Why is $\hat f(\xi)/\xi\in L^1(\mathbb R)$ when $f\in L^1$ is odd,$\hat f'(0)$ exists, and $\hat f(\xi)$ always has the same sign as $\xi$?

I need to prove:

If $f \in L^1(\mathbb{R})$ is an odd function such that its Fourier Transform $\hat{f}$ is differentiable at $\xi = 0$ and $\hat{f} \geq 0$ when $\xi \geq 0$, then $\hat{f}(\xi)/\xi \in L^1(\mathbb{R})$.

My work so far:

I have established (using only $f$ being odd) that we have $$\Gamma = \sup_{\alpha \geq 1} \left|\int_1^\alpha \frac{\hat{f}(\xi)}{\xi} \, d\xi\right| < \infty.$$ Now, $$\left\|\frac{\hat{f}(\xi)}{\xi}\right\| = \int_{-\infty}^{\infty} \left|\frac{\hat{f}(\xi)}{\xi}\right| \, d\xi= 2 \int_0^\infty \frac{\hat{f}(\xi)}{\xi} \, d\xi \\= 2 \left(\int_0^1 \frac{\hat{f}(\xi)}{\xi} \, d\xi + \int_1^\infty \frac{\hat{f}(\xi)}{\xi} \, d\xi\right) \leq 2\int_0^1 \frac{\hat{f}(\xi)}{\xi} \, d\xi+2\Gamma.$$ I am unsure how to proceed from here. I have not made use of assumption that $\hat{f}'(0)$ exists, but I am not sure how to use it.

As a follow-up: seems like a lot of questions on here have talked a lot about functions being odd and considering the Fourier Transform. Is there a particular reason, or is the theory for even functions simplified?

If $f$ is odd, then note that $$\int_{\mathbb R} f dx = \hat f(0)=0.$$ From the definition of differentiability, there exists $C\in\mathbb R$ such that $$ \lim_{h\to 0}\frac{\hat f(h)-\hat f(0)}h = C$$ as we have that $\hat f(0)=0$, this actually says that

$$ \lim_{h\to 0}\frac{\hat f(h) }h = C$$

And hence $\hat f(\xi)/\xi$ is a bounded function on the set $|\xi|<1$. The integrability follows.

(Its not clear to me where $\xi\ge0 \implies \hat f(\xi)\ge0$ would be used?)