Final step (Frullani's formula) The integral is:
$$I =\int_{0}^{\infty} \frac{\sin(\alpha x)\cos(\beta x)\cos(\gamma x)}{x}dx $$
My solution is:  
$$ I=\frac{1}{2}\int_{0}^{\infty}\frac{\sin((\alpha-\beta)x)\cos(\gamma x)}{x}dx + \frac{1}{2}\int_{0}^{\infty}\frac{\sin((\alpha+\beta)x)\cos(\gamma x)}{x}dx$$
By application of Frullani's formula, we have
$$ \int_{0}^{\infty}\frac{\sin((\alpha-\beta)x)\cos(\gamma x)}{x}dx = \frac{1}{2}\int_{0}^{\infty} \frac{\sin((\alpha - \beta -\gamma)x)-\sin((\beta - \gamma -\alpha)x)}{x}dx \\\qquad\quad= f(0)\ln\left(\frac{\beta - \gamma -\alpha}{\alpha - \beta -\gamma}\right) = 0$$
The same for: $$\int_{0}^{\infty}\frac{\sin((\alpha+\beta)x)\cos(\gamma x)}{x}dx$$ 
I'm not sure if $0$ is the right answer to this integral. Any advice would be much appreciated! 
 A: I'm pretty sure I saw this question the other day, but anyways. For a lot of parameters $\alpha$, $\beta$ and $\gamma$ the result is indeed $0$. But, for example, for $\alpha=\beta=\gamma=1$, you get
$$
\int_0^{+\infty}\frac{\sin x\cos^2x}{x}\,dx=\frac{\pi}{4}.
$$
So, a tip: Write
$$
\sin\alpha x\cos\beta x\cos\gamma x=\frac{1}{4}\bigl(\sin((\alpha+\beta+\gamma)x)+\sin((\alpha+\beta-\gamma)x)+\sin((\alpha-\beta+\gamma)x)+\sin((\alpha-\beta-\gamma)x)\bigr),
$$
and then use the fact (?) that
$$
\int_0^{+\infty}\frac{\sin ax}{x}\,dx=\frac{\pi}{2}\,\text{sign}\,a.
$$
A: Assuming $\alpha,\beta\in\mathbb{R}$, we have:
$$ I(\alpha,\beta)= \int_{0}^{+\infty}\frac{\sin(\alpha x)\cos(\beta x)}{x}\,dx =\frac{1}{2}\left[\int_{0}^{+\infty}\frac{\sin((\alpha+\beta)x)}{x}\,dx+\int_{0}^{+\infty}\frac{\sin((\alpha-\beta)x)}{x}\,dx\right]=\frac{\pi}{4}\left[\text{sign}(\alpha-\beta)+\text{sign}(\alpha+\beta)\right]$$
hence the given integral equals
$$\frac{\pi}{8}\left[\text{sign}(\alpha-\beta-\gamma)+\text{sign}(\alpha+\beta+\gamma)+\text{sign}(\alpha-\beta+\gamma)+\text{sign}(\alpha+\beta-\gamma)\right].$$
