I am a university student and today I face to a definite integral containing arctan(x). I can't solve it. As follow:
$$\int_{-1}^1 \frac{x^2 \arctan^2 (x) + \tan(x)}{1+x^2} \, dx.$$
Thanks for any help :)
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Following $\large\tt @Lucian$ comment: \begin{align} &2\int_{0}^{\pi/4}x^{2}\tan^{2}\pars{x}\,\dd x = -2\int_{0}^{\pi/4}x^{2}\,\dd x + 2\int_{0}^{\pi/4}x^{2}\sec^{2}\pars{x}\,\dd x \\[3mm]&=-\,{\pi^{3} \over 96} + 2\pars{\pi \over 4}^{2}\tan\pars{\pi \over 4} -4\int_{0}^{\pi/4}x\tan\pars{x}\,\dd x \\[3mm]&=-\,{\pi^{3} \over 96} + {\pi^{2} \over 8} -4\,{\pi \over 4}\ln\pars{\cos\pars{\pi \over 4}} +4\int_{0}^{\pi/4}\ln\pars{\cos\pars{x}}\,\dd x \\[3mm]&=-\,{\pi^{3} \over 96} + {\pi^{2} \over 8} + {\pi \over 2}\,\ln\pars{2} + 4\color{#c00000}{\int_{0}^{\pi/4}\ln\pars{\cos\pars{x}}\,\dd x} \end{align}
Felix Marin
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$$ \int_0^{\frac{\pi }{4}} \ln (\cos (x)) , dx = -\frac{1}{2}Catalan $$ Is this right? – 7sDream Jan 07 '14 at 11:05
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No, $\int_0^{\pi/4} \ln(\cos(x)); dx = \dfrac{1}{2} \text{Catalan} - \dfrac{\pi}{4} \ln(2)$. – Robert Israel Jan 07 '14 at 16:22
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Sign errors in last two terms of the third line. – Robert Israel Jan 07 '14 at 16:32
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Oh，Thanks for your help~ I will try to do it again by myself~and wolfram tell me what is Catalan. Thanks:) – 7sDream Jan 07 '14 at 17:21
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One term is obvious from symmetry. The other is not elementary: Maple gets
$$ \dfrac{\pi^2}{8} - \dfrac{\pi^3}{96} + \dfrac{\pi}{2} \ln(2) - 2 \;\text{Catalan}$$
Robert Israel
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@Lucian. I picked this one but I am really looking for the full result. I found the antiderivative but it is so awful ! At the same time, the result of the integral is so beautiful ! Thanks for helping. Cheers. – Claude Leibovici Jan 07 '14 at 09:29
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@ClaudeLeibovici: See Felix Marin’s answer on this thread, and then use an approach similar to Rob John’s to compute the value of the definite integral of $\ln\cos x$ between $0$ and $\dfrac\pi4$. – Lucian Jan 07 '14 at 09:59
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We can write $$\frac{x^2 \arctan^2 (x) + \tan(x)}{1+x^2}= \frac{x^2 \arctan^2 (x)}{1+x^2}+\frac{\tan(x)}{1+x^2} $$ Let $f(x)=\frac{x^2 \arctan^2 (x)}{1+x^2} \quad g(x)=\frac{\tan(x)}{1+x^2}$
$f(x)$ is an even function because $f(x)=f(-x)$, $g(x)$ is odd. The domain of the integral is symmetric, so we have $$\int_{-1}^{1} g(x) dx=0$$ $$\int_{-1}^{1} f(x)dx = 2\int_{0}^{1} f(x)dx$$
The whole integral is reduced to the solution of $$2\int_{0}^{1} \frac{x^2 \arctan^2 (x)}{1+x^2}dx$$
Matheman
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1Now notice that $\arctan'x=\dfrac1{1+x^2}$, and use substitution... Then read my comments on Robert Israel’s answer, and see Felix Marin’s response... :-) – Lucian Jan 07 '14 at 10:14
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