$\int (x\cos x-\sin x)/x^3\ dx$

To avoid manipulating singularities, start by considering $G_t(u) \; = \; \int_0^\infty \frac{\sin ux}{x}e^{-tx}\,dx \hspace{2cm} u \in \mathbb{R}\,,\,t > 0$ Then $G_t(0) = 0$ and $G_t'(u) \; = \; \int_0^\infty e^{-tx} \cos ux\,dx \; = \; \tfrac12\int_0^\infty e^{-tx}\big(e^{iux} + e^{-iux}\big)\,dx \; =\; \tfrac12\left(\frac{1}{t-iu} + \frac{1}{t+iu}\right) \; =\; \frac{t}{t^2+u^2}$ and hence $G_t(u) \; = \; \int_0^u \frac{t\,dv}{t^2 + v^2} \; = \; \tan^{-1}\tfrac{u}{t} \hspace{2cm} u \in \mathbb{R}\,,\,t > 0$ If we now define $F_t(u) \; = \; \int_0^\infty \frac{ux \cos ux - \sin ux}{x^3} e^{-tx}\,dx \hspace{2cm} u \in \mathbb{R}\,,\,t > 0$ then $F_t(0) = 0$ and $F_t'(u) \; = \; \int_0^\infty \frac{x \cos ux - ux^2 \sin ux - x\cos ux}{x^3}e^{-tx}\,dx \; = \; -u\int_0^\infty \frac{\sin ux}{x}e^{-tx}\,dx \; = \; -uG_t(u) \; = \; -u\tan^{-1}\tfrac{u}{t}$ so that $F_t(u) \; = \; -\int_0^u v \tan^{-1}\tfrac{v}{t}\,dv \hspace{2cm} u \in \mathbb{R}\,,\,t > 0$ Letting $t \to 0+$ gives $\int_0^\infty \frac{ux \cos ux - \sin ux}{x^3}\,dx \; = \; \lim_{t \to 0+}F_t(u) \; = \; -\tfrac12\pi\int_0^\infty\,v\,dv \; = \; -\tfrac14\pi u^2$ and hence the desired integral is, putting $u=1$ , equal to $\boxed{-\tfrac14\pi}$ . Alternatively, integration by parts enables us to evaluate $F_t(u)$ explicitly $F_t(u) \; = \; -\tfrac12u^2 \tan^{-1}\tfrac{u}{t} + t^2\big[\tfrac{u}{t} - \tan^{-1}\tfrac{u}{t}\big]$ and we can readily take the limit of this expression as $t \to 0+$ .

Adding the term $e^{-tx}$ ensures that all integrands are Lebesgue-integrable for all $t>0$ , which makes the process of differentiation under the integral sign valid. The validity of taking the limit as $t \to 0+$ follows from the Dominated Convergence Theorem.

Relevant wiki: Contour Integration

Define $I(\varepsilon, R) = \int_\varepsilon^R \frac{x\cos x - \sin x}{x^3}\,dx$ and note that we want to find $I := \displaystyle\lim_{\varepsilon\to0^+}\lim_{R\to\infty} I(\varepsilon, R)$ .

The function $\frac{x\cos x - \sin x}{x^3}$ is even, so we have $2I(\varepsilon, R) = \int_{-R}^{-\varepsilon} \frac{x\cos x - \sin x}{x^3}\,dx + \int_\varepsilon^R \frac{x\cos x - \sin x}{x^3}\,dx,$ and $\frac{(ix-1)e^{ix}}{x^3} = \frac{-\cos x - x\sin x}{x^3} + i\cdot \frac{x\cos x - \sin x}{x^3}$ so that for real values of $x$ , we have $\text{Im}\left(\frac{(ix-1)e^{ix}}{x^3}\right) = \frac{x\cos x - \sin x}{x^3}.$ Using this, we can write $2I(\varepsilon, R) = \text{Im}\left(\int_{-R}^{-\varepsilon} \frac{(ix-1)e^{ix}}{x^3}\,dx + \int_\varepsilon^R \frac{(ix-1)e^{ix}}{x^3}\,dx\right).$

