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, $\frac{1}{\phi}=\frac{\sqrt{5}-1}{2}$.


I evaluated it easy enough using parts. I arrived at


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.


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

  • $\begingroup$ 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. $\endgroup$ – Eric Naslund Jul 13 '11 at 15:15
  • $\begingroup$ One of those "secret source" problems! $\endgroup$ – 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!


Could we have noticed this from the beginning? Yes. Notice that your integral can be rewritten

$$\int_{-\infty}^0 [n(n+1)]^x dx$$

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

$$\int_{-\infty}^0 dx$$

which is clearly infinite.

  • $\begingroup$ Duh, I should have seen that. Thank you very much. $\endgroup$ – Cody Jul 13 '11 at 19:15

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