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Solve the integral: $$\int_0^{\pi/2} {\sin^2{x} \ln{\tan x} \,dx}$$ I have already found the answer to be $\frac{\pi}{4}$ by the method explained below, but I would like to know whether there is another way.

--- My method ---

Use a $u$-sub: $u=\tan x, \,du=\sec^2x \,dx$ $$\int_0^{\pi/2} {\sin^2{x} \ln{\tan x} \,dx}=\int_0^{\pi/2} {\frac{\tan^2{x}}{\sec^2{x}} \ln{\tan x} \,dx}=\int_0^{\pi/2} {\frac{\tan^2{x}}{\sec^4{x}} \ln{\left(\tan x\right)} \sec^2x \,dx}=\int_0^{\infty} {\frac{u^2}{\left(1+u^2\right)^2} \ln{u} \,du}$$

Now, just 'pull your luckiest rabbit out of your hat':

$$I(a) = \int_0^{\infty} {\frac{u^a \ln{u}}{\left(1+u^a\right)^2} \,du}$$

And use a 'reverse Feynman method':

$$\int{I(a) \,da} = \int_0^{\infty} {\int{ \left( \frac{u^a \ln{u}}{\left(1+u^a\right)^2} \,da\right)}\,du} = -\int_0^{\infty} {\frac{\,du}{1+u^a}} = -\frac{\pi}{a} \csc{\left(\frac{\pi}{a}\right)}$$

Now, $I(a)$ is just the derivative of the last expression:

$$I(a) = \frac{\,d}{\,da} {\left[ -\frac{\pi}{a} \csc{\left(\frac{\pi}{a}\right)} \right]} = \frac{1}{a} \left(\frac{\pi}{a}\csc{\left(\frac{\pi}{a}\right)}\right) \left(1-\frac{\pi}{a}\cot{\left(\frac{\pi}{a}\right)}\right) $$

Of course, the original integral was equivalent to $I(2)$, so:

$$\boxed{\int_0^{\pi/2} {\sin^2{x} \ln{\tan x} \,dx}=\frac{\pi}{4}}$$

Ant
  • 2,407

5 Answers5

12

Hint for an alternative method:

1) Substitute $t=\frac{\pi}{2}-x$ to get

$$I=\int_0^{\pi/2}\sin^2 x\log(\tan x) dx=-\int_0^{\pi/2}\cos^2 x\log(\tan x) dx$$

2) Write $2I$ as:

$$2I=\int_0^{\pi/2}(\sin^2 x-\cos^2 x)\log(\tan x)dx=x-\sin x \cos x \log(\tan x)\big|_{x=0}^{x=\pi/2}=\frac{\pi}{2}$$

3) So $I=\frac{\pi}{4}$

Momo
  • 16,027
  • Wow, that was neat! My method looks so contrived compared to that... thanks! – Ant Dec 30 '16 at 03:03
  • I didn't expect to come out that nice either. – Momo Dec 30 '16 at 03:23
  • Nice one. Can you tell me where you replacing terms with t and please explain me step 2. – Kanwaljit Singh Dec 30 '16 at 03:31
  • @Kanwaljit Singh $-\int_0^{\pi/2}\cos^2 x\log(\tan x) dx$ was initially $-\int_0^{\pi/2}\cos^2 t\log(\tan t) dt$. Then I replaced $t$ back with $x$ – Momo Dec 30 '16 at 03:35
  • @Kanwaljit Singh For the step 2, integrate by parts $u'=\sin^2 x+\cos^2 x$ and $v=\log(\tan x)$. Then $u=-\sin x\cos x$ and $uv'=1$ – Momo Dec 30 '16 at 03:37
4

