An interesting integral

Calculus Level 2

∫ 0 ∞ ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x d x = ? \large \int_0^\infty\frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}\, \mathrm dx = \, ?

The integral above has a closed form. Evaluate this integral and give your answer to three decimal places.


The answer is 6.283185307.

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Daniel Xiang by Daniel Xiang , 21, China

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3 solutions

Chew-Seong Cheong
Feb 12, 2018

Relevant wiki: Integration Tricks

I = ∫ 0 ∞ ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x d x = ∫ 0 ∞ ln ⁡ ( 1 + x 11 ) − ln ⁡ ( 1 + x 3 ) ( 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 11 ) − ln ⁡ ( 1 + x 3 ) ( 1 + x 2 ) ln ⁡ x + ln ⁡ ( 1 + 1 x 11 ) − ln ⁡ ( 1 + 1 x 3 ) x 2 ( 1 + 1 x 2 ) ln ⁡ 1 x ) d x = 1 2 ∫ 0 ∞ ( ln ⁡ ( 1 + x 11 ) − ln ⁡ ( 1 + x 3 ) ( 1 + x 2 ) ln ⁡ x + ln ⁡ ( 1 + x 11 ) − ln ⁡ ( 1 + x 3 ) − ln ⁡ x 11 + ln ⁡ x 3 ( 1 + x 2 ) ( − ln ⁡ x ) ) d x = 1 2 ∫ 0 ∞ ln ⁡ x 11 − ln ⁡ x 3 ( 1 + x 2 ) ln ⁡ x d x = 1 2 ∫ 0 ∞ 8 ln ⁡ x ( 1 + x 2 ) ln ⁡ x d x = 4 ∫ 0 ∞ 1 1 + x 2 d x = 4 tan ⁡ − 1 x ∣ 0 ∞ = 2 π ≈ 6.283 \begin{aligned} I & = \int_0^\infty \frac {\ln \frac {1+x^{11}}{1+x^3}}{\left(1+x^2\right)\ln x} dx \\ & = \int_0^\infty \frac {\ln (1+x^{11})- \ln (1+x^3)}{\left(1+x^2\right)\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^{11})- \ln (1+x^3)}{\left(1+x^2\right)\ln x} + {\color{#3D99F6}\frac {\ln \left(1+\frac 1{x^{11}}\right)- \ln \left(1+\frac 1{x^3}\right)}{x^2\left(1+\frac 1 {x^2} \right)\ln \frac 1x}} \right) dx \\ & = \frac 12 \int_0^\infty \left(\frac {\ln (1+x^{11})- \ln (1+x^3)}{\left(1+x^2\right)\ln x} + \frac {\ln (1+x^{11})- \ln (1+x^3) - \ln x^{11} + \ln x^3}{\left(1+ x^2 \right)(-\ln x)} \right) dx \\ & = \frac 12 \int_0^\infty \frac {\ln x^{11}-\ln x^3}{\left(1+x^2\right)\ln x} dx = \frac 12 \int_0^\infty \frac {8\ln x}{\left(1+x^2\right)\ln x} dx = 4 \int_0^\infty \frac 1 {1+x^2}dx \\ & = 4 \tan^{-1} x \ \bigg|_0^\infty = 2 \pi \approx \boxed{6.283} \end{aligned}

58 Helpful 0 Interesting 1 Brilliant 0 Confused
Daniel Xiang
Feb 11, 2018

for a function f : R → R f: \mathbb{R}\rightarrow\mathbb{R} , the integral

∫ 0 ∞ f ( x ) d x = ∫ 0 1 f ( x ) d x + ∫ 1 ∞ f ( x ) d x \displaystyle \int_0^\infty f(x)\mathrm dx = \int_0^1 f(x)\mathrm dx + \int_1^\infty f(x)\mathrm dx

and with the substitution u = 1 x \displaystyle u = \frac{1}{x} we have

∫ 1 ∞ f ( x ) d x = ∫ 0 1 1 u 2 f ( 1 u ) d u \displaystyle \int_1^\infty f(x)\mathrm dx = \int_0^1 \frac{1}{u^2}f\left(\frac{1}{u}\right)\mathrm du

