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This is an integral classified as "rational function multiplied by a trigonometric function", but I do not see how I could satisfy conditions of Jordan's lemma. Namely, it's obvious that singularities are $x = i$, $x = -i$ and $x = 0$, where all three are $\textit{first-order poles}$ (hence, $x = 0$ satisfies conditions for residues $\operatorname{Im}(z) > 0$ and $\operatorname{Im}(z) = 0$), but what contour should I choose and how do I calculate the integral? Integral is $$\int_{-\infty}^{\infty} \frac {x - \sin x} {x^3(x^2 + 1)}dx.$$ Also, I've searched online for it thoroughly, but I haven't found anything about it. Any help is appreciated. Thank you.

Travis Willse
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Nemanja Beric
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    Am I aloud to do that? Assistant told us we can only calculate "main values" of these integrals and iff singularities over real line are first order poles. – Nemanja Beric Jan 22 '18 at 21:06
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    Actually, I missed the singularity at the origin too. You'll need to add a contour that takes a semicircle around the origin to avoid it; but it's ok, because the numerator vanishes to third order, just like the denominator. The singularity is removable. –  Jan 22 '18 at 21:10
  • @user296602 There is no singularity at the origin. – José Carlos Santos Jan 22 '18 at 21:11
  • @JoséCarlosSantos There is a singularity at the origin, because of $1/z^3$; but it's a removable one. –  Jan 22 '18 at 21:12
  • @user296602 All right. I'll rephrase it: there is no non-removable singularity at the origin. – José Carlos Santos Jan 22 '18 at 21:13
  • Yes, you are right. But how do I then solve the integral? I cannot split it as sum, because the integrals I get do have singularities at the origin? – Nemanja Beric Jan 22 '18 at 21:13
  • @user296602 For $x\to 0$ the entire function goes to $1/6$. There is NO singularity at the origin. –  Jan 22 '18 at 21:13
  • By the way, just for the sake of curiosity. I calculated the integral and the result shall be

    $$\frac{\pi}{2} - \frac{\pi}{e}$$

     –  Jan 22 '18 at 21:16
  • Yes, that should be solution. Workbook claims it too. Do you have an advice? For both types, "rational function" and "rational function times trigonometric" we use upper semi-circle of radius R > 0. But I do not see how could that help me here. – Nemanja Beric Jan 22 '18 at 21:19

2 Answers2

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Think about the contour $C_1$ that starts at $z=\frac12$, follows the real axis from there to $z=-\frac12$, and then follows the circle $|z|=\frac12$ counterclockwise back to $z=\frac12$. There are no singularities except for the removable one at $z=0$ within or along the contour, so $$\oint_{C_1}\frac{z-\sin z}{z^3\left(z^2+1\right)}dz=0$$ Thus if $C_2$ runs form $z=-R$ to $z=R$ along the real axis, $$\begin{align}\int_{-R}^{R}\frac{x-\sin x}{x^3\left(x^2+1\right)}dx&=\int_{C_2}\frac{z-\sin z}{z^3\left(z^2+1\right)}dz\\ &=\int_{C_2}\frac{z-\sin z}{z^3\left(z^2+1\right)}dz+\int_{C_1}\frac{z-\sin z}{z^3\left(z^2+1\right)}dz\\ &=\int_{C_3}\frac{z-\frac{e^{iz}}{2i}}{z^3\left(z^2+1\right)}dz+\int_{C_3}\frac{\frac{e^{-iz}}{2i}}{z^3\left(z^2+1\right)}dz\end{align}$$ Where $C_3$ runs from $z=-R$ to $z=-\frac12$ along the real axis, from $z=-\frac12$ to $z=\frac12$ along the circle $|z|=\frac12$, and then to $z=R$ along the real axis. We close the first of the integrals above with a semicircle at $|z|=R$, enclosing the poles at $z=0$ and $z=i$ counterclockwise and the second below with a semicircle at $|z|=R$, enclosing the pole at $z=-i$ clockwise, and then take the limit as $R\rightarrow\infty$ of both integrals.

Thus we get $2\pi i$ times the residues at $z=0$ and $z=i$ of the first integrand and $-2\pi i$ times the residue at $z=-i$ of the second integrand.

$$\int_{-\infty}^{\infty}\frac{x-\sin x}{x^3\left(x^2+1\right)}dx=2\pi i\left(-\frac34i+\frac i2-\frac{e^{-1}}{4i}\right)-2\pi i\left(\frac{e^{-1}}{4i}\right)=\frac{\pi}2-\frac{\pi}e$$ As in the comment of @Artificial Intelligence above.

user5713492
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{-\infty}^{\infty}{x - \sin\pars{x} \over x^{3}\pars{x^{2} + 1}}\dd x} = \Im\int_{-\infty}^{\infty}{x\ic - \expo{\ic x} + 1 - x^{2}/2\over x^{3}\pars{x^{2} + 1}}\dd x \\[5mm] = & \ \Im\braces{2\pi\ic\on{Res}\bracks{{x\ic - \expo{\ic x} + 1 - x^{2}/2\over x^{3}\pars{x^{2} + 1}}, x = \ic}} \\[5mm] = & \ \Im\braces{2\pi\ic{\ \ic \times \ic - \expo{\ic\ \times\ \ic} + 1 - \ic^{2}/2\ \over \ic^{3}\pars{\ic + \ic}}} \\[5mm] = & \ \bbx{\color{#44f}{{\expo{} - 2 \over 2\expo{}}\,\pi}} \approx 0.4151\\ & \end{align}

Felix Marin
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