Show that $\int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \cos^2 x \, dx = \frac{\pi \log(6 + 4\sqrt{2})}{4} - \frac{\pi}{2}$

Question

Show that $$\int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \cos^2 x \, dx = \frac{\pi \log(6 + 4\sqrt{2})}{4} - \frac{\pi}{2}$$

My try : $$\Omega = \int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \cos^2 x \, dx = \int_{0}^{\infty} \frac{\log(1 + x^4)}{(1 + x^2)^2} \, dx$$ $$\int_{0}^{\infty} \frac{x^2 \left\{ \log(1 + x^4) - 4 \log(x) \right\}}{(1 + x^2)^2} \, dx$$

$$2\Omega = \int_{0}^{\infty} \frac{\log(1 + x^4)}{1 + x^2} \, dx - 4 \int_{0}^{\infty} \frac{x^2 \log(x)}{(1 + x^2)^2} \, dx$$

$$\Omega = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \log(1 + \tan^4 x) \, dx + 2 \int_{0}^{\infty} \frac{\log(x)}{(1 + x^2)^2} \, dx$$

Answer 1

Utilize $$\frac1{(x^2+1)^2}=\frac12\left(\frac x{x^2+1}\right)’+\frac12\frac1{x^2+1} $$ to continue with

\begin{align} \Omega = &\int_{0}^{\infty} \frac{\ln(1 + x^4)}{(1 + x^2)^2} \, dx\\ \overset{ibp}=&-2\int_0^\infty \frac{x^4}{(x^4+1)(x^2+1)}dx+ \frac12 \int_0^\infty \frac{\ln(x^4+1)}{x^2+1}dx \end{align} where $\int_0^\infty \frac{x^4}{(x^4+1)(x^2+1)}dx\overset{x\to \frac1x}=\frac12 \int_0^\infty \frac{1}{x^2+1}dx =\frac\pi4$ and \begin{align} &\int_0^{\infty} \frac{\ln (x^4+1)}{x^2+1} d x \\ =&\int_0^{\infty} \frac{\ln (x^2+\sqrt2 x+1)+\ln(x^2-\sqrt2 x+1)}{x^2+1} d x \\ =&\int_{-\infty}^{\infty} \frac{\ln (x^2+\sqrt2 x+1)}{x^2+1} \ d x\\ =& \int_{-\infty}^{\infty}\bigg( \int_0^{\pi/4}\frac{2x\cos t}{x^2+2x \sin t+1} dt+ {\ln (x^2+1)}\bigg) \frac{dx}{x^2+1}\\ =& \int_0^{\pi/4}\frac{-\pi\sin t}{1+\cos t} dt\ + {2\pi}\ln2 = \frac{\pi}2\ln \left(6+4\sqrt{2} \right) \end{align} As a result $$\Omega= \frac{\pi}4\ln(6 + 4\sqrt{2})- \frac{\pi}{2}$$

Answer 2

Let $$I=\int_0^{\frac\pi 2}\ln(1+\tan^4x)\cos^2x\,dx$$ $$J=\int_0^{\frac\pi 2}\ln(1+\tan^4x)\sin^2x\,dx$$ From Quanto's answer $$I+J=\pi\ln(2+\sqrt2)\tag1$$ On the other hand, by integration by parts $(IBP)$ we have $$I-J=\int_0^{\frac\pi 2}\ln(1+\tan^4x)\cos2x\,dx \stackrel{IBP}{=}-4\int_0^{\frac\pi 2}\frac{\tan^4x}{1+\tan^4x}\,dx \\=-2\pi+ \int_0^{\frac\pi 2}\frac{1}{1+\tan^4x}\,dx \\=-2\pi+\int_0^\infty\frac1{(u^4+1)(u^2+1)}du \\\stackrel{RT}{=}-2\pi+\pi=-\pi\tag2$$ where $RT$ stands for Residue theorem. From $(1)$ and $(2)$, $$I=\frac{\ln(2+\sqrt2)\pi}2-\frac\pi 2.$$

Answer 3

If you are allowed to use complex numbers, this is pretty nice.

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Lemma. For $\alpha \in \mathbb C$, $\Re\alpha>0$, we have $$\int_0^{\pi/2} \log(\sin^2 x + \alpha^2\cos^2x)dx = \pi \cdot \log\left(\frac{1+ \alpha}{2}\right).$$

(Proof below.)

