An integral containing a well known limit.

We will focus us on the integral $I = \int_0^\infty 1-x\sin\left(\frac{1}{x}\right)\mathrm{d}x.$ Substitute $x = \frac{1}{u}$ , then $\mathrm{d}x = -\frac{1}{u^2}\mathrm{d}u$ and so we notice that $I = -\int_\infty^0 \frac{u-\sin(u)}{u^3} \mathrm{d}u = \int_0^\infty \frac{u-\sin(u)}{u^3} \mathrm{d}u.$ We will now use integration by parts. We will differentiate the part $u-\sin(u)$ and integrate the part $\frac{1}{u^3}$ to get $I = \left. -\frac{u-\sin(u)}{2u^2} \right|_0^\infty + \int_0^\infty \frac{1-\cos(u)}{2u^2} \mathrm{d}u.$ We will repeat the last step with the new integral to get $I = \left. -\frac{u-\sin(u)}{2u^2} - \frac{1-\cos(u)}{2u} \right|_0^\infty + \frac{1}{2} \cdot \int_0^\infty \frac{\sin(u)}{u} \mathrm{d}u.$ We will show that $\left. \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u} \right|_0^\infty = 0.$ It is not difficult to see that $\lim_{u\to\infty} \left[ \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u}\right] = 0$ . This can be calculated with the squeeze theorem. We will focus us on the other required limit to evaluate. With l'Hôpital's rule we find that $\lim_{u\to 0} \frac{u-\sin(u)}{2u^2} = \lim_{u\to 0} \frac{1-\cos(u)}{4u} = \lim_{u\to 0} \frac{\sin(u)}{4} = 0,$ which shows us that indeed $\lim_{u\to 0} \left[ \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u}\right] = 0$ and so $\left. \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u} \right|_0^\infty = 0.$ We will now evaluate the integral $J = \int_0^\infty \frac{\sin(u)}{u} \mathrm{d}u.$ We will do this by using Feynman's trick: differentiation under the integral sign. Define $J(t) = \int_0^\infty \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u$ for $t \leq 0$ . Then $\frac{\mathrm{d}}{\mathrm{d}t} J(t) = \frac{\mathrm{d}}{\mathrm{d}t} \int_0^\infty \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u = \int_0^\infty \frac{\partial}{\partial t} \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u = \int_0^\infty e^{tu} \cdot \sin(u) \mathrm{d}u.$ By again using integration by parts, we can show that $J'(t) = 1 - t^2 \cdot J'(t)$ , from which follows that $J'(t) = \frac{1}{1+t^2}$ . We conclude that $J(t) = \int \frac{\mathrm{d} t}{1+t^2} = \arctan(t) + C.$ As for all $x \in (0,\infty)$ , $\left|\frac{\sin(x)}{x}\right| \leq 1,$ it follows that $0 \leq \lim_{t\to -\infty} |J(t)| = \lim_{t\to -\infty} \left|\int_0^{\infty} \frac{e^{tu} \cdot \sin(u)}{u}\mathrm{d}u\right| \leq \lim_{t\to -\infty} \int_0^{\infty} \left|\frac{e^{tu} \cdot \sin(u)}{u}\right|\mathrm{d}u \leq \lim_{t\to -\infty} \int_0^{\infty} e^{tu} \mathrm{d}u = \lim_{t\to -\infty} \left.\frac{e^{tu}}{t}\right|_{u=0}^{u=\infty} = \lim_{t\to -\infty} \frac{-1}{t} =0,$ so $\lim_{t\to -\infty} J(t) = \lim_{t\to -\infty} \arctan(t) + C = 0$ , which implies that $C = \frac{\pi}{2}$ . This means that $J = J(0) = \arctan(0) + \frac{\pi}{2} = \frac{\pi}{2}$ .

By using all the information we got before, we can conclude that $\frac{1}{\pi} \cdot \int_0^\infty \left[1-x\sin\left(\frac{1}{x}\right)\right]\mathrm{d}x = \frac{1}{\pi} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \boxed{\frac{1}{4}}.$