Evaluating the improper integral $\int_0^\infty \frac{x\cos x-\sin x}{x^3} \cos(\frac{x}{2}) \mathrm dx $ I've been working through the following integral and am stumped:
$$\int_0^\infty \frac{x\cos x-\sin x}{x^3}\cos\left(\frac{x}{2}\right)\mathrm dx$$
Given the questions in my class that have proceeded and followed this integral, I believe that this is some form of Fourier transform/integral. However, it doesn't look like any of the content surrounding it. That is, there is no $e^{-ikx}$ or $g(k)$ or anything else that I'm familiar with. 
I know that it's an even function, but that's about as far as I can get. If I try to split it over the subtraction, I get two non-converging integrals, so that wasn't much help either. I've been throwing lots of trig identities at it but nothing familiar has appeared yet.
Any help would be greatly appreciated.
Thank you.

Integration by parts can be performed for the indefinite integral, using relations

$$\dfrac{x\cos x-\sin x}{x^2} = \left(\dfrac{\sin x}{x}\right)',\quad
 \sin^3 z =\frac{3\sin z-\sin3z}4,\quad \cos^3z=\frac{3\cos
 z+\cos3z}4.$$

One can get
\begin{align}
&\int \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx 
= \int\dfrac1{4\sin \frac x2}\,\mathrm d\left(\dfrac{\sin x}{x}\right)^2  \\[4pt]
&=\dfrac1{4\sin \frac x2}\left(\dfrac{\sin x}{x}\right)^2
+\int\left(\dfrac{\sin x}{x}\right)^2\dfrac{\cos\frac x2}{8\sin^2 \frac x2}\,\mathrm dx
=\dfrac1{x^2}\sin\frac x2\,\cos^2 \frac x2
+\dfrac12\int\dfrac{\cos^3\frac x2}{x^2}\,\mathrm dx\\[4pt]
&=\dfrac1{x^2}\sin\frac x2
-\dfrac1{4x^2}\left(3\sin\frac x2-\sin \frac {3x}2\right)
+\dfrac18\int\dfrac{3\cos\frac x2+\cos\frac{3x}2}{x^2}\,\mathrm dx\\[4pt]
&=\dfrac1{4x^2}\left(\sin\frac x2+\sin \frac {3x}2\right)
-\dfrac18\int\left(3\cos\frac x2+\cos\frac{3x}2\right)\,\mathrm d\frac1x\\[4pt]
&=\dfrac1{8x^2}\left(2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos \frac {3x}2\right)
-\dfrac3{16}\int\frac1x\left(\sin\frac x2+\sin\frac{3x}2\right)\,\mathrm dx.
\end{align}
Since
$$\lim\limits_{x\to0}\dfrac{2\sin\frac x2+2\sin\frac{3x}2-3x\cos\frac {x}2-x\cos\frac {3x}2}{8x^2}
= \lim\limits_{x\to0}\dfrac{2\frac x2+2\frac{3x}2-3x-x+O(x^3)}{8x^2}=0,$$

$$\int\limits_{0}^{\infty} \dfrac{\sin x}x\,\mathrm dx =\frac\pi2,$$

then
$$\int\limits_{0}^{\infty} \dfrac{x\cos x-\sin x}{x^3}\,\cos\dfrac x2\, \mathrm dx
=-\frac3{16}\left(\int\limits_{0}^{\infty} \dfrac{\sin\frac x2}{\frac x2}\,\mathrm d\frac x2+\int\limits_{0}^{\infty} \dfrac{\sin \frac{3x}2}{\frac{3x}2}\,\mathrm d\frac{3x}2\right) = \color{green}{\mathbf{-\frac{3\pi}{16}}}.$$
