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I'm trying to evaluate:

$$\int_0^{\pi/ 4} \frac {\sin x} {x \cos^2 x} \mathrm d x$$

Mathematica gives the numerical approximation:

$0.959926156626593638859649248036004150970933774605514278777212260466184427508$

I cannot find a closed form though. Any ideas would be appreciated.

user85798
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    This is an elegant integral! – Jeff Faraci Apr 29 '14 at 20:52
  • @Integrals Do you think there exists a closed form? I might put a bounty on it... – user85798 Apr 30 '14 at 12:05
  • I do think a closed form exists for this integral. I have been trying on this one. However, my view is slightly bias as I have a firm belief that all integrals have a closed form:) – Jeff Faraci Apr 30 '14 at 13:04
  • @Oliver: Have you tried substituting the series of $\sin x$? – William Hilbert May 02 '14 at 08:05
  • @WilliamHilbert I have tried series expansions but they aren't very nice. – user85798 May 02 '14 at 14:35
  • So far, i've got this. And i don't know what to do next. And i forgot to write dx es. sorry for that. $$\int \frac{sinx}{xcos^{2}x}dx=\int \frac{sinx}{x}tan{}'x:dx=\frac{sin^{2}x}{xcosx}-\int \frac{xcosx-sinx}{x^{2}}tanx:dx=\frac{sin^{2}x}{xcosx}-\int \frac{sinx}{x}+\int \frac{1-cos^{2}x}{x^{2}cosx}=\frac{sin^{2}x}{xcosx}-Si(x)+Si(x)+\frac{cosx}{x}+ \int\frac{1}{x^{2}cosx}=\frac{1}{xcosx}+\int \frac{1}{x^{2}cosx}$$ – esege May 02 '14 at 21:02
  • @Oliver, I get something like $$\frac{4}{\pi}(\sqrt{2}-1) + \frac{4}{\pi^2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}\log\left(\frac{4n+3}{4n+1}\right)$$ It is not very nice and I don't have any idea how to simplify it further, is that what you get? – achille hui May 02 '14 at 21:22
  • @achillehui I got a further simplification: $$ \frac{4}{\pi}\left(\sqrt2+\sum^{\infty}{m=0}\frac{|E{2m}|(\pi/4)^{2m}}{(2m-1)(2m)!}\right).$$ where $E_{2m}$ are Euler numbers. – Chen Wang May 02 '14 at 23:40
  • @ChenWang, hmm.. this is the same expansion as the one in Felix's answer. Unluckily, the presence of the Euler numbers make it hard to resum it to known special functions... – achille hui May 03 '14 at 00:09

1 Answers1

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/4}{\sin\pars{x} \over x\cos^{2}\pars{x}}\,\dd x:\ {\large ?}}$

\begin{align} &\color{#00f}{\large% \int_{0}^{\pi/4}{\sin\pars{x} \over x\cos^{2}\pars{x}}\,\dd x}= \int_{0}^{\pi/4}{\sec\pars{x}\tan\pars{x} \over x}\,\dd x =\int_{x=0}^{x=\pi/4}\,{\dd\bracks{\sec\pars{x} -1}\over x} \\[3mm]&={4 \over \pi}\,\pars{\root{2} - 1} +\int_{0}^{\pi/4}{\sec\pars{x} -1 \over x^{2}}\,\dd x \\[3mm]&={4 \over \pi}\,\pars{\root{2} - 1} +\sum_{n = 1}^{\infty}\pars{-1}^{n}{E_{2n} \over \pars{2n}!} \int_{0}^{\pi/4}x^{2n - 2}\,\dd x \\[3mm]&=\color{#00f}{\large{4 \over \pi}\,\pars{\root{2} - 1} +\sum_{n = 1}^{\infty}\pars{-1}^{n}{E_{2n} \over \pars{2n}!\pars{2n - 1}} \pars{\pi \over 4}^{2n - 1}} \end{align} where $\ds{E_{n}}$ is an Euler Number.

Felix Marin
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    Nice approach. +1 – Jeff Faraci May 03 '14 at 14:52
  • Well I wouldn't call it a closed form, and that's kind of the definition of Euler numbers, but it's probably the best that exists. – user85798 May 04 '14 at 21:37
  • @Oliver Indeed, for a long time I struggled with several tricks. As you already said "it's probably the best that exists". Recently, I found that $\large \sec\left(x\right)$ Taylor series involves the Euler numbers. I saw many solutions in MSE that involves series. Thanks. – Felix Marin May 04 '14 at 21:49