I=∫π20sin2xln(sin2(tanx))dx(1)
Now, using definite integral property of ∫baf(x)dx=∫baf(a+b−x)dx I=∫π20cos2xln(sin2(cotx))dx(2)
After tanx=t substitution in (1) and cotx=m in (2), to get: I=∫π20t2(1+t2)2ln(sin2t)dt=∫π20m2(1+m2)2ln(sin2m)dm
After variable change and addition, I get:
2I=∫π20x2(1+x2)2ln(sin4x)dx⟹I2=∫π20x2(1+x2)2ln(sinx)dx
How could I proceed? Any other solutions which happen to be more efficient/simple?
Answer
I=∫π20cos2xln(sin2(tanx))dx
Use the substitution t=tan(x)
I=2∫∞0ln|sint|(1+t2)2dt
Then use the fourier expansion of log|sinx|
I=−2∑1k∫∞0cos(2kt)(1+t2)2dt−2∫∞0log(2)(1+t2)2dt
For the second integral
2∫∞0log(2)(1+t2)2
We can use the beta function to deduce that
2∫∞0log(2)(1+t2)2=π2log(2)
For the second integral
∫∞0cos(2kt)(1+t2)2dt=Re∫∞0e2kit(1+t2)2dt
It can be showing by complex analysis that
∫∞0cos(2kt)(1+t2)2dt=π4e−2k(1+2k)
Finally we have the integral
I=−π2∑e−2k(1+2k)k−π2log(2)
The sum can be computed using
−log(1−x)=∑xkk
Hence we deduce that
I=−π2{log(2e2−1)+2e2e2−1}
ADDENDUM
As required A proof using complex analysis for
∫∞0cos(2kt)(1+t2)2dt
This can be done by considering
F(z)=e2kiz(1+z2)2
The poles are z=±i, then consider half a circle in the upper half plane.
Note there is only one pole of order 2 inside the contour. Hence by the residue theorem
∫Ce2kiz(1+z2)2+∫0−∞e2kix(1+x2)2dx+∫∞0e2kix(1+x2)2dx=2πiRese2kiz(1+z2)2|z=i
The integral along the arc converges to 0
2∫∞0cos(2kx)(1+x2)2dx=2πiRese2kiz(1+z2)2|z=i
The pole of order 2 can be calculated using
lim
This simplifies to
2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{2}e^{-2k}(1+2k)
\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{4}e^{-2k}(1+2k)
For many other approaches see Calculating the integral \int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x without using complex analysis
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