# evaluate $\int _0 ^\infty \frac{1-\cos(ax)}{x^2}dx$

Im trying to evaluate for a given $a\in \mathbb R$$\int _0 ^\infty \frac{1-\cos(ax)}{x^2}dx$$ I have noticed that since$1-\cos(ax)$is analytic in$\mathbb C$, the integral $$\int _{C} \frac{1-\cos(az)}{z^2}dz$$where$C$is a simple closed contour around the point$z_0=0$, is by Cauchy's integral formula: $$(1-\cos(az_0))'=a\sin(az_0)=0$$which might hint that the solution lays within integration of half a circle in the positive (or negative) imaginary plane. I also think that since the function inside the integral is even, it is tempting to try and evaluate the real integral from$-\infty$to$\infty$, and correspondingly evaluate the complex integral for bigger contours. • Yes, since the integrand$f$is an even function, you can write$\int_0^{\infty}f(z)\; dz= \frac12\int_{-\infty}^{\infty}f(z)\; dz$and work with a half-annulus centered at the origin, if it helps. – MPW Feb 29 '16 at 14:35 • result will be$\frac{a\pi}{2}$– tired Feb 29 '16 at 14:50 • You do not need complex integration if you can work with the sine integral. $$a\left( \frac{\cos(a x)-1}{ax}+\mathrm{Si}(a x)\right)$$ is a primitive of the integrand. From$\mathrm{Si}(\pm \infty)=\pm\frac{\pi}{2}$you get for the integral the value$\frac{\pi}{2}|a|.$– gammatester Feb 29 '16 at 14:52 • What i would do: 1.) differentiate w.r.t$a$2.) using the classic integral$\int_R \text{sinc}(x)=\pi$3.) integrate back w.r.t$a$4.) use$I(0)=0$to determine the constant of integration – tired Feb 29 '16 at 14:57 ## 1 Answer Perhaps the following helps. Assume$\;a>0\;$and define : $$I(a):=\int_0^\infty\frac{1-\cos ax}{x^2}dx\implies I'(a)=\int_0^\infty\frac{\sin ax}x=\frac\pi2\implies I(a)=\frac{\pi a}2$$ With complex analysis: for very small real$\;\epsilon>0\;$: $$C_{\epsilon,R}:=[-R,-\epsilon]\cup\gamma_\epsilon=\{\epsilon e^{it}\;:\;\;0\le t<\pi\;\}\cup[\epsilon, R]\cup\gamma_R:=\{Re^{it}\;,\;\;0\le t<\pi\}\;,\;\;R\in\Bbb R^+$$ Define also $$f(z)=\frac{1-e^{iaz}}{z^2}$$ then, as$\;f\;$is analytic on and within$\;C_{\epsilon,R}\;$, we get $$\oint_{C_{\epsilon,R}}f(z)\,dz=0$$ Yet, as the function has a simple pole (check this) at$\;z=0\;$, and $$\text{Res}_{z=0}(f)=\lim_{z\to0}\,(z\,f(z))\stackrel{\text{l'H}}=-ia$$ we can use this to obtain $$\lim_{\epsilon\to0}\int_{\gamma_\epsilon}f(z)\,dz=-\pi i(ia)=\pi a$$ and also $$\left|\int_{\gamma_R}f(z)\,dz\right|\le\pi R\max_{0\le t<\pi}\frac{1+e^{-aR\sin t}}{R^2}\xrightarrow[R\to\infty]{}0$$ so $$0=\oint_{C_{\epsilon,R}}f(z)\,dz=\int_{-R}^{-\epsilon}\frac{1-e^{iax}}{x^2}dx+\int_{\gamma_\epsilon}f(z)\,dz+\int_\epsilon^R\frac{1-e^{iax}}{x^2}dx+\int_{\gamma_R}f(z)\,dz$$ $$\xrightarrow[\epsilon\to0,\,R\to\infty]{}\int_{-\infty}^0\frac{1-\cos ax-i\sin ax}{x^2}-\pi ia+\int_0^\infty\frac{1-e^{iax}}{x^2}dx +0\implies$$ $$\int_{-\infty}^\infty\frac{1-\cos ax-i\sin ax}{x^2}dx=\pi a$$ and comparing real parts and dividing by two (even function):: $$\int_0^\infty\frac{1-\cos ax}{x^2}=\frac{\pi a}2$$ If$\;a<0\;$something very similar (in fact, I'd say identical) is done above, so the result for general$\;a\;$should, in my opinion, be$\;\frac{\pi |a|}2\;$, but I'm not quite sure so you better check this. • Awesome answer! And I can confirm it should be$|a|$, and the whole procedure holds for$a\in\mathbb{R}\$. ^^ – Turing Feb 29 '16 at 16:24
• @1over137 Thank you very much. That confirmation surely can save some time to the asker. – DonAntonio Feb 29 '16 at 16:28