Question

$\red{ \rm\int\limits_{0}^{ \frac{\pi}{2}} l {n}^{2} \bigg( \frac{ {e}^{ - {x}^{2} } }{ \cos(x) } (1 + \cos(4x)) \bigg ) dx }$​

Answer 1

First we rewrite

$\dfrac{1 + \cos(4x)}{\cos(x)} = 2\cos^2(2x)$

then expand the integrand as

$\displaystyle \ln^2(2) - 2 \ln(2) x^2 + x^4 \\\\ {} + 2\ln(2) \ln(\cos^2(2x)) - 2\ln(2) \ln(\cos(x)) - 2x^2 \ln(\cos^2(2x) + 2x^2 \ln(\cos(2x)) \\\\ {} + \ln^2(\cos^2(2x)) + \ln^2(\cos(x)) - 2 \ln(\cos(x)) \ln(\cos^2(2x))$

We'll use the following identity:

$\displaystyle \cos(2kx) = \frac{e^{i2kx} + e^{-i2kx}}2 \\\\ \sum_{k=1}^\infty \frac{\cos(2kx)}k = \frac12 \left(\sum_{k=1}^\infty \frac{(e^{i2x})^k}k + \frac{(e^{-i2x})^k}k\right) \\\\ \sum_{k=1}^\infty \frac{\cos(2kx)}k = -\frac12 \left(\ln(1-e^{i2kx}) + \ln(1 - e^{-i2kx})\right) \\\\ \implies \ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$

as well as the fact that for any integer n,

$\displaystyle \int_0^{\frac\pi2} \cos(2nx) \, dx = 0$

Consult the attachments for the integrals of the non-trivial terms.

Putting everything together, the end result is then

$\displaystyle \int_0^{\frac\pi2} \ln^2\left(\frac{e^{-x^2}}{\cos(x)}(1+\cos(4x))\right) \, dx \\\\ = \boxed{\frac{\pi^5}{160} + \frac{\pi^3}4 - \frac{11\pi}{16} \zeta(3)}$