Cosine integral question In the books i am asked to show that $\int \limits_{0}^{\infty} \frac{\cos(x) - \cos(y x)}{x} dx = \ln (y) $ for all $y>0$ ?.
I did solve the question $\int \limits_{0}^{\infty} \frac{\cos(z x) - \cos(y x)}{x^2} dx = \frac{\pi}{2} (y-z)$, i tried to find a connection between them but with no success  
 A: Here's a proof from scratch using some nice tricks to have in one's arsenal. First note that integration by parts shows that
$$
I(y) = \int_{0}^{\infty} \frac{\cos(x) - \cos(xy)}{x} dx = \int_{0}^{\infty}  \bigg( \sin (x)-\frac{\sin (x y)}{y} \bigg) \frac{dx}{x^2}.
$$
(The fact that the lower boundary term in the integration by parts vanishes is due to $\lim_{x\to0} \frac{\sin x}x = 1$.) The new integral converges absolutely, which allows for more flexibility.
Now let's differentiate under the integral sign:
\begin{align*}
I'(y) &= \int_{0}^{\infty} \frac\partial{\partial y} \bigg( \sin (x)-\frac{\sin (x y)}{y} \bigg) \frac{dx}{x^2} \\
&= \int_{0}^{\infty} \frac{\sin (x y)-x y \cos (x y)}{y^2} \frac{dx}{x^2} \\
&= -\frac{\sin (x y)}{x y^2} \bigg|_0^\infty = \frac1y
\end{align*}
(again using $\lim_{x\to0} \frac{\sin x}x = 1$). Together with the obvious value $I(1)=0$, this establishes that $I(y) = \ln(y)$.
A: I'm not sure if the two integrals are related, but the first one can be solved directly. Note that by change of variables
$$\begin{align*}
I(y)&=\lim_{\epsilon\to 0} \int_\epsilon^\infty \frac{\cos (x)-\cos(yx)}x dx
\\&=\lim_{\epsilon\to 0} \int_\epsilon^\infty \frac{\cos (x)}xdx-\int_\epsilon^\infty \frac{\cos (yx)}xdx
\\&=\lim_{\epsilon\to 0} \int_\epsilon^\infty \frac{\cos (x)}xdx-\int_{y\epsilon}^\infty \frac{\cos (t)}tdt\tag{$t=yx$}
\\&=\lim_{\epsilon\to 0} \int_\epsilon^{y\epsilon} \frac{\cos (x)}xdx
\\&=\lim_{\epsilon\to 0} \int_1^y \frac{\cos (\epsilon u)}udu.\tag{$\epsilon u=x$}
\end{align*}$$ 
We can interchange the limit and the integral (either by LDCT or joint continuity of the integrand $(\epsilon,u)\mapsto \frac{\cos(\epsilon u)}{u}$) to obtain
$$
I(y)=\int_1^y \frac {du}u=\ln(y).
$$
A: This is one of the formulas of Froullani
Theorem If $f：[0,+\infty)\to [0,+\infty)$ is continuous at $x\in [0,+\infty)$,and $\int_1^{+\infty} \frac{f(x)}{x} \mathrm{d}x$ is convergent.Then
$$\int_0^{+\infty}\frac{f(ax)-f(bx)}{x}\mathrm{d}x＝f(0)\log{\frac{b}{a}}$$
for $b>a>0$.
