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I found this nice result.

Prove that

$$\int_{0}^{\infty}\sin{x}\arctan\left({\frac{1}{x}}\right)\,\mathrm dx=\frac{\pi }{2} \left(\frac{e-1}e\right)$$

enter image description here

I tried some methods but I can't evaluate it.

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Here is another way to evaluate the integral. Notice that $$\int_0^1 \frac{x}{x^2+y^2}\mathrm dy=\arctan\left(\frac{1}{x}\right)$$ We also have $$\int_{0}^{\infty}\frac{{x}\sin{x}}{x^2+{y^2}}\,\mathrm dx=\frac{\pi}{2}e^{-y}$$ Hence \begin{align} \int_{0}^{\infty}\sin{x}\arctan\left({\frac{1}{x}}\right)\,\mathrm dx&=\int_{0}^{\infty}\sin{x}\int_0^1 \frac{x}{x^2+y^2}\mathrm dy\,\mathrm dx\\ &=\int_0^1 \int_{0}^{\infty}\frac{x\sin{x}}{x^2+y^2}\mathrm dx\,\mathrm dy\\ &=\frac{\pi}{2}\int_0^1 e^{-y}\,\mathrm dy\\ &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\pi }{2} \left(\frac{e-1}e\right)}} \end{align}

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  • $\begingroup$ +1 Actually, this solution is the same as that of Integrator's, but with a slightly different presentation. $\endgroup$ – M.N.C.E. Dec 15 '14 at 13:55
  • $\begingroup$ @M.N.C.E. Indeed, you're right. Thanks. $\endgroup$ – Venus Dec 15 '14 at 13:57
  • $\begingroup$ +1. I was thinking again in my answer to changed to a simpler one. However, I found that your answer is the simplest one. $\endgroup$ – Felix Marin Dec 19 '14 at 13:36
  • $\begingroup$ @Venus, how did you convert the single integral into a double integral? I am interested in this method, but cant find a lead... $\endgroup$ – Amad27 Dec 21 '14 at 8:21
  • $\begingroup$ @Amad27 Double integral method is essentially a differentiation under integral sign with the bounds. I learned this method here on MSE. You may find it this technique in Dirichlet integral. $\endgroup$ – Venus Dec 21 '14 at 9:53
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Consider following parametric integral

$$I(\alpha)=\int_{0}^{\infty}\sin{x}\arctan\left({\dfrac{\alpha}{x}}\right)\,\mathrm dx$$ We have $I(0)=0$ and $I(1)$ yields required Integral.

Differentiating wrt $\alpha$, we get $$I'(\alpha)=\int_{0}^{\infty}\frac{{x}\sin{x}}{x^2+{\alpha^2}}\,\mathrm dx=\frac{\pi}{2}e^{-\alpha}$$

$I'(\alpha)$Integrating wrt $\alpha$, we get $$I(\alpha)=-\frac{ \pi}{2} e^{-\alpha}+c$$

$$I(0)=-\frac{\pi}{2} +c=0\implies c=\frac\pi2$$ Hence, $$I(\alpha)=\int_{0}^{\infty}\sin{x}\arctan\left({\dfrac{\alpha}{x}}\right)\,\mathrm dx=\frac{ \pi }{2}\Big(1- e^{-\alpha}\Big)$$

$$I(1)=\frac{\pi }{2} \left(1- \frac1e\right)=\frac{\pi }{2} \left(\frac{e-1}e\right)$$

$$\large \int_{0}^{\infty}\sin{x}\arctan\left({\dfrac{1}{x}}\right)\,\mathrm dx=\frac{\pi }{2} \left(\frac{e-1}e\right)$$

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    $\begingroup$ Poor me I am very late for this one. Nice answer anyway, (+1). $\endgroup$ – Venus Dec 15 '14 at 12:07
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With an integration by parts, we have $$ \int_{0}^{\infty}\sin{x}\arctan{\dfrac{1}{x}}\,\mathrm dx=\left.-\cos{x}\arctan{\dfrac{1}{x}}\right|_0^{\infty}-\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x=\frac{\pi}{2}-\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$ then we use the standard integral $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x =\frac{\pi}{2}e^{-1}$$ proved here to get the anounced result: $$ \frac{\pi}{2}\left(1-e^{-1}\right).$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}\sin\pars{x}\arctan\pars{1 \over x}\,\dd x ={\pi \over 2}\,{\expo{} - 1 \over \expo{}}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}\sin\pars{x}\arctan\pars{1 \over x}\,\dd x} =\half\int_{-\infty}^{\infty}\sin\pars{x}\arctan\pars{1 \over x}\,\dd x \\[5mm]&=\half\,\Im\ \overbrace{% \int_{-\infty}^{\infty}\expo{\ic x}\arctan\pars{1 \over x}\,\dd x} ^{\ds{\dsc{\ic x}\ =\ \dsc{t}\ \imp\ \dsc{x}\ =\ \dsc{-\ic t}}} =\half\,\Im\int_{-\infty\ic}^{\infty\ic}\expo{t}\arctan\pars{\ic \over t} \,\pars{-\ic}\,\dd t \\[5mm]&=\half\,\Im\int_{-\infty\ic}^{\infty\ic}\expo{t} \,{\rm arctanh}\pars{1 \over t}\,\dd t ={1 \over 4}\,\Im\int_{-\infty\ic}^{\infty\ic} \expo{t}\ln\pars{t + 1 \over t - 1}\,\dd t \\[5mm]&={1 \over 4}\,\Im\int_{-\infty\ic}^{\infty\ic}\expo{t} \int_{-1}^{1}{\dd\xi \over t + \xi} ={1 \over 4}\,\Im\int_{-1}^{1}\int_{-\infty\ic}^{\infty\ic}\expo{t} {\dd t \over t + \xi}\,\dd\xi \\[5mm]&={1 \over 4}\,\Im\int_{-1}^{1}\pars{-\int_{-\infty}^{0}\expo{t} {\dd t \over t + \xi + \ic 0^{+}}-\int_{0}^{-\infty}\expo{t} {\dd t \over t + \xi - \ic 0^{+}}}\,\dd\xi \\[5mm]&={1 \over 4}\,\Im\int_{-1}^{1} \int_{-\infty}^{0}\expo{t}\pars{% {1 \over t + \xi - \ic 0^{+}} - {1 \over t + \xi + \ic 0^{+}}}\,\dd t\,\dd\xi \\[5mm]&={1 \over 4}\,\Im\int_{-1}^{1} \int_{-\infty}^{0}\expo{t}\bracks{2\pi\ic\,\delta\pars{t + \xi}}\,\dd t\,\dd\xi ={\pi \over 2}\int_{-1}^{1}\expo{-\xi}\Theta\pars{\xi}\,\dd\xi ={\pi \over 2}\int_{0}^{1}\expo{-\xi}\,\dd\xi \\[5mm]&={\pi \over 2}\pars{-\expo{-1} + 1} =\color{#66f}{\large{\pi \over 2}\,{\expo{} - 1 \over \expo{}}} \end{align}

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