Beautiful integral involving a product of cosine functions
Today we show the proof of this beautiful result posted by @SrinivasR1729 on Twitter:
∫2π0cos(1a+x)cos(x)dx=πcos(1a)
∫4π0cos(1a+x)cos(1a2+x)cos(2x)=πcos(a+1a2)
∫8π0cos(1a+x)cos(1a2+x)cos(1a3+x)cos(3x)=πcos(a2+a+1a3)
∫16π0cos(1a+x)cos(1a2+x)cos(1a3+x)cos(1a4+x)cos(3x)=πcos(a3+a2+a+1a4)
∫2mπ0[cos(1a+x)cos(1a2+x)⋯cos(1am+x)]cos(mx)=πcos(am−1+am−2+am−3+...+1am)
Proof
Recall that under the assumptions of the residue theorem with winding numbers, if γ is a closed curve: ∮γfdz=2πin∑i=1{Res(f,zi)I(γ,zi)}
where I(γ,zi) is the winding number of γ with respect to zi.
Also recall that the circle γ(t)=eit0≤t≤2πn has an index I(γ,0)=n with respect to 0
Case n=1 (the circle winds one time around 0)
The fact that x varies from 0 to 2π suggest we can consider x as an argument of a point z on the unit circle C centered at the origin; hence we write z=eix0≤x≤2π⟹dx=dziz Then
I=∫2π0cos(1a+x)cos(x)=∫2π0(ei(1a+x)+e−i(1a+x)2)(eix+e−ix2)dx=∫2π0(eiaeix+e−iae−ix2)(eix+e−ix2)dx=122∮C(eiaz+e−iaz)(z+1z)izdz=122i∮C{eiaz+1z(eia+e−ia)+e−iaz3}dz
Hence, if g(z)=eiaz+1z(eia+e−ia)+e−iaz3
⟹Res(g,0)=eia+e−ia
Therefore z=0 is the only singularity of g and:
I=∫2π0cos(1a+x)cos(x)dx=2πi(122i)Res(g,0)I(C,0)⏟=1=π2(eia+e−ia)=πcos(1a)
∫2π0cos(1a+x)cos(x)dx=πcos(1a)
Case n = 2 (the circle winds 2 times around 0)
I=∫4π0cos(1a+x)cos(1a2+x)cos(2x)dx=∫4π0(ei(1a+x)+e−i(1a+x)2)(ei(1a2+x)+e−i(1a2+x)2)(ei2x+e−i2x2)dx=∫4π0(eiaeix+e−iae−ix2)(eia2eix+e−ia2e−ix2)(ei2x+e−i2x2)dx=123i∮C(eiaz+e−iaz)(eia2z+e−ia2z)(z2+1z2)zdz(z=eix)=123i∮C(eiaz2+e−ia)(eia2z2+e−ia2)(z4+1)z5⏟g(z)dz
It is easy to see that
Res(g,0)=eiaeia2+e−iae−ia2=ei(1a+1a2)+e−i(1a+1a2)
Therefore, again z=0 is the only singularity of g and
I=∫4π0cos(1a+x)cos(1a2+x)cos(2x)dx=2πi(123i)Res(g,0)I(C,0)⏟=2=22πi(123i)(eiaeia2+e−iae−ia2)=π2[ei(1a+1a2)+e−i(1a+1a2)]=πcos(1a+1a2)=πcos(a+1a2)
∫4π0cos(1a+x)cos(1a2+x)cos(2x)=πcos(a+1a2)
Case n = m (the circle winds 2m−1 times around 0)
I=∫2mπ0(m∏j=1cos(1aj+x))cos(mx)=∫2mπ0(m∏j=1ei(1aj+x)+e−i(1aj+x)2)(eimx+e−imx2)dx=∫2m−12π0(m∏j=1eiajeix+e−iaje−ix2)(eimx+e−imx2)dx=12m+1∮C[m∏j=1(eiajz+e−iajz)](zm+1zm)dziz(z=eix)=1i2m+1∮C[∏mj=1(eiajz2+e−iaj)](z2m+1)z2m+1dz
Therefore if
g(z)=[∏mj=1(eiajz2+e−iaj)](z2m+1)z2m+1
Then
Res(g,0)=m∏j=1eiaj+m∏j=1e−iaj=ei(1a+1a2+...+1am)+e−i(1a+1a2+...+1am)
Hence
I=∫2mπ0(m∏j=1cos(1aj+x))cos(mx)=2πi(1i2m+1)Res(g,0)I(C,0)⏟=2m−1=π2(ei(1a+1a2+...+1am)+e−i(1a+1a2+...+1am))=πcos(1a+1a2+...+1am)=πcos(am−1+am−2+am−3+...+1am)
Hence
∫2mπ0[cos(1a+x)cos(1a2+x)⋯cos(1am+x)]cos(mx)=πcos(am−1+am−2+am−3+...+1am)
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