + \frac 1{x^3} + \frac 1{x^2} + \frac 1x + 1\right) dx \\ & = \int_0^1 \log \left(\frac {1+x + x^2 + x^3 + x^4 + x^5}{x^5}\right) dx \\ & = \int_0^1 \log \left(\frac {1- x^6}{x^5(1-x)}\right) dx \\ & = {\color{#3D99F6}\int_0^1 \log(1- x^6) \ dx} - 5 \int_0^1 \log x \ dx - \int_0^1 \log (1-x) \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6} x \log (1-x^6) \bigg|_0^1 + \int_0^1 \frac {6x^6}{1- x^6} dx} - 5 (x\log x- x)\bigg|_0^1 - (1-x) \log(1-x) - x \bigg|_0^1 \\ & = {\color{#3D99F6} x \log (1-x^6) \bigg|_0^1 - 6\int_0^1 dx + 6\int_0^1 \frac 1{1- x^6} dx} + 6 \\ & = {\color{#3D99F6} x \log (1-x^6) \bigg|_0^1 - 6 - 6\int_0^1\frac 1{(x-1)(x^2+x+1)(1+x)(x^2-x+1)} dx} + 6 \\ & = x \log (1-x^6) \bigg|_0^1 - 6\int_0^1 \frac 16 \left(\frac 1{x-1} - \frac 1{x+1} + \frac {x-2}{x^2-x+1} - \frac {x+2}{x^2+x+1} \right) dx \\ & = {\color{#3D99F6}x \log (1-x^6) \bigg|_0^1 - \log(x-1)\bigg|_0^1} + \log(x+1)\bigg|_0^1 - \frac 12 \int_0^1 \frac {2x-1-3}{x^2-x+1} dx + \frac 12 \int_0^1 \frac {2x+1+3}{x^2+x+1} dx \\ & = {\color{#3D99F6} \lim_{x \to 1} \left(x\log (1-x^6) - \log (1-x) \right)} + \log 2 + \frac 12 \log \left(\frac {x^2+x+1}{x^2-x+1}\right) \bigg|_0^1 + \frac 32 \int_0^1 \frac 1{x^2-x+1} dx + \frac 32 \int_0^1 \frac 1{x^2+x+1} dx \\ & = {\color{#3D99F6} \lim_{x \to 1} \log \left(\frac {1-x^6}{1-x}\right)} + \log 2 + \frac 12 \log 3 + 2 \int_0^1 \frac 1{\frac 43\left(x-\frac 12\right)^2+1} dx + 2 \int_0^1 \frac 1{\frac 43\left(x+\frac 12\right)^2+1} dx \\ & = {\color{#3D99F6} \lim_{x \to 1} \log \left(\sum_{n=0}^5 x^n\right)} + \log 2 + \frac 12 \log 3 + \sqrt 3 \int_{-\frac 1{\sqrt 3}}^{\frac 1{\sqrt 3}} \frac 1{x^2+1} dx +\sqrt 3 \int_{\frac 1{\sqrt 3}}^{\sqrt 3} \frac 1{x^2+1} dx \\ & = {\color{#3D99F6} \log 6} + \log 2 + \frac 12 \log 3 + \sqrt 3 \int_{-\frac 1{\sqrt 3}}^{\sqrt 3} \frac 1{x^2+1} dx \\ & = \frac 12 \left(2\log 6 + 2 \log 2 + \log 3\right) + \sqrt 3 \tan^{-1} x\ \bigg|_{-\frac 1{\sqrt 3}}^{\sqrt 3} \\ & = \frac 12 \left(\log \left(6^2 \cdot 2^2 \cdot 3\right) \right) + \sqrt 3 \left(\tan^{-1} \sqrt 3 + \tan^{-1} \frac 1{\sqrt 3}\right) \\ & = \frac 12 \log 432 + \sqrt 3 \left(\tan^{-1} \frac {\sqrt 3 + \frac 1{\sqrt 3}}{1 - \sqrt 3 \times \frac 1{\sqrt 3}} \right) \\ & = \frac 12 \log 432 + \sqrt 3 \times \frac \pi 2 \\ & = \frac 12 (\pi \sqrt 3 + \log 432) \end{aligned}

Therefore, a + b = 3 + 432 = 435 a+b = 3+432 = \boxed{435}

a = 3 , b = 432 a=3, b=432

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