Using Integration by Parts $$ \begin{align} &\int_0^{\pi/2}\sin^2(x)\log(\tan(x))\,\mathrm{d}x\tag{1}\\ &=\int_0^{\pi/2}\sin^2(x)\log(\sin(x))\,\mathrm{d}x -\int_0^{\pi/2}\sin^2(x)\log(\cos(x))\,\mathrm{d}x\tag{2}\\ &=\int_0^{\pi/2}\cos^2(x)\log(\cos(x))\,\mathrm{d}x -\int_0^{\pi/2}\sin^2(x)\log(\cos(x))\,\mathrm{d}x\tag{3}\\ &=\int_0^{\pi/2}\cos(2x)\log(\cos(x))\,\mathrm{d}x\tag{4}\\ &=\frac12\int_0^{\pi/2}\log(\cos(x))\,\mathrm{d}\sin(2x)\tag{5}\\ &=\frac12\int_0^{\pi/2}\sin(2x)\tan(x)\,\mathrm{d}x\tag{6}\\ &=\int_0^{\pi/2}\sin^2(x)\,\mathrm{d}x\tag{7}\\ &=\int_0^{\pi/2}\cos^2(x)\,\mathrm{d}x\tag{8}\\[3pt] &=\frac\pi4\tag{9} \end{align} $$ Explanation:
$(2)$: $\log(\tan(x))=\log(\sin(x))-\log(\cos(x))$
$(3)$: substitute $x\mapsto\frac\pi2-x$ in first integral
$(4)$: $\cos(2x)=\cos^2(x)-\sin^2(x)$
$(5)$: prepare to integrate by parts
$(6)$: integrate by parts
$(7)$: $\frac12\sin(2x)\tan(x)=\sin^2(x)$
$(8)$: substitute $x\mapsto\frac\pi2-x$
$(9)$: average $(7)$ and $(8)$

Using the Trigonometric Series for $\boldsymbol{\log(\cos(x))}$

Using the series for $\log(1+x)$, $$ \begin{align} \log(2\cos(x)) &=\log\left(e^{ix}+e^{-ix}\right)\\ &=ix+\log\left(1+e^{-2ix}\right)\\ &=-ix+\log\left(1+e^{2ix}\right)\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(2kx) \end{align} $$ Thus, using the orthogonality of $\{\cos(kx)\}$ $$ \begin{align} &\int_0^{\pi/2}\sin^2(x)\log(\tan(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\sin^2(x)\log(2\sin(x))\,\mathrm{d}x -\int_0^{\pi/2}\sin^2(x)\log(2\cos(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\cos^2(x)\log(2\cos(x))\,\mathrm{d}x -\int_0^{\pi/2}\sin^2(x)\log(2\cos(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\cos(2x)\log(2\cos(x))\,\mathrm{d}x\\ &=\int_0^{\pi/2}\cos(2x)\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(2kx)\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{4k}\int_{-\pi}^\pi\cos(x)\cos(kx)\,\mathrm{d}x\\[3pt] &=\frac\pi4 \end{align} $$

robjohn
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  • +1. As an observation, you may directly calculate $\int_0^{\pi/2}\cos(2x)\log(2\cos(x)),\mathrm{d}x$, there is no need to complicate with the series. – Momo Dec 30 '16 at 05:39
  • @Momo: I've added a second approach – robjohn Dec 30 '16 at 06:22
2

This is not an answer but it is put for your curiosity.

Surprizing or not, the antiderivative can be computed. Considering $$I=\int {\sin^2(x)} \ln(\tan (x)) \,dx$$ a CAS led to $$8I=4 i \text{Li}_2\left(e^{2 i x}\right)-i \text{Li}_2\left(e^{4 i x}\right)-2 \sin (2 x) \log (\tan (x))+4 x \left(2 \tanh ^{-1}\left(e^{2 i x}\right)+\log (\tan (x))+1\right)$$ where appears the polylogarithm function.

