Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1.
$$
What an amazing result this is! I tried to write
$$
I=\int_0^{\pi/4} \log \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx
$$
and played around enough here to realize it probably isn't the best idea.
Now back to the original integral I, we can possibly change variables $y=\tan x$ and
re-writing the original integral to obtain
$$
\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}.
$$
Simplifying this we have
$$
I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy
$$
Another change of variables $y=e^{-t}$ and we have
$$
I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt
$$
but this is where I am stuck…How can we calculate I? Thanks.
[Math] Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$
calculuscomplex-analysisdefinite integralsintegrationreal-analysis
Related Solutions
Introduce variables $t$ and $y$ such that $t = \tan x = e^{-y}$, we have
$$\begin{align} \int_0^{\pi/4}\frac{\log\tan x}{\cos 2x} dx &= \int_0^1\frac{\log t}{\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}\\ &= -\int_0^\infty \frac{y}{1 - e^{-2y}} e^{-y} dy = -\int_0^\infty \sum_{k=0}^\infty y e^{-(2k+1)y} dy\\ &= -\sum_{k=0}^\infty\frac{1}{(2k+1)^2} = -\left[\sum_{k=1}^\infty\frac{1}{k^2} - \sum_{k=1}^\infty\frac{1}{(2k)^2}\right]\\ &= -\left(1-\frac14 \right)\zeta(2) = -\frac{\pi^2}{8} \end{align} $$
We make use of the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{a^{2} - (x + n\pi)^{2}} = \frac{\cot(x+a) - \cot(x-a)}{2a}, \quad a > 0 \text{ and } x \in \Bbb{R}. $$
Then for $\alpha, \beta > 0$ it follows that
\begin{align*} I := \mathrm{PV}\int_{0}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} &= \frac{1}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} \, dx \\ &= \frac{\alpha}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - x^{2}} \, dx \\ &= \frac{\alpha}{2} \sum_{n=-\infty}^{\infty} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \, dx \\ &= \frac{\alpha}{2} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \right) \log\cos^{2}x \, dx \\ &= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx, \end{align*}
where interchanging the order of integration and summation is justified by Tonelli's theorem applied to the summation over large indices $n$. Then
\begin{align*} I &= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx \\ &= \frac{1}{2\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\left|2\cos x\right| \, dx \tag{1} \end{align*}
Here, we exploited the following identity to derive (1).
$$ \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot(x+a) \, dx = 0 \quad \forall a \in \Bbb{R}. $$
Now with the substitution $z = e^{2ix}$ and $\omega = e^{2i\alpha\beta}$, it follows that
\begin{align*} I &= \frac{1}{2\beta} \Re \mathrm{PV} \int_{|z|=1} \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \log(1 + z) \, \frac{dz}{z}. \tag{2} \end{align*}
Now consider the following unit circular contour $C$ with two $\epsilon$-indents $\gamma_{\omega,\epsilon}$ and $\gamma_{\bar{\omega},\epsilon}$.
Then the integrand of (2)
$$ f(z) = \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \frac{\log(1 + z)}{z} $$
is holomorphic inside $C$ (since the only possible singularity at $z = 0$ is removable) and has only logarithmic singularity at $z = -1$. So we have
$$ \oint_{C} f(z) \, dz = 0. $$
This shows that
\begin{align*} I &= \frac{1}{2\beta} \Re \lim_{\epsilon \downarrow 0} \left( \int_{-\gamma_{\omega,\epsilon}} f(z) \, dz + \int_{-\gamma_{\bar{\omega},\epsilon}} f(z) \, dz \right) \\ &= \frac{1}{2\beta} \Re \left( \pi i \mathrm{Res}_{z=\omega} f(z) + \pi i \mathrm{Res}_{z=\bar{\omega}} f(z) \right) \\ &= \frac{1}{2\beta} \Re \left( - \pi i \log(1 + \omega) + \pi i \log(1 + \bar{\omega}) \right) \\ &= \frac{\pi}{\beta} \arg(1 + \omega) = \frac{\pi}{\beta} \arctan(\tan (\alpha \beta)). \end{align*}
In particular, if $\alpha\beta < \frac{\pi}{2}$ then we have
$$ I = \pi \alpha. $$
But due to the periodicity of $\arg$ function, this function draws a scaled saw-tooth function for $\alpha > 0$. Of course, $I$ is an even function of both $\alpha$ and $\beta$, so the final result is obtained by even extension of this saw-tooth function.
Best Answer
Integrate by parts; then you get that
$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$
The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and get
$$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$
A little manipulation leads us to
$$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$
The LHS is just $[\alpha I(\alpha)]'$, so the solution is
$$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$