Integral of $\frac{\cos(2x) − \cos(x)}{x^2}$

Question

Will you please solve $$ \int_{0}^{\infty}{\frac{\cos 2x− \cos x}{x^2}dx}$$ using indented contour. I tried like in $$\frac{\sin x}{x}$$ but couldn't figure out.

Answer 1

Let your integral be $J$. By the usual evenness argument, look at $$ I = \int_{\gamma_{R,\varepsilon}} \frac{e^{2i z}-e^{iz}}{z^2} \, dz, $$ where $C$ is the contour made out of $\gamma_1=[\varepsilon,R]$, $\gamma_2$ a semicircle of radius $R$ in the upper half-plane centred at the origin, traversed anticlockwise, $\gamma_3 = [-R,-\varepsilon]$, and $\gamma_4$ is the semicircle of radius $\varepsilon$ around the origin in the upper half-plane, tranversed in the negative direction.

Obviously the integral over $\gamma_1 \cup \gamma_3 $ tends to double the integral we want, $J$, by evenness. The integral over $\gamma_2$ vanishes by Jordan's lemma. The contour contains no poles, and hence $I=0$, or $$ 2J = -\int_{\gamma_4} \frac{e^{2i z}-e^{iz}}{z^2} \, dz $$ At this point we use a series expansion: $$ 2J = -\int_{\gamma_4} \frac{1+2iz-1-iz+O(z^2)}{z^2} \, dz = -\int_{\gamma_4} \left( \frac{i}{z} + O(1) \right) \, dz $$ Clearly the $O(1)$ term will disappear as $\varepsilon \to 0$. For the last bit, set $z= \varepsilon e^{i(\pi-\theta)}$, so $dz/z = -i \, d\theta $, $$ J = -\frac{1}{2}\int_0^{\pi} i (-i) \, d\theta = -\frac{\pi}{2}. $$

Answer 2

We have $$\cos(2x) - \cos(x) = (1-2\sin^2(x)) - (1-2\sin^2(x/2)) = 2\left(\sin^2(x/2) - \sin^2(x)\right)$$ Hence, the integral is $$\dfrac{I}2 = \int_0^{\infty} \dfrac{\sin^2(x/2)}{x^2}dx - \int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx = \int_0^{\infty} \dfrac{\sin^2(t)}{4t^2} \cdot (2dt) - \int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx$$ This gives us $$I = -\int_0^{\infty} \dfrac{\sin^2(x)}{x^2}dx$$ And from the post here and other posted linked from that, we have $$\boxed{\color{blue}{I = - \dfrac{\pi}2}}$$