(Only hints, please)
Let $$f(x)=\begin{cases} x^3+1 & x >0\\ 0 & x = 0\\ x^2-1 & x <0 \end{cases}$$
I am trying to show this function satisfies Dini’s condition at $x=0$. In our case here, that is there exists $\delta >0$ such that $$\int_{-\delta}^{\delta} \frac{f(x+t)-f(x)}{t}dt$$ is finite.
Naturally, if $f$ is differentiable at $x$ then Dini’s condition holds. Further, if even the right and left derivatives exist, we may conclude.
However, while certainly on $x>0$, $f’(x)=3x^2 \to 0$ as $x\to 0^+$ coincides with the behavior on $x<0$, with $f’(x)=2x\to 0$ as $x\to 0^-$, we also can compute by definition $$\lim_{h\to 0^+} \frac{f(h)-f(0)}{h}$$ $$=\lim_{h\to 0^+} \frac{h^3+1}{h}=\infty,$$ no? So, the right hand derivative does not exist at $x=0$? Now, if I try to compute the integral directly to show Dini’s condition holds for some $\delta$, I get some blow ups about any neighborhood of zero I take.
So my question is, since I assume this problem was correctly written, what am I doing wrong and what are some hints to proceed?
Edit: Here is my direct calculation. With $x=0$ and $f(0)=0$, the relevant integral expression becomes $$\int_{-\delta}^\delta \frac{f(t)}{t}dt.$$ We split this up and use the piecewise definition of $f$, to get $$\int_{-\delta}^0 \frac{t^2-1}{t} dt+\int_0^\delta \frac{t^3+1}{t}dt$$ $$=\int_{-\delta}^0 t-\frac{1}{t} dt+\int_0^\delta t^2+\frac{1}{t}dt.$$ But both of these integrals diverge, since $\int \frac{1}{t}=\log |t|+C$ blows up as we approach zero, no? Or am I being too quick to judge?
Edit 2 My professor said to consider principal value integration and use the fact that $1/t$ is an odd function hence the integral over $(-\delta, \delta)$ vanishes in the principal value sense. This I understand and can prove. However, I do not see how to clinch the problem with this hint, unless the piecewise function was changed to $f(x)=x^2+1$ on $x<0$. For otherwise, when I split up the integral and isolate the integrand $1/t$, I cannot recombine them into an integral over the whole interval because of a sign difference.