A curious Golden ratio integral via a quartic equation

Algebra Level 2

∫ 0 1 φ x 2 − x − φ x 2 − x − 1 φ d x = 1 φ + log ⁡ ( 3 + α 3 − α ) α \large \int_0^\frac 1\varphi \frac {x^2-x-\varphi}{x^2-x-\frac 1\varphi}\ dx = \frac 1\varphi + \frac {\log \left(\frac {3+\alpha}{3-\alpha}\right)}\alpha

The equation above holds true for real number α \alpha . Compute α 4 + 2 α 2 \alpha^4 + 2\alpha^2 .

Notation: φ \varphi denotes the golden ratio .


The answer is 19.

Srinivasa Raghava by Srinivasa Raghava , 101, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Apr 28, 2020

I = ∫ 0 1 φ x 2 − x − φ x 2 − x − 1 φ d x = ∫ 0 1 φ x 2 − x − 1 φ − 1 x 2 − x − 1 φ d x = ∫ 0 1 φ ( 1 − 1 x 2 − x − 1 φ ) d x = x ∣ 0 1 φ − ∫ 0 1 φ 1 ( x − 1 + 4 φ − 3 2 ) ( x − 1 − 4 φ − 3 2 ) d x Let α = 4 φ − 3 . = 1 φ − 2 α ∫ 0 1 φ ( 1 2 x − 1 − α − 1 2 x − 1 + α ) d x = 1 φ − 1 α [ ln ⁡ ∣ 2 x − 1 − α ∣ − ln ⁡ ∣ 2 x − 1 + α ∣ ] 0 1 φ = 1 φ + 1 α [ ln ⁡ ∣ 2 x − 1 + α ∣ ∣ 2 x − 1 − α ∣ ] 0 1 φ = 1 φ + 1 α ln ⁡ ( 3 + α 3 − α ) \begin{aligned} I & = \int_0^\frac 1\varphi \frac {x^2-x-\blue \varphi}{x^2-x-\frac 1\varphi} dx \\ & = \int_0^\frac 1\varphi \frac {x^2-x- \blue{\frac 1\varphi- 1}}{x^2-x-\frac 1\varphi} dx \\ & = \int_0^\frac 1\varphi \left(1 - \frac 1{x^2-x-\frac 1\varphi} \right) dx \\ & = x \ \bigg|_0^\frac 1\varphi - \int_0^\frac 1 \varphi \frac 1{\left(x-\frac {1+\sqrt{4\varphi-3}}2 \right)\left(x- \frac {1 -\sqrt{4\varphi-3}} 2\right)} dx & \small \blue{\text{Let }\alpha = \sqrt{4\varphi-3}.} \\ & = \frac 1 \varphi - \frac 2 \alpha \int_0^\frac 1 \varphi \left(\frac 1{2x-1-\alpha} - \frac 1{2x-1+\alpha} \right) dx \\ & = \frac 1 \varphi \blue - \frac 1 \alpha \bigg[\ln |2x-1-\alpha| - \ln |2x-1+\alpha| \bigg]_0^\frac 1\varphi \\ & = \frac 1 \varphi \red + \frac 1 \alpha \left[\ln \frac {|2x-1+\alpha|}{|2x-1-\alpha|} \right]_0^\frac 1\varphi \\ & = \frac 1\varphi + \frac 1\alpha \ln \left(\frac {3+\alpha}{3-\alpha} \right) \end{aligned}

Therefore α = 4 φ − 3 \alpha = \sqrt{4\varphi - 3} and

α 2 + 2 α 2 = ( 4 φ − 3 ) 2 + 2 ( 4 φ − 3 ) = 16 φ 2 − 24 φ + 9 + 8 φ − 6 = 16 φ + 16 − 24 φ + 9 + 8 φ − 6 = 19 \begin{aligned} \alpha^2 + 2\alpha^2 & = (4\varphi - 3)^2 + 2(4\varphi -3) \\ & = \blue{16\varphi^2} - 24\varphi + 9 + 8 \varphi - 6 \\ & = \blue{16 \varphi + 16} - 24\varphi + 9 + 8 \varphi - 6 \\ & = \boxed{19} \end{aligned}

0 Helpful 1 Interesting 2 Brilliant 0 Confused

Very well done.

Srinivasa Raghava - 1 year, 1 month ago

Log in to reply

Glad that you like it.

Chew-Seong Cheong - 1 year, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...