Evaluate $$ \int_{0}^{\pi/2} x^2 \ln^2(2\cos x) \ln^2(2\sin x) \, dx $$
(Collective) attempt so far:
Let the integral be $I$. Let $A = \ln(2\cos x)$ and $B = \ln(2\sin x)$. The product is $AB$. We can express $AB$ in terms of $A+B$ and $A-B$. $$A+B = \ln(2\cos x) + \ln(2\sin x) = \ln(4\sin x\cos x) = \ln(2\sin(2x))$$ $$A-B = \ln(2\cos x) - \ln(2\sin x) = \ln(\cot x)$$
From these, we can write $A = \frac{(A+B)+(A-B)}{2}$ and $B = \frac{(A+B)-(A-B)}{2}$. The product is $AB = \frac{1}{4}((A+B)^2 - (A-B)^2)$. The term in the integrand is $\ln^2(2\cos x)\ln^2(2\sin x) = (AB)^2$. So, $(AB)^2 = \frac{1}{16} \left( \ln^2(2\sin(2x)) - \ln^2(\cot x) \right)^2$
The integral becomes: $$ I = \frac{1}{16} \int_{0}^{\pi/2} x^2 \left( \ln^2(2\sin(2x)) - \ln^2(\cot x) \right)^2 dx $$ Expanding the square, we get: $$ I = \frac{1}{16} \int_{0}^{\pi/2} x^2 \left( \ln^4(2\sin(2x)) - 2\ln^2(2\sin(2x))\ln^2(\cot x) + \ln^4(\cot x) \right) dx $$ This splits the integral into three parts: $$I = \frac{1}{16} (I_1 - 2I_2 + I_3)$$ where $$I_1 = \int_0^{\pi/2} x^2 \ln^4(2\sin(2x)) \, dx$$ $$I_2 = \int_0^{\pi/2} x^2 \ln^2(2\sin(2x)) \ln^2(\cot x) \, dx$$ $$I_3 = \int_0^{\pi/2} x^2 \ln^4(\cot x) \, dx$$
These three integrals are themselves very difficult. We have not been able to make any progress further than this. According to WA the solution is $1.094$. But we are interested in a closed form solution.
Edit: Here is the link to the file
import numpy as np
from scipy.integrate import quad
pi = np.pi
def integrand1(x):
if x == 0 or x == pi/2:
return 0.0
return x2 * np.log(2 * np.sin(2 * x))4
def integrand2(x):
if x == 0 or x == pi/2:
return 0.0
return x2 * (np.log(2 * np.sin(2*x))2) * (np.log(1/np.tan(x))**2)
def integrand3(x):
if x == 0 or x == pi/2:
return 0.0
return x2 * np.log(1/np.tan(x))4
numerical_I1, error_I1 = quad(integrand1, 0, pi/2, limit=200)
numerical_I2, error_I2 = quad(integrand2, 0, pi/2, limit=200)
numerical_I3, error_I3 = quad(integrand3, 0, pi/2, limit=200)
print("--- Numerical Results ---")
print(f"Approximation for I₁: {numerical_I1:.5f} (Estimated error: {error_I1:.2e})")
print(f"Approximation for I₂: {numerical_I2:.5f} (Estimated error: {error_I2:.2e})")
print(f"Approximation for I₃: {numerical_I3:.5f} (Estimated error: {error_I3:.2e})")
Running this script produces the following output:
Approximation for I₁: 14.73628 (Estimated error: 2.17e-07)
Approximation for I₂: 27.06450 (Estimated error: 9.97e-08)
Approximation for I₃: 56.89675 (Estimated error: 4.81e-07)
