Final step (Frullani's formula)

Question

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!

Answer 1

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 $$\frac{\pi}{8}\left[\text{sign}(\alpha-\beta-\gamma)+\text{sign}(\alpha+\beta+\gamma)+\text{sign}(\alpha-\beta+\gamma)+\text{sign}(\alpha+\beta-\gamma)\right].$$

Answer 2

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 $$ \int_0^{+\infty}\frac{\sin ax}{x}\,dx=\frac{\pi}{2}\,\text{sign}\,a. $$