  • For $x=0$, the above expression leads to $I=\frac{i \pi ^2}{16}$
  • For $x=\frac \pi 2$, the above expression leads to $I=\frac{\pi }{4}+\frac{i \pi ^2}{16}$ and hence the result for the integral between $0$ and $\frac\pi 2$
  • For $x=\frac \pi 4$, the above expression leads to $I=-\frac{C}{2}+\frac{\pi }{8}+\frac{i \pi ^2}{16}$ and hence another "interesting" result $$\int_0^{\frac\pi 4} {\sin^2(x)} \ln(\tan (x)) \,dx=\frac{\pi }{8}-\frac{C}{2}$$ where appears Catalan number.
  • For small values of $\epsilon$, Taylor expansion of $I$ gives $$\int_0^{\epsilon} {\sin^2(x)} \ln(\tan (x)) \,dx=\frac{1}{9} (3 \log (\epsilon)-1)\,\epsilon^3+\frac{1}{75} (6-5 \log (\epsilon))\, \epsilon^5+ \left(\frac{2 \log (\epsilon)}{315}-\frac{5}{882}\right)\,\epsilon^7+O\left(\epsilon^9\right)$$
Claude Leibovici
  • 260,315
1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{\pi/2}\sin^{2}\pars{x}\ln\pars{\tan\pars{x}}\,\dd x = \left.\totald{}{\mu}\int_{0}^{\pi/2}\sin^{2}\pars{x}\tan^{\mu}\pars{x}\,\dd x \,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.\totald{}{\mu}\int_{0}^{\pi/2}\sin^{\mu + 2}\pars{x} \bracks{1 - \sin^{2}\pars{x}}^{-\pars{\mu + 1}/2}\,\cos\pars{x}\,\dd x \,\right\vert_{\ \mu\ =\ 0} \\[5mm] \stackrel{t\ \equiv\ \sin\pars{x}}{=} &\,\,\, \left.\totald{}{\mu}\int_{0}^{1}t^{\mu + 2} \pars{1 - t^{2}}^{-\pars{\mu + 1}/2}\,\dd t \,\right\vert_{\ \mu\ =\ 0} \,\,\,\stackrel{t^{2}\ \mapsto\ t}{=} \left.{1 \over 2}\totald{}{\mu}\int_{0}^{1}t^{\mu/2 + 1/2} \pars{1 - t}^{-\pars{\mu + 1}/2}\,\dd t \,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ \left.{1 \over 2}\totald{}{\mu} {\Gamma\pars{\mu/2 + 3/2}\Gamma\pars{-\mu/2 + 1/2} \over \Gamma\pars{2}} \,\right\vert_{\ \mu\ =\ 0} \\[5mm] = &\ {1 \over 2}\totald{}{\mu}\bracks{% {\mu + 1 \over 2}\,\Gamma\pars{\mu + 1 \over 2}\Gamma\pars{1 - \mu \over 2}} _{\ \mu\ =\ 0} \\[5mm] = &\ {1 \over 4}\totald{}{\mu}\bracks{% \pars{\mu + 1}{\pi \over \sin\pars{\pi\bracks{\mu + 1}/2}}}_{\ \mu\ =\ 0} = {\pi \over 4}\totald{}{\mu}\bracks{% {\mu + 1 \over \cos\pars{\pi\mu/2}}}_{\ \mu\ =\ 0} = \bbx{\ds{\pi \over 4}} \end{align}

Felix Marin
  • 89,464
-1

Using double angle formula and IBP, we have $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \sin ^2 x \ln (\tan x) d x = & \int_0^{\frac{\pi}{2}} \frac{1-\cos 2 x}{2} \ln (\tan x) d x \\ = & \frac{1}{2} \int_0^{\frac{\pi}{2}} \ln (\tan x) d x-\frac{1}{4} \int_0^{\frac{\pi}{2}} \ln (\tan x)d(\sin 2x) \\ = & -\frac{1}{4}\left([\sin 2 x\ln(\tan x)]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}} \frac{\sin 2 x}{\tan x} \sec ^2 x d x\right) \\ = & \frac{1}{4} \int_0^{\frac{\pi}{2}} 2 d x \\ = & \frac{\pi}{4} \end{aligned} $$

Lai
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