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I need some help in evaluating $ \displaystyle \int_0^{\infty} dx \frac{2 \sin x \cos^2 x}{x e^{x \sqrt{3}}}$

The original question: Evaluate $\displaystyle \int_0^{\infty} dx \frac{e^{- x \sqrt 3}}{x} (1 - \sin x)(1 + 2 \sin x - \cos 2x)$

Using $\cos 2x = 1 - 2\sin^2 x$ and $1 - \sin^2 x = \cos^2 x$ I was able to get it into the above form. However, I do not know how to proceed. I would like some guidance rather than a full answer, please.

MT_
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2 Answers2

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#c00000}{% \int_{0}^{\infty}{2\sin\pars{x}\cos^{2}\pars{x} \over x \expo{x\root{3}}}\,\dd x} =\int_{0}^{\infty}\expo{-x\root{3}}\, {\sin\pars{x} + \sin\pars{x}\cos\pars{2x} \over x }\,\dd x \\[3mm]&=\int_{0}^{\infty}\expo{-x\root{3}}\, {\sin\pars{x} + \bracks{\sin\pars{x + 2x} + \sin\pars{x - 2x}}/2 \over x }\,\dd x \\[3mm]&=\half\int_{0}^{\infty}\expo{-x\root{3}}\, {\sin\pars{3x} + \sin\pars{x} \over x }\,\dd x \\[3mm]&=\color{#c00000}{\half\int_{0}^{\infty} \pars{\expo{-\root{3}x/3} + \expo{-\root{3}x}}\, {\sin\pars{x} \over x }\,\dd x}\tag{1} \end{align}

With $\mu > 0$: \begin{align} &\color{#00f}{\int_{0}^{\infty}\expo{-\mu x}\,{\sin\pars{x} \over x }\,\dd x} =\int_{0}^{\infty}\expo{-\mu x}\,\half\int_{-1}^{1}\expo{\ic kx}\,\dd k\,\dd x =\half\int_{-1}^{1}\dd k\int_{0}^{\infty}\expo{\pars{\ic k - \mu}x}\,\dd x \\[3mm]&=\half\int_{-1}^{1}{-1 \over \ic k - \mu}\,\dd k =\half\int_{-1}^{1}{\ic k + \mu \over k^{2} + \mu^{2}}\,\dd k =\int_{0}^{1}{\mu \over k^{2} + \mu^{2}}\,\dd k =\color{#00f}{\arctan\pars{1 \over \mu}} \end{align}

By replacing this result in $\pars{1}$ we find: \begin{align}&\color{#00f}{\large% \int_{0}^{\infty}{2\sin\pars{x}\cos^{2}\pars{x} \over x \expo{x\root{3}}}\,\dd x} =\half\bracks{\arctan\pars{\root{3}} + \arctan\pars{\root{3} \over 3}} \\[3mm]&=\half\pars{{\pi \over 3} + {\pi \over 6}} = \color{#00f}{\large{\pi \over 4}} \approx 0.7854 \end{align}

Felix Marin
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I would consider the Laplace transform

$$F(p)=\int_0^{\infty} dx \, \frac{\sin{x}}{x} \cos^2{x} \, e^{-p x}$$

Then

$$\begin{align}F'(p) &= -\int_0^{\infty} dx \, \sin{x} \cos^2{x} \, e^{-p x}\\ &= -\frac14 \int_0^{\infty} dx \,(\sin{3 x}+\sin{x}) e^{-p x}\\ &= -\frac14 \left (\frac{3}{p^2+9}+\frac1{p^2+1} \right )\end{align}$$

Thus

$$F(p) = -\frac14 \left (\arctan{\frac{p}{3}}+\arctan{p} \right ) +C$$

The integration constant is

$$\begin{align}C &= \int_0^{\infty} dx \frac{\sin{x}}{x} \cos^2{x}\\ &= \frac14 \int_0^{\infty} dx \frac{\sin{3 x}}{x} + \frac14 \int_0^{\infty} dx \frac{\sin{x}}{x} \\ &= \frac14 \left ( \frac{\pi}{2} + \frac{\pi}{2}\right )\\ &= \frac{\pi}{4}\end{align}$$

so that

$$F(p) = \frac{\pi}{4} - \frac14 \left (\arctan{\frac{p}{3}}+\arctan{p} \right ) $$

and your answer is $2 F(\sqrt{3})$ (which accounts for the factor of two in the original integral):

$$\begin{align}2 F(\sqrt{3}) &= \frac{\pi}{2} - \frac12 \left (\arctan{\frac{\sqrt{3}}{3}}+\arctan{\sqrt{3}} \right )\\ &= \frac{\pi}{2} - \frac12 \left (\frac{\pi}{6} + \frac{\pi}{3} \right ) \\ &= \frac{\pi}{4}\end{align}$$

Ron Gordon
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