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If you perform the change of variable $x\mapsto \pi/2-x$ your integral becomes \begin{eqnarray} \mathcal I &=& \int_0^{\pi/2} \cos^2x \log(1+\tan^4x)dx=\\ &=&\int_0^{\pi/2}\sin^2x \log\left(1+\frac1{\tan^4x}\right)dx=\\ &=&\int_0^{\pi/2}\sin^2x\log(1+\tan^4x)dx-\int_0^{\pi/2}\sin^2x\log(\tan^4x)dx=\\ &=&\int_0^1\log(1+\tan^4x)dx - \mathcal I - 4\int_0^{\pi/2}\sin^2 \log (\tan x)dx. \end{eqnarray} From the above equation we get \begin{eqnarray}\mathcal I &=& \frac12 \int_0^{\pi/2}\log(1+\tan^4x)dx - 2\int_0^{\pi/2}\sin^2x\log (\tan x)dx=\\ &=&\frac12\underbrace{ \int_0^{\pi/2}\log(\sin^4x + \cos^4x) dx}_{\mathcal A} +\\ &&-\frac12 \underbrace{\int_0^{\pi/2}\log(\cos^4x)dx}_{\mathcal B} - 2\underbrace{\int_0^{\pi/2}\sin^2x\log (\tan x)dx}_{\mathcal C} \end{eqnarray} For $\mathcal A$ we can use the Lemma to get \begin{eqnarray} \mathcal A &=& 2\Re\left\{ \int_0^{\pi/2}\log(\sin^2x + i\cos^2x)dx\right\}=\\ &=&2\Re\pi \log\left(\frac{\sqrt 2 + 1+i}{2\sqrt 2}\right)=\\ &=&2\pi \log\left|\frac{\sqrt 2 + 1+i}{2\sqrt 2}\right|=\\ &=& \pi \log\left(\frac{2+\sqrt 2}4\right) \end{eqnarray} Integral $\mathcal B$ is a classical result $$\mathcal B = 4\int_0^{\pi/2}\log(\cos x) = -2\pi\log2.$$ As for $\mathcal C$, perform the change of variable $x\mapsto \pi/2-x$ to obtain \begin{eqnarray} \mathcal C &=& \int_0^{\pi/2}\sin^2x \log(\tan x)dx = \\ &=& -\int_0^{\pi/2}\cos^2x \log(\tan x)dx, \end{eqnarray} So that \begin{eqnarray}\mathcal C &=&-\frac12\int_0^{\pi/2}\cos2x \log(\tan x)dx= \\ &\stackrel{IBP}{=}&\frac14\int_0^{\pi/2}\sin2x \cdot \frac{\cos x}{\sin x}\cdot \frac1{\cos^2x}dx=\frac{\pi}4. \end{eqnarray} In the end, letting $\mathcal I = \frac12\mathcal A - \frac12\mathcal B -2\mathcal C$ yields $$\boxed{\mathcal I = \frac{\pi}2\log(2+\sqrt 2) -\frac{\pi}2}.$$

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Proof of the Lemma

Let $$\mathcal J(\alpha) = \int_0^{\pi/2} \log(\sin^2 x + \alpha^2\cos^2x)dx,$$ where $\alpha = r\cdot e^{i\theta}$, with $-\frac{\pi}2<\theta< \frac{\pi}2$. Observe that, if $\omega = \rho \cdot e^{i\theta}$, then \begin{eqnarray} \int_0^r\frac{2\omega e^{i\theta} \cos^2x}{\sin^2 x+ \omega^2\cos^2x}d\rho &=& \log(\sin^2x + \alpha^2\cos^2x) - \log(\sin^2x), \end{eqnarray} so that \begin{eqnarray} \mathcal J(\alpha) &=& \underbrace{\int_0^{\pi/2}\int_0^r \frac{2\omega e^{i\theta}\cos^2x}{\sin^2 x+ \omega^2\cos^2x}d\rho dx}_{\mathcal H(\alpha)} +\int_0^{\pi/2}\log(\sin^2x)dx=\\ &=& \mathcal H(\alpha) - \pi \log 2.\tag{1}\label{1} \end{eqnarray} We rewrite $\mathcal H(\alpha)$ as follows. \begin{eqnarray} \mathcal H(\alpha) &=& \int_0^r 2\omega e^{i\theta}\int_0^{\infty}\frac1{\omega^2+t^2}\cdot \frac1{1+t^2}dt d\rho=\\ &=&\int_0^r\frac{2\omega e^{i\theta}}{1-\omega^2}\int_0^\infty\left(\frac1{\omega^2+t^2}-\frac1{1+t^2}\right)dt d\rho=\\ &=& \int_0^1\frac{2\omega e^{i\theta}}{1-\omega^2}\left[\frac1{\omega}\arctan\left(\frac{t}{\omega}\right) - \arctan t\right]_0^\infty d\rho=\\ &=&\frac{\pi}2\int_0^r\frac{2\omega e^{i\theta}}{1-\omega^2}\left(\frac1{\omega}-1\right)d\rho=\\ &=& \pi\int_0^r\frac{e^{i\theta}}{1+\omega}d\rho=\pi\log(1+\alpha). \end{eqnarray} Use this result in \eqref{1} to prove the Lemma.