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I would like to show that : $$\displaystyle\int_0^\infty \frac{x^{\frac\pi5-1}}{1+x^{2\pi}} \mathrm dx=\phi$$ ($\phi$ is golden ratio)

Apparently it comes from a more general form : $$\int_0^\infty \frac{x^{\pi/k-1}}{1+x^{2\pi}}\mathrm dx=\frac{1}{2}\csc\Big( \frac{\pi}{2k}\Big)$$

Indeed for $k=5$ we find the right value, but I couldn't manage to find the solution.

My attempt :

Let $x=u^{\frac{1}{\pi}}$, and $\mathrm{d}x = \frac{1}{\pi} u^{\frac{1}{\pi}-1} \mathrm{d}u$. We have : $$\displaystyle\int_0^\infty \frac{x^{\frac\pi5-1}}{1+x^{2\pi}} \mathrm dx= \frac{1}{\pi} \displaystyle\int_0^\infty \frac{u^{\frac{1}{5\pi}-1}}{1+u^{2}} \mathrm du $$

And then I tried integration by parts but nothing was conclusive. Any idea ?

LexLarn
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  • I have a feeling using Ramanujans master theorem on the more general integral and substituting will work. I will try when I am home. – bob Aug 23 '23 at 15:33
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    You made a mistake: the numerator in the last integral should be $u^{\frac{1}{5}-1}=u^{-\frac{4}{5}}$. Then you can find its evaluation here: https://math.stackexchange.com/questions/180998/evaluating-this-integral-for-different-values-of-a-constant. – Gonçalo Aug 23 '23 at 15:40
  • Formula $(1)$ from this answer may be of use here. – robjohn Aug 23 '23 at 15:53
  • This is just part of a standard representation of the beta function, no? – FShrike Aug 23 '23 at 16:05

1 Answers1

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The integral can be written as $$\int_0^\infty \frac{x^{a-1}}{1+x^b}dx$$

If you substitute $x=\tan^{2/b}u$, you will get

$$\frac{dx}{1+x^b}=\frac{2}{b} \tan^{2/b-1}udu $$

Then your integral will be

$$\int_0^\infty \frac{x^{a-1}}{1+x^b} = \int_0^{\pi/2} \frac{2}{b} \tan^{2a/b-1}udu $$

This can be turned into Beta function $$\int_0^{\pi/2} \frac{2}{b} \tan^{2a/b-1}udu = \frac{1}{b} B(a/b,1-a/b) $$

This simplifies to

$$ \frac{1}{b} \Gamma(a/b) \Gamma(1-a/b)$$

Using the Euler reflection formula it follows

$$ \frac{1}{b}\Gamma(a/b) \Gamma(1-a/b)$$

$$ = \frac{\pi}{b} \csc(\frac{\pi a}{b})=\frac{1}{2}\csc(\frac{\pi}{10})$$

Which will give the golden ratio.

Mico Mrkaic
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