Fun integral posted by @integralsbot
Today we evaluate this fun integral posted by @integralsbot ∫π0ln(1−2acosx+a2)1−2acosx+a2dx=2πln(1−a2)1−a2|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(1−2acosx+a2)x∈(0,2π),a2≤1
The proof is almost straight: take principal branch of the ln(z) function, then:
∞∑k=1akcos(kx)k=12∞∑k=1akk(eikx+e−ikx)=12∞∑k=1(aeix)kk+12∞∑k=1(ae−ix)kk=−12ln(1−aeix)−12ln(1−ae−ix)=−12ln((1−aeix)(1−ae−ix))=−12ln(1−a(e−ix+eix)+a2)=−12ln(1−2acos(x)+a2)
Back to the integral:
∫π0ln(1−2acosx+a2)1−2acosx+a2dx=−2∫π0∞∑k=1akcos(kx)k(1−2acosx+a2)dx=−2∞∑k=1akk∫π0cos(kx)1−2acosx+a2dx
Now note that
f(x)=cos(kx)1−2acosx+a2
is an even function.
Hence ∫π0cos(kx)1−2acosx+a2dx=12∫π−πcos(kx)1−2acosx+a2dx=12ℜ(∫π−πeixk1−2acosx+a2dx)
If we make the substitution
cosx=z+z−12
eix=z
dx=dzzi
Our integral is transformed in a contour integral round the unit complex circle:
∫π−πeixk1−2acosx+a2dx=i∮|z|=1zk(z−a)(az−1)dz
The function f(z)=zk(z−a)(az−1) 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(z−a)(az−1)dz=2πak1−a2
Therefore
∫π0ln(1−2acosx+a2)1−2acosx+a2dx=−2∞∑k=1akk∫π0cos(kx)1−2acosx+a2dxfrom (2) =−∞∑k=1akk∫π−πcos(kx)1−2acosx+a2dx=−∞∑k=1akkℜ(∫π−πeixk1−2acosx+a2dx)from (3) =−∞∑k=1akkℜ(i∮|z|=1zk(z−a)(az−1)dz)from (4)=−∞∑k=1akkℜ(2πak1−a2)from (5)=2π1−a2(−∞∑k=1a2kk)=2πln(1−a2)1−a2
Hence, we can conclude
∫π0ln(1−2acosx+a2)1−2acosx+a2dx=2πln(1−a2)1−a2|a|<1
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