Proof involving definite integral equality involving continuity criterion 
Let $f:[0,1]\to\Bbb R$ be continuous and $\{a,b\}\subset\Bbb{R}_+$ such that $0<a<b\leq1$. Prove that $$\int_0^1 {f\!\left(ax^n\right) - f\!\left(bx^n\right) \over x}\,dx = \frac1n\left( f(0) \ln \frac ba - \int_a^b{f(x)\over x}\,dx\right)$$

I have been trying this below problem for an hour. I’ve thought to take $f(0)=0$, but the problem asks for a proof. I have also used the substitutions $ax^n=z$ and $bx^n=k$.
Am I doing something wrong?
 A: First, enforcing the substitution $x\mapsto x^{1/n}$ reveals
$$\begin{align}
\int_0^1 \frac{f(ax^n)-f(bx^n)}{x}\,dx=\frac1n \int_0^1 \frac{f(ax)-f(bx)}{x}\,dx
\end{align}$$

Second, it is straightforward to show that  
$$\begin{align}
\int_\epsilon^1 \frac{f(ax)}{x}\,dx-\int_\epsilon^1 \frac{f(bx)}{x}\,dx&=\int_{a\epsilon}^a \frac{f(x)}{x}\,dx-\int_{b\epsilon}^b \frac{f(x)}{x}\,dx\\\\
&=\int_{a\epsilon}^{b\epsilon}\frac{f(x)}{x}\,dx-\int_a^b \frac{f(x)}{x}\,dx\\\\
&=f(0)\log(b/a) +\int_{a\epsilon}^{b\epsilon}\frac{f(x)-f(0)}{x}\,dx-\int_a^b \frac{f(x)}{x}\,dx\\\\
\end{align}$$

Third, we have the estimate
$$\left|\int_{a\epsilon}^{b\epsilon}\frac{f(x)-f(0)}{x}\,dx\right|\le \int_{a\epsilon}^{b\epsilon} \frac{|f(x)-f(0)|}{x}\,dx\le \sup_{x\in[a\epsilon,b\epsilon]}|f(x)-f(0)| \log(b/a)$$
which approaches $0$ as $\epsilon \to 0$ since $f$ is continuous.

Finally, putting it together reveals 
$$\int_0^1 \frac{f(ax^n)-f(bx^n)}{x}\,dx=\frac1n\left(f(0)\log(b/a)-\int_a^b \frac{f(x)}{x}\,dx \right)$$
as was to be shown!
