# How is this integral related to the Golden Ratio?

I ran across an interesting integral and I am wondering how in the world it could relate to the Golden Ratio, $\frac{1}{\phi}$.

The problem says the solution must include the Golden Ratio, $\frac{1}{\phi}=\frac{\sqrt{5}-1}{2}$.

$\int\limits_{-\infty}^{0}n^{x}(n+1)^{x}dx$

I evaluated it easy enough using parts. I arrived at

$\frac{1}{ln(n^{2}+n)}$.

But, it escapes me is how this can be written in terms of the aforementioned Golden Ratio.

I found something in Excursions in Calculus by Robert Young that relates a logarithmic spiral to the Golden Ratio, but it seems rather iffy.

$\frac{1}{ln(\beta)}=\frac{\pi}{2ln(\phi)}$

Replacing beta with $n^{2}+n$ and solving for n gives a solution, but I doubt if it is correct.

I notice that n and n+1 could be somehow related to the Fibonacci sequence?.

Any thoughts are appreciated. Thanks

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It feels like we are missing part of the question. Is $n$ an arbitrary integer, or is there something else going on? As is, I think any way of relating $\frac{1}{n^2+n}$ to the golden ratio will be artificial. – Eric Naslund Jul 13 '11 at 15:15
One of those "secret source" problems! – GEdgar Jul 13 '11 at 15:24

You have an expression for the integral, which is $I(n) = 1/\log n(n+1)$. A good thing to check is whether this integral is finite or not. When would it not be finite? Precisely when the denominator is zero, or when

$$\log n(n+1) = 0$$

which occurs when

$$n(n+1) = 1 \quad \Rightarrow\quad n^2 + n - 1 = 0$$

Replacing $n\to 1/m$ gives the equation

$$m^2 - m - 1 = 0$$

which has the well-known positive solution $m=1.618\dots$, the golden ratio.

So $n=1/\phi$ is the magic number (the 'secret sauce' as GEdgar put it) which means that your integral is not finite!

$$\int_{-\infty}^0 [n(n+1)]^x dx$$
which is all fine and dandy as long as $n(n+1) > 1$. If you're at the boundary, $n(n+1)=1$ then you are integrating
$$\int_{-\infty}^0 dx$$