Show that; $\displaystyle\int_0^\infty\frac{f(bx)-f(ax)}{x}dx=(l-f(0))\ln\frac{b}{a}$

Question

Assume that $f\in C^1([0,\infty))$, $f'$ is monotonic on $[0,\infty)$ and $\lim\limits_{x\to\infty} f(x)=l$ is finite. If $a,b>0$, prove that; $$\displaystyle\int_0^\infty\frac{f(bx)-f(ax)}{x}dx=(l-f(0))\ln\frac{b}{a}$$

The trick is clear for me, I have to write is as double integral, i.e. the function $\frac{f(bx)-f(ax)}{x}$ as an integral, obviously with limits $a$ and $b$. For example

$\displaystyle\int_a^bf'(xy)dy$$\quad$(Here is $f$ differentiated with respect to $y$)

works.

Now if I assume that, $\displaystyle\int_0^\infty\Big(\displaystyle\int_a^bf'(xy)dy\Big)dx=\displaystyle\int_a^b\Big(\displaystyle\int_0^\infty f'(xy)dx\Big)dy$

and $\displaystyle\int_0^\infty f'(xy)dx=\frac{l-f(0)}{y}$ $(\bigstar)$

then $\displaystyle\int_a^b\frac{l-f(0)}{y}=(l-f(0))\ln\frac{b}{a}$ gives the correct result.

MY QUESTIONS

1)Why is the line with $(\bigstar)$ correct ? Because $f'(xy)$ is differentiated w.r. to $y$ but if we integrate w.r.to $x$ why do we get $f$ again ?

2)Why can we change the order of integration, We have to make use of the monotonicity of $f'$, I guess $\lim\limits_{x\to\infty}f'(x)=0$, but this is not sufficient.

Answer 1

Ad 1): $f'(xy)$ is the derivative of $f$, evaluated at the point $xy$. Since $f$ is a function of a single variable, "differentiated w.r.t" some variable is not a distinction to be made, you can differentiate $f$ only with respect to its single real argument. You can differentiate the composition $F(x,y) = f(xy)$ with respect to $x$ or $y$, and the results are $\frac{\partial F}{\partial x}(x,y) = f'(xy)\cdot y,\; \frac{\partial F}{\partial y}(x,y) =f'(xy)\cdot x$. From which you can see that

$$f'(xy)= \frac{1}{y}\frac{\partial F}{\partial x}(x,y)= \frac{1}{x}\frac{\partial F}{\partial y}(x,y),$$

which gives you the respective results when integrating with respect to $x$ or $y$.

Ad 2): The monotonicity of $f'$ together with $\lim\limits_{x\to\infty} f'(x) = 0$ (which follows from $\lim\limits_{x\to\infty} f(x) = l$) imply that $f'$ has constant sign, either it is always non-negative, or always non-positive. Tonelli's theorem then assures that the interchange of order of integration is allowed.