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Wednesday, August 11, 2021

Golden ratio

Integral involving the 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: 0xπ511+x2πdx=ϕ
In the proof we made use of the integral representation of the secx function, a trick that we have used extensively in other occasions (link 1, link 2)

Proof


I=0xπ511+x2πdx=1π0y451+y2dw(yxπ)
Recall the integral representation of sec(w) (proof below): sec(w)=2π0y2wπ1+y2dy
Therefore if w=2π5 I=1π0y451+y2dy=sec(2π5)2=1+52=ϕ
Hence 0xπ511+x2πdx=ϕ


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 11+y2=10ty2dt
Therefore 0ya1+y2dy=010yaty2dtdy=100yaty2dydt=12100wa212twdwdt(wy2)
The integral 0wa212twdw is the Mellin transform of tw at a+12 Recall that tw=ewln(t)=n=0lnn(t)wnn!=n=0(ln(t))n(wn)n!
Therefore, by Ramanujan's master theorem 0wa212twdw=Γ(a+12)[ln(t)]a+12
Therefore 12100wa212twdwdt=1210Γ(a+12)[ln(t)]a+12dt=12Γ(a+12)10[ln(t)]a+12dt=12Γ(a+12)10va+12evdv(vln(t))=12Γ(a+12)Γ(1a2)
Finally, recall the Euler's reflection formula Γ(z)Γ(1z)=πsinπz=πcscπz
Therefore 0ya1+y2dy=12Γ(a+12)Γ(1a2)=π2csc(π1+a2)=π2sec(πa2)
If we let x=πa2 then a=2xπ. Then 2π0y2xπ1+y2dy=secx

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