(Note that u \displaystyle u is a dummy variable)

applying this, we have

∫ 0 ∞ ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x d x = ∫ 0 1 ( ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x ) d x + ∫ 0 1 ( ln ⁡ 1 + x − 11 1 + x − 3 ( 1 + x − 2 ) x 2 ln ⁡ x − 1 ) d x = ∫ 0 1 ( ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x − ln ⁡ x − 8 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x ) d x = ∫ 0 1 ln ⁡ 1 + x 11 1 + x 3 + 8 ln ⁡ x − ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x d x = 8 ∫ 0 1 d x 1 + x 2 = 8 [ arctan ⁡ x ] 0 1 = 2 π \begin{aligned}\int_0^\infty\frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}\mathrm dx&=\int_0^1\left(\frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}\right)\mathrm dx+\int_0^1\left(\frac{\ln\frac{1+x^{-11}}{1+x^{-3}}}{(1+x^{-2})x^2\ln x^{-1}}\right)\mathrm dx\\&=\int_0^1\left(\frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}-\frac{\ln x^{-8}\frac{1+x^{11}}{1+x^{3}}}{(1+x^{2})\ln x}\right)\mathrm dx\\&=\int_0^1\frac{\ln\frac{1+x^{11}}{1+x^3}+8\ln x-\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x}\mathrm dx\\&=8\int_0^1\frac{\mathrm dx}{1+x^2}=8\bigg[\arctan x\bigg]_0^1=\boxed{2\pi}\end{aligned}

31 Helpful 0 Interesting 0 Brilliant 0 Confused

So this is approached best by U subbing? I'm still struggling with this problem. Help!!!

Laquita Jackson - 3 years, 3 months ago

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Yes, you can use the substitution u = 1 x \displaystyle u = \frac{1}{x} to show that

∫ 1 ∞ f ( x ) d x = ∫ 1 1 ∞ f ( 1 u ) d ( 1 u ) = − ∫ 1 0 1 u 2 f ( 1 u ) d u = ∫ 0 1 1 u 2 f ( 1 u ) d u \displaystyle \int_1^\infty f(x)\mathrm dx = \int_1^{\frac{1}{\infty}} f\left( \frac{1}{u}\right) \mathrm d \left( \frac{1}{u}\right) = -\int_1^0 \frac{1}{u^2}f\left(\frac{1}{u}\right)\mathrm du = \int_0^1 \frac{1}{u^2}f\left(\frac{1}{u}\right)\mathrm du

In definite integrals the variable is only a dummy variable, so

∫ 0 1 1 u 2 f ( 1 u ) d u = ∫ 0 1 1 x 2 f ( 1 x ) d x \displaystyle \int_0^1 \frac{1}{u^2}f\left(\frac{1}{u}\right)\mathrm du = \int_0^1 \frac{1}{x^2}f\left(\frac{1}{x}\right)\mathrm dx

By ∫ 0 ∞ f ( x ) d x = ∫ 0 1 f ( x ) d x + ∫ 1 ∞ f ( x ) d x \displaystyle \int_0^\infty f(x)\mathrm dx = \int_0^1 f(x)\mathrm dx + \int_1^\infty f(x)\mathrm dx we know that

∫ 0 ∞ f ( x ) d x = ∫ 0 1 ( f ( x ) + 1 x 2 f ( 1 x ) ) d x \displaystyle \int_0^\infty f(x)\mathrm dx = \int_0^1 \left( f(x) + \frac{1}{x^2} f\left(\frac{1}{x}\right)\right) \mathrm dx

Substituding f ( x ) = ln ⁡ 1 + x 11 1 + x 3 ( 1 + x 2 ) ln ⁡ x \displaystyle f(x) = \frac{\ln\frac{1+x^{11}}{1+x^3}}{(1+x^2)\ln x} and you can finish the rest :)

Daniel Xiang - 3 years, 3 months ago

i struggle with motivation in solving complex integrals without backstory on how they might pertain to something someone encountered in the path towards some goal

Charlie Recchia - 3 years, 1 month ago
Swetank Sharan
Feb 22, 2018

Better to solve this by putting x=1/t and then we can solve it easily

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