$\int_0^1 \frac{{f}(x)}{x^p} $ exists and finite $\implies f(0) = 0 $ Need some help with this question please.
Let $f$ be a continuous function and 
let the improper inegral 
$$\int_0^1 \frac{{f}(x)}{x^p} $$ 
exist and be finite for any $ p \geq 1 $.
I need to prove that 
$$f(0) = 0 $$
In this question, I really wanted to use somehow integration by parts and/or the Fundamental theorem of calculus.
Or even maybe use Lagrange Means value theorem,
but couldn't find a way to se it up.
I'll really appreciate your help on proving this.
 A: Assume $f(0) > 0$. Since $f$ is continuous, we can find $\epsilon > 0$ so that $\forall x \in [0, \epsilon] : f(x) > 0$. Since $[0, \epsilon]$ is compact, we can find $m > 0$ so that $\forall x \in [0, \epsilon] : f(x) > m$. Thus:
$$
\int_0^\epsilon \frac{f(x)}{x^p}dx \ge m \int_0^\epsilon \frac{dx}{x^p}
$$
But the integral on the RHS diverges for all $p \ge 1$.
What's left is to show that this implies that the integral on $[0, 1]$ diverges too, and handle the case $f(0) < 0$. Both should be easy.
A: $\int_{0}^{1} \frac{f(x)}{x^p}dx=\int_{0}^{\delta}\frac{f(x)}{x^p}dx+\int_{\delta}^{1}\frac{f(x)}{x^p}dx$.
We know that $\int_{\delta}^{1}\frac{f(x)}{x^p}dx$ is bounded for given $\delta>0$.
For near $x=0$, $f(x)$ is almost constant, called $c$, because $f$ is continuous.
For small $\delta>0$,
$\int_{0}^{\delta} \frac{f(x)}{x^p}dx$ is almost $\int_{0}^{\delta} \frac{c}{x^p}dx=c\int_{0}^{\delta} \frac{1}{x^p}dx$
Thus $c$ should be $0$.
A: Hint: Let
$$
g(t) = \int_t^1 \frac{f(x)}{x^p}dx
$$
Now investigate properties of $g(t)$ around $t=0$.
