Golden ratio

Integral that relates the secant function and the Golden ratio

Today we repost here the solution to this integral posted on Twitter by @integralsbot: \[\int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \phi \] In the proof we made use of the integral representation of the $\sec x$ function, a trick that we have used extensively in other occasions (link 1, link 2)

Proof


\[ I= \int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \frac{1}{\pi}\int_{0}^{\infty} \frac{y^{\frac{4}{5}}}{1+y^2} dw \quad (y \mapsto x^{\pi}) \] Recall the integral representation of $\displaystyle \sec(w)$ (proof below): \[\sec(w) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2w}{\pi}}}{1+y^2} dy \] Therefore if $ \displaystyle w= \frac{2\pi}{5}$ \[ I = \frac{1}{\pi}\int_{0}^{\infty} \frac{y^{\frac{4}{5}}}{1+y^2} dy = \frac{\sec\left(\frac{2\pi}{5}\right)}{2} = \frac{1+\sqrt{5}}{2} = \phi \] Hence \[ \boxed{\int_{0}^{\infty} \frac{x^{\frac{\pi}{5}-1}}{1+x^{2\pi}} dx = \phi }\]

Appendix

We reproduce here the proof of the integral representation of the sec function again:
The proof can be obtained trough contour integration in a branch cut but here we will use a more straightforward approach with the help of the Ramanujan's master theorem

Let $|a|<1$

Recall \[\frac{1}{1+y^2} = \int_{0}^{1} t^{y^2} dt \] Therefore \begin{align*} \int_{0}^{\infty}\frac{y^a}{1+y^2}dy =& \int_{0}^{\infty}\int_{0}^{1} y^a t^{y^2} dtdy\\ =&\int_{0}^{1} \int_{0}^{\infty} y^a t^{y^2} dydt\\ =& \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt \quad (w \mapsto y^2) \end{align*} The integral $\displaystyle \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw$ is the Mellin transform of $\displaystyle t^w$ at $\displaystyle \frac{a+1}{2}$ Recall that \[t^w = e^{w\ln(t)} = \sum_{n=0}^{\infty} \frac{\ln^n(t)w^n}{n!} = \sum_{n=0}^{\infty} \frac{(-\ln(t))^n(-w^n)}{n!} \] Therefore, by Ramanujan's master theorem \[ \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw = \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} \] Therefore \begin{align*} \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt =& \frac{1}{2}\int_{0}^{1} \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}v^{-\frac{a+1}{2}}e^{-v} dv \quad (v\mapsto -ln(t))\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) \end{align*} Finally, recall the Euler's reflection formula \[\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin \pi z} = \pi \csc \pi z \] Therefore \[\int_{0}^{\infty}\frac{y^a}{1+y^2}dy = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \frac{\pi}{2} \csc\left(\pi \frac{1+a}{2}\right)=\frac{\pi}{2} \sec\left(\pi \frac{a}{2}\right)\] If we let $\displaystyle x = \pi \frac{a}{2}$ then $\displaystyle a=\frac{2x}{\pi}$. Then \[\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}\]