Integration Bee

Calculus Level 4

0 ln ( 1 + x n ) ln 2 ( 1 + x 2 ) ln x d x = 2 π \int_{0}^{\infty} \frac{\ln(1+x^n)-\ln2}{(1+x^2)\ln x} dx=2\pi n = ? n=?


The answer is 8.

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2 solutions

Chew-Seong Cheong
Jan 29, 2019

I = 0 ln ( 1 + x n ) ln 2 ( 1 + x 2 ) ln x d x Using 0 f ( x ) d x = 0 f ( 1 x ) x 2 d x = 1 2 0 ( ln ( 1 + x n ) ln 2 ( 1 + x 2 ) ln x + ln ( 1 + 1 x n ) ln 2 x 2 ( 1 + 1 x 2 ) ln 1 x ) d x = 1 2 0 ( ln ( 1 + x n ) ln 2 ( 1 + x 2 ) ln x + ln ( x n + 1 x n ) ln 2 x 2 ( x 2 + 1 x 2 ) ( ln x ) ) d x = 1 2 0 ( ln ( 1 + x n ) ln 2 ( 1 + x 2 ) ln x ln ( 1 + x n ) ln x n ln 2 ( 1 + x 2 ) ln x ) d x = 1 2 0 ln x n ( 1 + x 2 ) ln x d x = 1 2 0 n ln x ( 1 + x 2 ) ln x d x = 1 2 0 n 1 + x 2 d x = n 2 tan 1 x 0 = n π 4 \begin{aligned} I & = \int_0^\infty \frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x) \ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2} dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} + \frac {\ln \left(1+ \frac 1{x^n}\right) - \ln 2}{x^2\left(1+\frac 1{x^2} \right)\ln \frac 1x} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} + \frac {\ln \left(\frac {x^n+1}{x^n}\right) - \ln 2}{x^2\left(\frac {x^2+1}{x^2} \right) (-\ln x)} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^n) - \ln 2}{(1+x^2)\ln x} - \frac {\ln (1+x^n) - \ln x^n - \ln 2}{(1+x^2)\ln x} \right) dx \\ & = \frac 12 \int_0^\infty \frac {\ln x^n}{(1+x^2) \ln x} dx = \frac 12 \int_0^\infty \frac {n \ln x}{(1+x^2) \ln x} dx \\ & = \frac 12 \int_0^\infty \frac n{1+x^2} dx = \frac n2 \tan^{-1} x \ \bigg|_0^\infty = \frac {n\pi}4 \end{aligned}

Therefore, n π 4 = 2 π n = 8 \dfrac {n\pi}4 = 2\pi \implies n = \boxed 8 .

Albert Yiyi
Jan 29, 2019

hint1: let t = 1 x t = \frac{1}{x}

hint2: 0 n ln t ln ( 1 + t n ) + ln 2 ( 1 + t 2 ) ln t d t \int_{0}^{\infty} \frac{n \ln t - \ln(1+t^n) + \ln2}{(1+t^2)\ln t} dt

hint3: 1 4 n π = 2 π \frac{1}{4} n \pi = 2\pi

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