Tuesday, 7 March 2017

calculus - How to evaluate I=intfracpi20sin2xln(sin2(tanx))dx



I=π20sin2xln(sin2(tanx))dx(1)
Now, using definite integral property of baf(x)dx=baf(a+bx)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)dxI2=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=20ln|sint|(1+t2)2dt



Then use the fourier expansion of log|sinx|



I=21k0cos(2kt)(1+t2)2dt20log(2)(1+t2)2dt



For the second integral



20log(2)(1+t2)2




We can use the beta function to deduce that



20log(2)(1+t2)2=π2log(2)



For the second integral



0cos(2kt)(1+t2)2dt=Re0e2kit(1+t2)2dt



It can be showing by complex analysis that




0cos(2kt)(1+t2)2dt=π4e2k(1+2k)



Finally we have the integral



I=π2e2k(1+2k)kπ2log(2)



The sum can be computed using



log(1x)=xkk




Hence we deduce that




I=π2{log(2e21)+2e2e21}







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.




enter image description here



Note there is only one pole of order 2 inside the contour. Hence by the residue theorem
Ce2kiz(1+z2)2+0e2kix(1+x2)2dx+0e2kix(1+x2)2dx=2πiRese2kiz(1+z2)2|z=i



The integral along the arc converges to 0



20cos(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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