On wiki, http://en.wikipedia.org/wiki/List_of_definite_integrals , appears some integral which is not trivial for me: $$\int_{0}^{\infty}\frac{f(ax)-f(bx)}{x}dx=\left(f(0)-f(\infty)\right)\log\left(\frac{b}{a}\right)$$
Can someone please tell me what are the steps to prove it?
Let $\phi(xy)$ be given. Note that $\frac{\partial \phi}{\partial x}=y\phi'(xy)$ and $\frac{\partial \phi}{\partial y}=x\phi'(xy)$. Then note
$$\int_a^b\int_c^d\phi'(xy)dxdy=\int_a^b\int_c^d \frac1y \frac{\partial \phi} {\partial x}dxdy=\int_a^b\frac1y\left(\phi(dy)-\phi(cy)\right)dy\tag 1$$
$$\int_a^b\int_c^d\phi'(xy)dxdy=\int_a^b\int_c^d \frac1x \frac{\partial \phi}{\partial y}dxdy=\int_c^d\frac1x\left(\phi(bx)-\phi(ax)\right)dx\tag 2$$
Now, assume that $\lim_{x\to \infty}\phi(x)=\phi(\infty)$. Letting $c=0$ and $d \to \infty$ yields from $(1)$
$$\int_a^b\frac1y\left(\phi(by)-\phi(ay)\right)dy\to (\phi(\infty)-\phi(0))\log(b/a)\tag 3$$
and from $(2)$
$$\int_c^{d}\frac1x\left(\phi(bx)-\phi(ax)\right)dx\to \int_0^{\infty}\frac1x\left(\phi(bx)-\phi(ax)\right)dx\tag 4$$
Equating $(3)$ and $(4)$ gives the coveted result
$$\int_0^{\infty}\frac1x\left(\phi(bx)-\phi(ax)\right)dx=(\phi(\infty)-\phi(0))\log(b/a)$$
