Integral that relates the secant function and the Golden ratio
Today we repost here the solution to this integral posted on Twitter by @integralsbot: ∫∞0xπ5−11+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=∫∞0xπ5−11+x2πdx=1π∫∞0y451+y2dw(y↦xπ)
Recall the integral representation of sec(w) (proof below):
sec(w)=2π∫∞0y2wπ1+y2dy
Therefore if w=2π5
I=1π∫∞0y451+y2dy=sec(2π5)2=1+√52=ϕ
Hence
∫∞0xπ5−11+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=∫∞0∫10yaty2dtdy=∫10∫∞0yaty2dydt=12∫10∫∞0wa2−12twdwdt(w↦y2)
The integral ∫∞0wa2−12twdw 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
∫∞0wa2−12twdw=Γ(a+12)[−ln(t)]−a+12
Therefore
12∫10∫∞0wa2−12twdwdt=12∫10Γ(a+12)[−ln(t)]−a+12dt=12Γ(a+12)∫10[−ln(t)]−a+12dt=12Γ(a+12)∫10v−a+12e−vdv(v↦−ln(t))=12Γ(a+12)Γ(1−a2)
Finally, recall the Euler's reflection formula
Γ(z)Γ(1−z)=πsinπz=πcscπz
Therefore
∫∞0ya1+y2dy=12Γ(a+12)Γ(1−a2)=π2csc(π1+a2)=π2sec(πa2)
If we let x=πa2 then a=2xπ. Then
2π∫∞0y2xπ1+y2dy=secx
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