solution to the tricky integral

Posted: December 26, 2013 in Mathematics

\displaystyle\int_{\frac{1+\sqrt5}{2}}^{\frac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}}\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1}\;dx

I believe I saw this integral 5 or 6 years ago on AoPS forum.  If my memory serves me right, it was a question from a Japanese university entrance exam.  (I could be wrong, since it’s long time ago… and I am getting the “old people memory disorder” recently…)  The first time when I saw it, I was like WTF?  The integrand already looked hard not to mention the upper limit.  It took me about a week to finally solve it.  (And I had spent a lot of time sitting in the bathroom and thinking hard on it.)  And last month when I post this on my blog, Koopa Koo solved it in a night, or maybe minutes after he saw this.  So this is the difference between a real mathematician and an amateur like me.

Anyway, let’s begin.

Actually the thing that attracted me to solve this integral is the lower limit, it is the famous golden ratio.  I just think that everything that has something to do with the golden ratio is beautiful.  As we all know (if you don’t know then now you know), golden ratio is one of the solution to the equation

x^2-x-1=0

and hence,

x-\dfrac{1}{x}=1

So if we make a substitution of

w=x-\dfrac{1}{x}, \quad dw=\left(1+\dfrac{1}{x^2}\right)\;dx

the lower limit will simplify to 1, what about the upper limit?  Certainly I don’t want to let x equal to that huge expression and then plug it into x-1/x and simplify, that’s not my style coz I am lazy.  When I look at it,

\dfrac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}

It is definitely not the golden ratio although it has a denominator of 2 and some radicals in the numerator and it’s nested too.  How about try to let b=-(1+\sqrt2) and because the expression looks somehow similar to the solution of the quadratic equation.  The quadratic equation states that

\mbox{If } ax^2+bx+c=0\mbox{ and } a\neq0, \mbox{ then } x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

and let’s try to square 1+\sqrt2

(1+\sqrt2)^2=3+2\sqrt2

This is nice because 7+2\sqrt2 = 3+2\sqrt2+4 and we can let

a=1,\quad b=-(1+\sqrt2), \quad c=-1

hence we have

x^2-(1+\sqrt2)x-1=0

x-\dfrac{1}{x}=1+\sqrt2

So the upper limit is 1+\sqrt2

Now the integrand, first we divide top and bottom by x^3

\displaystyle\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1} = \frac{\left(1+\frac{1}{x^2}\right)\left(x+2-\frac{1}{x}\right)}{x^3+14-\frac{1}{x^3}}

=\displaystyle\frac{\left(1+\frac{1}{x^2}\right)\left(x+2-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)+14}\qquad\mbox{(Difference of cube formula)}

=\displaystyle\frac{\left(1+\frac{1}{x^2}\right)\left(x-\frac{1}{x}+2\right)}{\left(x-\frac{1}{x}\right)\left(\left(x-\frac{1}{x}\right)^2+3\right)+14}\qquad\mbox{(Completing the square)}

then with the following substitution,

w=x-\dfrac{1}{x}, \quad dw=\left(1+\dfrac{1}{x^2}\right)\;dx

\displaystyle\int_{\frac{1+\sqrt5}{2}}^{\frac{1+\sqrt2+\sqrt{7+2\sqrt2}}{2}}\frac{(x^2+1)(x^2+2x-1)}{x^6+14x^3-1}\;dx

=\displaystyle\int_1^{1+\sqrt2}\frac{w+2}{w(w^2+3)+14}\;dw

=\displaystyle\int_1^{1+\sqrt2}\frac{w+2}{w^3+3w+14}\;dw

=\displaystyle\int_1^{1+\sqrt2}\frac{dw}{w^2-2w+7}

=\displaystyle\int_1^{1+\sqrt2}\frac{dw}{(w-1)^2+6}

=\displaystyle\frac{1}{6}\int_1^{1+\sqrt2}\frac{dw}{\left(\frac{w-1}{\sqrt6}\right)^2+1}

=\displaystyle\frac{\sqrt6}{6}\tan^{-1}\frac{w-1}{\sqrt6}\Bigg\vert_1^{1+\sqrt2}

=\displaystyle\frac{\sqrt6}{6}\left(\tan^{-1}\frac{1}{\sqrt3}-\tan^{-1}0\right)

=\dfrac{\sqrt6\pi}{36}

🙂

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Comments
  1. I couldn’t resist commenting. Perfectly written!

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