Using the expression on the right-side to direct us, we consider the contour integral $\int_{\gamma} \frac{(iz-1)e^{iz}}{z^3}\,dz$ where $\gamma = \gamma_{-} \cup (-\gamma_{\varepsilon}) \cup \gamma_{+} \cup (-\gamma_{R})$ has four pieces defined by $\begin{aligned} \gamma_{-} &= \text{ Line segment from } z=-R \text{ to } z=-\varepsilon \\ \gamma_{+} &= \text{ Line segment from } z=\varepsilon \text{ to } z=R \\ \gamma_\varepsilon &= \text{ Upper half of the circle } |z|=\varepsilon, \text{ traversed counterclockwise} \\ \gamma_R &= \text{ Upper half of the circle } |z|=R, \text{ traversed clockwise} \end{aligned}$ Since the integrand is analytic everywhere except $z=0$ , which lies outside $\gamma$ , the Cauchy Goursat Theorem tells us that $\int_\gamma \frac{(iz-1)e^{iz}}{z^3}\,dz = 0,$ so that $\begin{aligned} \int_{-R}^{-\varepsilon} \frac{(ix-1)e^{ix}}{x^3}\,dx + \int_\varepsilon^R \frac{(ix-1)e^{ix}}{x^3}\,dx &= \int_{\gamma_{-} \cup \gamma_{+}} \frac{(iz-1)e^{iz}}{z^3}\,dz \\ &= \int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz + \int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz \end{aligned}$ $\begin{aligned} \implies \quad I &= \frac{1}{2} \lim_{\varepsilon\to 0^+}\lim_{R\to\infty} 2I(\varepsilon, R) \\ &= \frac{1}{2} \lim_{\varepsilon\to 0^+}\lim_{R\to\infty} \text{Im}\left(\int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz + \int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz\right) \\ &= \frac{1}{2} \text{Im}\left(\left(\lim_{\varepsilon\to 0^+}\int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz\right) + \left(\lim_{R\to\infty}\int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz\right)\right) \end{aligned}$

Therefore, we compute each of these two integrals separately:

• For $\gamma_\varepsilon$ : We use $e^{iz} = \sum_{n=0}^\infty \frac{i^n z^n}{n!}$ to write $\frac{(iz-1)e^{iz}}{z^3} = (iz-1)\sum_{n=0}^\infty \frac{i^n z^{n-3}}{n!} = -\frac{1}{z^3} - \frac{1}{2z} + \sum_{n=0}^\infty (n+2) i^{n-1} \frac{z^n}{(n+3)!} = -\frac{1}{z^3} - \frac{1}{2z} + f(z)$ where $f$ is entire, so it has a continuous primitive $F$ defined everywhere. If we define $\ln z$ to be a branch of the complex logarithm on the complex plane minus the negative imaginary axis, then we have $\begin{aligned} \int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz &= \int_{\gamma_\varepsilon} \left(-\frac{1}{z^3} - \frac{1}{2z} + f(z)\right)\,dz \\ &= \frac{1}{2z^2} - \frac{1}{2}\ln z + F(z) \Big|_\varepsilon^{-\varepsilon} \\ &= \left(\frac{1}{2(-\varepsilon)^2} - \frac{1}{2\varepsilon^2}\right) - \frac{1}{2}\left(\ln(-\varepsilon) - \ln\varepsilon\right) + F(-\varepsilon) - F(\varepsilon) \\ &= 0 - \frac{\pi i}{2} + F(-\varepsilon) - F(\varepsilon) \end{aligned}$ so that, since $F$ is continuous, $\lim_{\varepsilon\to 0^+} \int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz = -\frac{\pi i}{2} + \lim_{\varepsilon\to 0^+} \left(F(-\varepsilon) - F(\varepsilon)\right) = -\frac{\pi i}{2}$
• For $\gamma_R$ : We note that $\left|\int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz \right| \leq (\pi R) \cdot \max_{z\in \gamma_R} \left|\frac{(iz-1)e^{iz}}{z^3}\right| \leq \frac{\pi(R+1)}{R^2} \cdot \max_{z\in \gamma_R} \left(e^{-\text{Im}(z)}\right) = \frac{\pi(R+1)}{R^2} \xrightarrow{R\to\infty} 0$ so that $\lim_{R\to\infty} \int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz = 0.$

Now combining all our results, we find the answer: $I = \frac{1}{2} \text{Im}\left(\left(\lim_{\varepsilon\to 0^+}\int_{\gamma_\varepsilon} \frac{(iz-1)e^{iz}}{z^3}\,dz\right) + \left(\lim_{R\to\infty}\int_{\gamma_R} \frac{(iz-1)e^{iz}}{z^3}\,dz\right)\right) = \frac{1}{2} \text{Im}\left(-\frac{\pi i}{2} + 0\right) = -\frac{\pi}{4} \approx \boxed{-0.785398163}$