Problem of the Week #294 - May 4, 2020

Status
Not open for further replies.

Staff member
Last edited:

Euge

MHB Global Moderator
Staff member
Congratulations to GJA , Ackbach , and Dhamnekar Winod for their correct solutions. You can read GJA 's solution below.

We will show $$\int_{0}^{\infty}\frac{x-\sin(x)}{x^{3}}dx = \frac{\pi}{4}$$ by considering the auxiliary integral $$\int_{-\infty}^{\infty}\frac{x+ie^{ix}}{x^3}dx\qquad (1);$$ the motivation for which comes from noting that $f(x) = \dfrac{x-\sin(x)}{x^{3}}$ is an even function and $\Re(x+ie^{ix}) = -\sin(x)$. We evaluate (1) using the multicolored contour, $C$, shown below, where the inner and outer semicircles have radii $\varepsilon$ and $R$, respectively.

Since the integrand in (1) is analytic/holomorphic on and inside $C$ when extended to $\mathbb{C}$, the residue theorem tells us $$\int_{C}\frac{z+ie^{iz}}{z^3}dz = 0\qquad (2).$$ Next, we observe that $$\left |\int_{\text{Blue}}\frac{z+ie^{iz}}{z^{3}}dz\right |\leq\int_{\text{Blue}}\frac{|z|+1}{|z|^3}dz=\frac{\pi R(R+1)}{R^{3}}\longrightarrow 0\,\,\,\text{as}\,\,\, R\rightarrow \infty.$$ Hence, in the limit as $R\rightarrow\infty$, (2) becomes $$\int_{-\infty}^{-\varepsilon}\frac{x+ie^{ix}}{x^3}dx+\int_{\text{Green}}\frac{z+ie^{iz}}{z^{3}}dz + \int_{\varepsilon}^{\infty}\frac{x+ie^{ix}}{x^3}dx = 0.$$ Using the substitution $x\mapsto -x$ in the first integral immediately above and the identity $\sin(x) = \dfrac{e^{ix}-e^{-ix}}{2i}$, the previous equation yields $$\int_{\varepsilon}^{\infty}\frac{x-\sin(x)}{x^{3}}dx = -\frac{1}{2}\int_{\text{Green}}\frac{z+ie^{iz}}{z^{3}}dz\qquad (3).$$ Along the green contour we have $z = \varepsilon e^{i\theta}$ from $\theta = \pi$ to $\theta = 0$; hence, $$\int_{\text{Green}}\frac{z+ie^{iz}}{z^{3}}dz = \int_{\pi}^{0}\frac{\varepsilon e^{i\theta}+ie^{i\varepsilon e^{i\theta}}}{\varepsilon^{3}e^{3i\theta}}i\varepsilon e^{i\theta}d\theta = -\frac{i}{\varepsilon^{2}}\int_{0}^{\pi}\frac{\varepsilon e^{i\theta}+ie^{i\varepsilon e^{i\theta}}}{e^{2i\theta}}d\theta.$$ Expanding $e^{i\varepsilon e^{i\theta}}$ in a power series the above becomes $$\frac{1}{\varepsilon^{2}}\int_{0}^{\pi}\frac{1-\dfrac{\varepsilon^{2}e^{2i\theta}}{2}+\mathcal{O}(\varepsilon^{3})}{e^{2i\theta}}d\theta = \frac{1}{\varepsilon^{2}}\left[\int_{0}^{\pi}e^{-2i\theta}d\theta-\frac{\varepsilon^{2}}{2}\int_{0}^{\pi}d\theta + \mathcal{O}(\varepsilon^{3})\int_{0}^{\pi}e^{-2i\theta}d\theta\right],$$ where $\mathcal{O}$ is taken to mean for $\varepsilon\rightarrow 0$. Hence, taking $\varepsilon\rightarrow 0$ in (3) we obtain $$\int_{0}^{\infty}\frac{x-\sin(x)}{x^{3}}dx = \left(-\frac{1}{2}\right)\left(-\frac{\pi}{2}\right) = \frac{\pi}{4}.$$

Status
Not open for further replies.