Evaluate \[ \int_{-\infty}^{\infty}\frac{x-\sin x}{x^3}\,dx. \]
Solution
1. The integrand has a removable singularity at \(0\), is integrableon \(\RR\), and is even.
The Taylor expansion \[ \sin x = x-\frac{x^3}{6}+O(x^5) \] gives \[ \frac{x-\sin x}{x^3} = \frac16+O(x^2) \] as \(x\to0\). Thus the singularity at \(0\) is removable.
For \(|x|\ge1\), \[ \left| \frac{x-\sin x}{x^3} \right| \le \frac1{x^2}+\frac1{|x|^3}, \] so the integral converges absolutely at infinity.
Finally, both \(x-\sin x\) and \(x^3\) are odd, so their quotient is even.
2. Put\[ J \coloneqq \int_0^\infty\frac{x-\sin x}{x^3}\,dx. \]Then\[ J = \frac12 \int_0^\infty\frac{1-\cos x}{x^2}\,dx. \]
Perform integration by parts on \([\varepsilon,R]\) with \[ u=x-\sin x, \qquad dv=\frac{dx}{x^3}. \] Then \[ du=(1-\cos x)\,dx, \qquad v=-\frac1{2x^2}, \] so \[ \int_\varepsilon^R\frac{x-\sin x}{x^3}\,dx = \left[ -\frac{x-\sin x}{2x^2} \right]_\varepsilon^R + \frac12 \int_\varepsilon^R\frac{1-\cos x}{x^2}\,dx. \]
As \(R\to\infty\), \[ \frac{R-\sin R}{R^2}\to0. \] As \(\varepsilon\to0^+\), \[ \frac{\varepsilon-\sin\varepsilon}{\varepsilon^2} = O(\varepsilon) \to0. \] Taking the limits gives the claim.
3. One has\[ \int_0^\infty\frac{1-\cos x}{x^2}\,dx = \int_0^\infty\frac{\sin x}{x}\,dx. \]
Again integrate by parts on \([\varepsilon,R]\), now with \[ u=1-\cos x, \qquad dv=\frac{dx}{x^2}. \] Then \[ du=\sin x\,dx, \qquad v=-\frac1x, \] and hence \[ \int_\varepsilon^R\frac{1-\cos x}{x^2}\,dx = \left[ -\frac{1-\cos x}{x} \right]_\varepsilon^R + \int_\varepsilon^R\frac{\sin x}{x}\,dx. \] The boundary term at infinity tends to \(0\) because \(1-\cos R\) is bounded, and the boundary term at \(0\) tends to \(0\) because \[ 1-\cos\varepsilon = O(\varepsilon^2). \] Taking limits proves the identity.
4. The half-line integral is\[ J=\frac{\pi}{4}. \]
The standard Dirichlet integral gives \[ \int_0^\infty\frac{\sin x}{x}\,dx = \frac{\pi}{2}. \] Combining this with steps <1>2–<1>3 yields \[ J = \frac12\cdot\frac{\pi}{2} = \frac{\pi}{4}. \]
5. Therefore\[ \boxed{ \int_{-\infty}^{\infty} \frac{x-\sin x}{x^3}\,dx = \frac{\pi}{2}. } \]
By step 1 the integrand is even, so \[ \int_{-\infty}^{\infty} \frac{x-\sin x}{x^3}\,dx = 2J. \] Now apply step 4.
6. Q.E.D.
Step 5 gives the requested value.