Question

Let
\[\begin{align*}\mathcal{I}&=\int_0^{\pi/2}\tan x\ln(\sin x)\,\mathrm{d}x\\[2ex]
\mathcal{J}&=\int_0^{\pi/2}x\sec^2x\ln(\sin x)\,\mathrm{d}x\\[2ex]
\mathcal{K}&=\int_0^{\pi/2}x\csc^2x\ln(\cos x)\,\mathrm{d}x\end{align*}\]
Show that\[4\mathcal{I}=2\mathcal{J}=\mathcal{K}\]assuming no knowledge of the fact that \(\mathcal{I}=-\dfrac{\pi^2}{24}\).

Answer 1

I'm thinking there's a useful method via integration by parts, but it's eluding me.

Answer 2

The original problem was showing that the LHS evaluates to \(-\dfrac{\pi^2}{24}\). Integrating by parts with
\[\begin{matrix}u=\tan x\ln(\sin x)&&&\mathrm{d}v=\mathrm{d}x\\[1ex]
\mathrm{d}u=(1+\sec^2x\ln(\sin x))\,\mathrm{d}x&&&v=x\end{matrix}\]gave
\[\begin{align*}\mathcal{I}&=\bigg[x\tan x\ln(\sin x)\bigg]_0^{\pi/2}-\int_0^{\pi/2}x(1+\sec^2x\ln(\sin x))\,\mathrm{d}x\\[1ex]
&=-\frac{\pi^2}{8}-\int_0^{\pi/2}x\sec^2x\ln(\sin x)\,\mathrm{d}x\end{align*}\]where \(\mathcal{I}\) denotes the LHS integral. Since \(\mathcal{I}=-\dfrac{\pi^2}{24}\), that would suggest that the remaining integral is \(2\mathcal{I}\).

Answer 3

I'm going to call the integral with secant \(\mathcal{J}\).

Another interesting relationship: Replacing \(x\) with \(\dfrac{\pi}{2}-x\) in the LHS gives
\[\mathcal{I}=\int_0^{\pi/2}\cot x\ln(\cos x)\,\mathrm{d}x\]and integrating by parts with an analagous setup gives
\[\begin{matrix}u=\cot x\ln(\cos x)&&&\mathrm{d}v=\mathrm{d}x\\[1ex]\mathrm{d}u=-(1+\csc^2x\ln(\cos x))\,\mathrm{d}x&&&v=x\end{matrix}\]so that
\[\begin{align*}\mathcal{I}&=\bigg[x\cot x\ln(\sin x)\bigg]_0^{\pi/2}+\int_0^{\pi/2}x(1+\csc^2x\ln(\cos x)\,\mathrm{d}x\\[1ex]
&=\frac{\pi^2}{8}+\underbrace{\int_0^{\pi/2} x\csc^2x\ln(\cos x)\,\mathrm{d}x}_\mathcal{K}\end{align*}\]which in turn means that this new integral is equivalent to \(\mathcal{K}=4\mathcal{I}\).

Answer 4

I think it might be better to just try to evaluate \(\mathcal{I}\) directly. I'm not entirely convinced there's a way to express \(\mathcal{I}\) explicitly in terms of \(\mathcal{J}\) or \(\mathcal{K}\).

To that end, we can set \(\sin x=e^t\), so \(\cos x\,\mathrm{d}x=e^t\,\mathrm{d}t\), or \(\mathrm{d}x=\dfrac{e^t}{\sqrt{1-e^{2t}}}\,\mathrm{d}t\). Then
\[\begin{align*}\mathcal{I}&=\int_{-\infty}^0 \frac{e^t}{\sqrt{1-e^{-2t}}}\times t\times\frac{e^t}{\sqrt{1-e^{-2t}}}\,\mathrm{d}t\\[1ex]
&=\int_{-\infty}^0\frac{te^{2t}}{1-e^{2t}}\,\mathrm{d}t\\[1ex]
&=-\frac{1}{2}\int_{-\infty}^0t(1+\coth t)\,\mathrm{d}t\\[1ex]
&=\frac{1}{2}\int_0^\infty t(1-\coth t)\,\mathrm{d}t\end{align*}\]which seems more approachable.

Answer 5

Probably using the series for \(\coth\) will make short work of this.

Answer 6

Maybe not, the series is only valid for \(0<|t|<\pi\), hmm...

Answer 7

Let \(u = \sin x \implies du = \cos x dx = \sqrt{1-u^2}dx\\ \tan x =\dfrac{u}{\sqrt{1-u^2}} \)

\[\int_0^{\pi/2}\tan x\ln(\sin x)\,\mathrm{d}x=\int_0^{1} \dfrac{u\ln u}{1-u^2}\,\mathrm{d}u \]

Letting \(t = 1-u^2\) we get
\[\dfrac{1}{4} \int\limits_0^1 \dfrac{\ln(1-t)}{t}\, du\]

Answer 8

Nice, I bet there's a special log-related function defined by that integral.

Answer 9

Series expansion might suggest a polylogarithm.

Answer 10

yeah

\[\dfrac{\log(1-x)}{x} =- \sum\limits_{n=0}^{\infty} \dfrac{x^n}{n+1}\]

Answer 11

http://www.wolframalpha.com/input/?i=taylor+series+ln%281-x%29%2Fx

Answer 12

I don't know if this makes my \(\coth\) integral any easier, but I rewrote it as a double integral and changed the order of integration:
\[\newcommand{\csch}{\text{csch}}
\begin{align*}\mathcal{I}&=\frac{1}{2}\int_0^\infty t(1-\coth t)\,\mathrm{d}t\\[1ex]
&=\frac{1}{2}\int_0^\infty t\int_t^\infty \csch^2s\,\mathrm{d}s\,\mathrm{d}t\\[1ex]
&=\frac{1}{2}\int_0^\infty\int_0^s t\csch^2s\,\mathrm{d}t\,\mathrm{d}s\\[1ex]
&=-\frac{1}{4}\int_0^\infty s^2\csch^2s\,\mathrm{d}s\end{align*}\]

Answer 13

I wonder if these \(\csch\) and \(\coth\) integrals are just transformed versions of \(\mathcal{J}\) and \(\mathcal{K}\)?