Friday, December 3, 2021

Integral of the day XII

Fun integral

Fun integral posted by @integralsbot


Today we evaluate this fun integral posted by @integralsbot π0ln(12acosx+a2)12acosx+a2dx=2πln(1a2)1a2|a|<1
To proof this result we will use Fourier series and contour integration.

Proof

Firt, we need the following Fourier series: k=1akcos(kx)k=12ln(12acosx+a2)x(0,2π),a21
The proof is almost straight: take principal branch of the ln(z) function, then: k=1akcos(kx)k=12k=1akk(eikx+eikx)=12k=1(aeix)kk+12k=1(aeix)kk=12ln(1aeix)12ln(1aeix)=12ln((1aeix)(1aeix))=12ln(1a(eix+eix)+a2)=12ln(12acos(x)+a2)
Back to the integral: π0ln(12acosx+a2)12acosx+a2dx=2π0k=1akcos(kx)k(12acosx+a2)dx=2k=1akkπ0cos(kx)12acosx+a2dx
Now note that f(x)=cos(kx)12acosx+a2
is an even function.

Hence π0cos(kx)12acosx+a2dx=12ππcos(kx)12acosx+a2dx=12(ππeixk12acosx+a2dx)
If we make the substitution cosx=z+z12
eix=z
dx=dzzi
Our integral is transformed in a contour integral round the unit complex circle: ππeixk12acosx+a2dx=i|z|=1zk(za)(az1)dz


The function f(z)=zk(za)(az1) has two poles at z=a and z=1a.

The conditions in (1) assure that z=a with |a|<1 is the only pole inside the unit circle.

Then, by the Cauhcy integral formula i|z|=1zk(za)(az1)dz=2πak1a2
Therefore π0ln(12acosx+a2)12acosx+a2dx=2k=1akkπ0cos(kx)12acosx+a2dxfrom (2) =k=1akkππcos(kx)12acosx+a2dx=k=1akk(ππeixk12acosx+a2dx)from (3) =k=1akk(i|z|=1zk(za)(az1)dz)from (4)=k=1akk(2πak1a2)from (5)=2π1a2(k=1a2kk)=2πln(1a2)1a2
Hence, we can conclude π0ln(12acosx+a2)12acosx+a2dx=2πln(1a2)1a2|a|<1

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