Thursday, August 19, 2021

Residue theorem IV

Residue theorem for winding numbers

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(am1+am2+am3+...+1am)

Proof

Recall that under the assumptions of the residue theorem with winding numbers, if γ is a closed curve: γfdz=2πini=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)=eit0t2πn 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=eix0x2πdx=dziz
Then
I=2π0cos(1a+x)cos(x)=2π0(ei(1a+x)+ei(1a+x)2)(eix+eix2)dx=2π0(eiaeix+eiaeix2)(eix+eix2)dx=122C(eiaz+eiaz)(z+1z)izdz=122iC{eiaz+1z(eia+eia)+eiaz3}dz
Hence, if g(z)=eiaz+1z(eia+eia)+eiaz3
Res(g,0)=eia+eia
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+eia)=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)+ei(1a+x)2)(ei(1a2+x)+ei(1a2+x)2)(ei2x+ei2x2)dx=4π0(eiaeix+eiaeix2)(eia2eix+eia2eix2)(ei2x+ei2x2)dx=123iC(eiaz+eiaz)(eia2z+eia2z)(z2+1z2)zdz(z=eix)=123iC(eiaz2+eia)(eia2z2+eia2)(z4+1)z5g(z)dz
It is easy to see that Res(g,0)=eiaeia2+eiaeia2=ei(1a+1a2)+ei(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+eiaeia2)=π2[ei(1a+1a2)+ei(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 2m1 times around 0)

I=2mπ0(mj=1cos(1aj+x))cos(mx)=2mπ0(mj=1ei(1aj+x)+ei(1aj+x)2)(eimx+eimx2)dx=2m12π0(mj=1eiajeix+eiajeix2)(eimx+eimx2)dx=12m+1C[mj=1(eiajz+eiajz)](zm+1zm)dziz(z=eix)=1i2m+1C[mj=1(eiajz2+eiaj)](z2m+1)z2m+1dz
Therefore if g(z)=[mj=1(eiajz2+eiaj)](z2m+1)z2m+1
Then Res(g,0)=mj=1eiaj+mj=1eiaj=ei(1a+1a2+...+1am)+ei(1a+1a2+...+1am)
Hence I=2mπ0(mj=1cos(1aj+x))cos(mx)=2πi(1i2m+1)Res(g,0)I(C,0)=2m1=π2(ei(1a+1a2+...+1am)+ei(1a+1a2+...+1am))=πcos(1a+1a2+...+1am)=πcos(am1+am2+am3+...+1am)
Hence 2mπ0[cos(1a+x)cos(1a2+x)cos(1am+x)]cos(mx)=πcos(am1+am2+am3+...+1am)


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