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I'm working through a contour integral question, which is rounded off by finding the integral:

$$\int^{\infty}_{0} \frac{x-\sin(x)}{x^3} dx$$

I have already shown that the residue at $0$ of the function

$$f(z)=\frac{1+iz-e^{iz}}{z^3}$$

on $\mathbb{C} - \{0\}$ is $\frac{1}{2}$, and that

$$\int_{\gamma_R} f(z) dz \longrightarrow 0$$ as $R \longrightarrow 0$ where $\gamma_R:[0,\pi]\rightarrow \mathbb{C}$ is given by $\gamma_R(t)=Re^{it}$.

Problem

How do I progress from here to finding the required integral? It's an odd function, so clearly all that is left is to integrate from $R$ to $-R$ along the real axis and half it to find the integral, but I can't see how to pull out the '$x-\sin(x)$' and replace '$z^3$' with '$x^3$'.

An explanation of how to finish this off would be much appreciated. Thanks in advance.

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Is the integral you're trying to evaluate really from 0 to $\pi$, or should it be to $\infty$? –  Harald Hanche-Olsen May 27 '12 at 19:14
    
@HaraldHanche-Olsen: Sorry, it should be infinity. I'll correct it. –  Mathmo May 27 '12 at 19:20
    
Okay. I suppose you have already observed that the function you wish to integrate is an even function, and moreover, it is the imaginary part of $f(x)$, when $x$ is real? –  Harald Hanche-Olsen May 27 '12 at 19:27

2 Answers 2

up vote 3 down vote accepted

Since the integrand is even, $$ \int^{\infty}_{0} \frac{x-\sin(x)}{x^3}\,dx=\frac12\int^{\infty}_{-\infty} \frac{x-\sin(x)}{x^3}\,dx. $$ Take $\epsilon>0$ small and $R>0$ large. The integral of $f(z)$ along the closed path formed by the segment $[\epsilon,R]$, the semi-circumference $\gamma_R$ counterclockwise, the segment $[-R,-\epsilon]$ and the semi-circumference $\gamma_\epsilon$ of radius $\epsilon$ (in the upper half-plane) clockwise is equal to $0$. Then $$ \Bigl(\int_{-R}^{-\epsilon}+\int_{\epsilon}^{R}\Bigr)f(x)\,dx+\int_{\gamma_R}f(z)\,dz+\int_{\gamma_\epsilon}f(z)\,dz=0. $$ As $R\to\infty$ and $\epsilon\to0$, $$ \Bigl(\int_{-R}^{-\epsilon}+\int_{\epsilon}^{R}\Bigr)f(x)\,dx\to\int^{\infty}_{-\infty} \frac{1-\cos(x)}{x^3} dx+i\int^{\infty}_{-\infty} \frac{x-\sin(x)}{x^3}\,dx. $$ You already know that $\lim_{R\to\infty}\int_{\gamma_R}f(z)\,dz=0$. All is left is to find $\lim_{\epsilon\to0}\int_{\gamma_\epsilon}f(z)\,dz$.

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Thank you! Very very helpful work. –  Mathmo May 27 '12 at 20:26
    
Julian, can $\lim_{\epsilon\to0}\int_{\gamma_\epsilon}f(z)\,dz$ be computed by hand? I reduced it to $\lim_{\varepsilon\to 0} \frac{1}{\varepsilon^2}\int_0^\pi\frac{1+i\varepsilon^{it}-e^{i\varepsilon\ \cdot e^{it}}}{e^{2it}}dt$ but that is indeterminate. –  The Substitute Dec 4 at 14:49
1  
For $z$ close to $0$ $$1+i\,z-e^{iz}=\frac{z^2}{2}+O(z^3).$$ –  Julián Aguirre Dec 4 at 15:04

Since the integrand has no singularities and the it vanishes on $x+it$ for $t\in[-1,0]$ as $|x|\to\infty$, we get $$ \begin{align} \int_0^\infty\frac{x-\sin(x)}{x^3}\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{x-\sin(x)}{x^3}\mathrm{d}x\\ &=\frac12\int_{-\infty-i}^{\infty-i}\frac{x-\sin(x)}{x^3}\mathrm{d}x\\ &=\frac12\int_{-\infty-i}^{\infty-i}\frac{\mathrm{d}x}{x^2} -\frac1{4i}\int_{-\infty-i}^{\infty-i}\frac{e^{ix}}{x^3}\mathrm{d}x +\frac1{4i}\int_{-\infty-i}^{\infty-i}\frac{e^{-ix}}{x^3}\mathrm{d}x\\ &=\frac12\int_{\gamma^+}\frac{\mathrm{d}z}{z^2} -\frac1{4i}\int_{\gamma^+}\frac{e^{iz}}{z^3}\mathrm{d}z +\frac1{4i}\int_{\gamma^-}\frac{e^{-iz}}{z^3}\mathrm{d}z\\ &=0-\frac\pi2\left(-\frac12\right)+0\\[6pt] &=\frac\pi4 \end{align} $$ $\gamma^+$ follows $(-\infty-i)$ to $(\infty-i)$ then circles back counter-clockwise through the upper half-plane along the circle centered at $-i$.

$\gamma^-$ follows $(-\infty-i)$ to $(\infty-i)$ then circles back clockwise through the lower half-plane along the circle centered at $-i$.

The residue of $\dfrac1{z^2}$ at $z=0$ is $0$ and its integral is $0$ along the circular arc of $\gamma^+$.

The residue of $\dfrac{e^{iz}}{z^3}$ at $z=0$ is $-\dfrac12$ and its integral is $0$ along the circular arc of $\gamma^+$.

$\dfrac{e^{-iz}}{z^3}$ has no residues in $\gamma^-$ and its integral is $0$ along the circular arc of $\gamma^-$.

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