I've been reading a post on Quora about lesser known techniques of integration and I'm just curious if there's also a novel way to integrate this type of integral without resorting to complex analysis.
$$ \int^\infty_0 \frac {\cos (ax)\,dx}{x^2}, a \geq 0 $$
Answer
The given integral is not converging, so I assume you wanted to study:
$$f(a)=\int_{0}^{+\infty}\frac{1-\cos(ax)}{x^2}\,dx$$
that is an even function, hence we can assume $a\geq 0$ WLOG. Integration by parts then gives:
$$ f(a) = a\int_{0}^{+\infty}\frac{\sin(ax)}{x}\,dx $$
and by replacing $x$ with $\frac{z}{a}$ we get:
$$ f(a)=a\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = \frac{\pi}{2}\,a $$
leading to:
$$\forall r\in\mathbb{R},\qquad \int_{0}^{+\infty}\frac{1-\cos(rx)}{x}\,dx = \frac{\pi}{2}|r|.$$
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