## Archive for December, 2013

Due to the holiday season, all my students had gone to vacation so I have a day off playing facebook and saw this integral. I decided to type this up since I haven’t updated the blog recently.

Solution:

integration by parts,

then apply long division onto the integrand,

Now, here is the hard part of the whole problem. We need to integrate this

Here we need to break it into two integrals,

The first integral, let the substitution

Now the harder one, the second integral,

By partial fraction we have,

Hence the integral becomes,

Therefore,

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

and hence,

So if we make a substitution of

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,

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 and because the expression looks somehow similar to the solution of the quadratic equation. The quadratic equation states that

and let’s try to square

This is nice because and we can let

hence we have

So the upper limit is

Now the integrand, first we divide top and bottom by

then with the following substitution,

🙂