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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.

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  • $\begingroup$ 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. $\endgroup$ – MPW Feb 29 '16 at 14:35
  • $\begingroup$ result will be $\frac{a\pi}{2}$ $\endgroup$ – tired Feb 29 '16 at 14:50
  • $\begingroup$ 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 $\frac{\pi}{2}|a|.$ $\endgroup$ – gammatester Feb 29 '16 at 14:52
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    $\begingroup$ 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 $\endgroup$ – tired Feb 29 '16 at 14:57
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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.

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    $\begingroup$ Awesome answer! And I can confirm it should be $|a|$, and the whole procedure holds for $a\in\mathbb{R}$. ^^ $\endgroup$ – Turing Feb 29 '16 at 16:24
  • $\begingroup$ @1over137 Thank you very much. That confirmation surely can save some time to the asker. $\endgroup$ – DonAntonio Feb 29 '16 at 16:28

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