Are there other methods to tackle the integral $ \int_0^\theta \cot x \ln (\sec x) d x$?

Question

Inspired by the post, $$\int_0^{\pi/2} \cot x \ln(\sec x) dx= \frac{\pi^2}{24}, $$ I want to generalise it as:

$$ \int_0^\theta \cot x \ln (\sec x) d x=-\frac{1}{8}\left(2\operatorname{Li_2}(-\tan^2 \theta)+ \ln^2 (\sec^2 \theta )\right) $$

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$$ \begin{aligned} \int_0^\theta \cot x \ln (\sec x) d x = & \frac{1}{2} \int_0^{\tan \theta} \frac{\ln \left(1+t^2\right)}{t} \cdot \frac{d t}{1+t^2},\quad \textrm{ where } t=\tan x \\ = & \frac{1}{4} \int_0^{\tan^2 \theta} \frac{\ln (1+t)}{t(1+t)} d t \quad (\textrm{ Via } t^2 \to t)\\ = & \frac{1}{4}\left[\int_0^{\tan^2 \theta} \frac{\ln (t+1)}{t} d x-\int_0^{\tan^2\theta} \frac{\ln (t+1)}{t+1} d x\right] \\ = & \frac{1}{4}\left[-\operatorname{Li_2}(-t)-\frac{1}{2} \ln ^2(t+1)\right]_0^{\tan^2 \theta}\\=& -\frac{1}{8}\left(2\operatorname{Li_2}(-\tan^2\theta)+ \ln^2 (\sec^2 \theta )\right) \end{aligned} $$

My question:

Is there other method to tackle the integral? Your comments and alternative methods are highly appreciated>

Answer 1

Note that the derived result in the post is incorrect, as contrasted below

$$ \int_0^\theta \cot x \ln (\sec x) d x = \int_0^\theta (\csc x\sec x-\tan x)\ln (\sec x) d x \\ =-\frac{1}{4}\operatorname{Li_2}(-\tan^2 \theta)-\frac12\ln^2 (\sec \theta ) $$

Answer 2

The Taylor expansion proposed by Openspace in the comment section turns out to be a quite straightforward approach leading to the "cleanest" result to my opinion. Indeed, one has : $$ \begin{align} I(\theta) &= -\int_0^\theta \cot x \ln(\cos x) \,\mathrm{d}x \\ &= -\frac{1}{2} \int_0^\theta \frac{\cos x}{\sin x} \ln(1-\sin^2x) \,\mathrm{d}x \\ &= \frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} \int_0^\theta \cos x \sin^{2n-1}x \,\mathrm{d}x \\ &= \frac{1}{4} \sum_{n=1}^\infty \frac{\sin^{2n}\theta}{n^2} \\ &= \frac{1}{4}\mathrm{Li}_2(\sin^2\theta) \end{align} $$

Answer 3

If we use the tangent half-angle subsitution, the antiderivative write $$\int \frac{1-t^2}{t \left(1+t^2\right)}\log \left(\frac{1+t^2}{1-t^2}\right)\,dt$$ $$\frac{1-t^2}{t \left(1+t^2\right)}=\frac{1}{t}-\frac{1}{t-i}-\frac{1}{t+i}$$ Expand the logarithm to face simple integrals

Answer 4

$$I=\int \cot (x) \log (\sec (x))\,dx$$ $$\sec(x)=t \quad \implies \quad x=\cos ^{-1}\left(\frac{1}{t}\right) \quad \implies \quad dx=\frac{dt}{t \sqrt{t^2-1}}$$ $$I=\int \frac{\log (t)}{(t-1) t (t+1)} \,dt$$ $$\frac{1}{(t-1) t (t+1)} =\frac{1}{2 (t+1)}+\frac{1}{2 (t-1)}-\frac{1}{t}$$ $$I=\frac{1}{2} \log (t) \log\left(\frac{t+1}{t}\right)-\frac 12 (\text{Li}_2(1-t)+\text{Li}_2(-t))$$ Back to $x$ $$J=\int_0^\theta \cot (x) \log (\sec (x))\,dx$$ $$J=\frac{1}{2} (\log (\sec (\theta )) \log (\cos(\theta )+1)-\text{Li}_2(1-\sec (\theta))+ \text{Li}_2(-\sec (\theta )))+\frac{\pi^2}{24}$$

Answer 5

$$ \begin{aligned} \int_0^\theta \cot x \ln \left(\cos ^2 x\right) d x & =-\frac{1}{2} \int_0^\theta \cot x \ln \left(\cos ^2 x\right) d x \\ & =-\frac{1}{4} \int_0^\theta \frac{\ln \left(1-\sin ^2 x\right)}{\sin ^2 x} d\left(\sin ^2 x\right) \\ & =-\frac{1}{4}\left[-\operatorname{Li_2}\left(\sin ^2 x\right)\right]_0^\theta \\ & =\frac{1}{4} \operatorname{Li_2} \left(\sin ^2 \theta\right) \end{aligned} $$

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As a bonus, we have obtained an identity that $$\operatorname{Li_2} \left(\sin ^2 \theta\right)= -\operatorname{Li_2}(-\tan^2 \theta)-\frac 12\ln^2 (\sec^2 \theta ) \blacksquare $$

Answer 6

$$\int_0^\theta\cot x\ln(\sec x)dx=-\int_0^\theta\cot x\ln(\cos x)dx$$ Consider the parametrized integral $$I(t)=-\int_0^\theta\cot x\cos^tx\,dx$$ differentiating with respect to $t$ gives: $$\begin{align*} \frac{dI}{dt}&=-\int_0^\theta\cot x\frac{\partial}{\partial t}\cos^tx\,dx \\ &=-\int_0^\theta\cot x\cos^tx\ln(\cos x)dx \end{align*}$$ From this we can deduce that $$\int_0^\theta\cot x\ln(\sec x)dx=\frac{dI}{dt}{\huge|}_{t=0}$$ Expressing $I(t)$ in a more manageable form $$I(t)=\int_0^\theta \frac{\cos^{t+1}x}{\sin x}dx$$ making the substitution $\begin{cases}u=\sin^2x \\ du=2\sin x\cos x\,dx\end{cases}$ $$I(t)=\frac12\int_0^{\sin^2\theta}\frac{(1-u)^{t/2}}{u}du$$ $$\begin{align*} \frac{dI}{dt}{\huge|}_{t=0}&=\frac12\int_0^{\sin^2\theta}\frac{\partial}{\partial t}\frac{(1-u)^{t/2}}{u}du \\ &=\frac14\int_0^{\sin^2\theta}\frac{(1-u)^{t/2}}{u}\ln(1-u)du \\ &=\frac14\int_0^{\sin^2\theta} \frac{\ln(1-u)}{u}du \\ &=\frac14\operatorname{Li}_2(\sin^2\theta)\end{align*}$$

Using the dilogarithm identity $\operatorname{Li}_2(1-z)+\operatorname{Li}_2(1-1/{z})=-\frac12\ln^2 z$, for $|z|\leq1$. Setting $z=\cos^2\theta$, we get $$\frac14\operatorname{Li}_2(\sin^2\theta)+\frac14\operatorname{Li}_2\left(1-\frac{1}{\cos^2\theta}\right)=-\frac18\ln^2(\cos^2\theta) \\ \frac14\operatorname{Li}_2(\sin^2\theta)=-\frac14\operatorname{Li}_2(-\tan^2\theta)-\frac18\ln^2(\sec^2\theta)$$