Fee, Phi, Fo, Sum

Time to return to integrals for a while and practice my LaTeX skills. I came across an interesting video on YouTube recently that investigated the following integral:

$$\int_0^{\infty} \frac{1}{(1+x^{\phi})^{\phi}}\, \text{d}x$$ Figure 1 shows that the result is 1 using GeoGebra, using 1000 as the limit of integration rather than infinity, because the program doesn't seem to cope with the latter. 

Figure 1

The program however, gives no clue as to how this result was arrived at, although it looks to be likely given the appearance of the area under the curve. Symbolab isn't much help. See Figure 2.

Figure 2

The online integral calculator wasn't any help either. See Figures 3 and 4.

Figure 3

Figure 4

So to show why the integral is equal to 1, I'll basically follow the steps as outlined in the video. Let's remember that $\phi$ is the solution to the equation: $$\begin{align}x^2-x-1&=0\\ \text{where }x&=\frac{1+\sqrt{5}}{2}=\phi\\ \text{also } 1&=\phi^2-\phi\\ \text{and } \frac{1}{\phi}&=\phi-1 \end{align}$$ To integrate, the following substitution is used: $$\begin{align}u&=x^{\phi}\\ \text{d}u&=\phi x^{\phi-1} \text{d}x\\ \frac{\text{d}u}{\phi x^{\phi-1}}&=\text{d}x \end{align}$$ The limits of integration don't need to be changed because $u=0$ when $x=0$ and $u \rightarrow \infty$ as $x \rightarrow \infty$. So the integral becomes: $$\int_0^{\infty} \frac{\text{d} u}{(1+u)^{\phi} \, \phi \, x^{\phi-1}}$$ However $\dfrac{u}{x}=x^{\phi-1}$ because $u=x^{\phi}$ and so the integral becomes: $$\frac{1}{\phi} \, \int_0^{\infty} \frac{x}{(1+u)^{\phi} \, u} \text{d}u$$ Because $u=x^{\phi}$, we can write $u^{1/\phi}=x$, thus the integral now becomes: $$\frac{1}{\phi} \, \int_0^{\infty} \frac{u^{1/\phi}}{(1+u)^{\phi} \, u} \text{d}u$$ Now we saw earlier that $\dfrac{1}{\phi}=\phi-1$ and we can use this fact in transforming the integral even further. We can now write it as: $$\begin{align} \frac{1}{\phi} \, \int_0^{\infty} \frac{u^{\phi-1}}{(1+u)^{\phi} \, u} \text{d}u &= \frac{1}{\phi} \, \int_0^{\infty} \frac{u^{\phi-1-1}}{(1+u)^{\phi}} \text{d}u \\ &= \frac{1}{\phi} \, \int_0^{\infty} \frac{u^{\phi-1-1}}{(1+u)^{\phi-1+1}} \text{d}u \end{align}$$ Now this last transformation of the integral may seem strange but it's usefulness becomes apparent once we bear in mind the beta function, defined as: $$\beta(x,y)=\int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}} \text{d}u=\frac{\Gamma(x) \, \Gamma(y)}{\Gamma(x+y)}$$ In this beta function, if we let $x=\phi-1$ and $y=1$, our integral now becomes: $$\begin{align} \frac{1}{\phi} \beta(\phi-1,1)&=\frac{1}{\phi} \, \frac{\Gamma(\phi-1) \, \Gamma(1)}{\Gamma(\phi)}\\ &=\frac{1}{\phi} \, \frac{(\phi-2)! \, 0!}{(\phi-1)!}\\&= \frac{1}{\phi} \, \frac{1}{\phi-1} \\ &= \frac{1}{\phi^2-\phi}\\ &=1 \end{align}$$ Thus we have confirmed that the integral does indeed evaluate to 1. The transformation of the gamma function to the factorial is achieved via the fact that $\Gamma(x)=(x-1)!$.

Here is the actual video embedded into this blog. It covers precisely the same steps and my main purpose in creating this blog is simply to prevent my LaTeX skills from becoming too rusty.