I learned on a book that, if $f(x)$ has Fourier transform $F[f(x)] = \hat{f}(\lambda)$, and the Fourier transform of $\int_{-\infty}^x f(\xi)\ d\xi$ exists, $\hat{f}(0)=0$, then $$ F[\int_{-\infty}^x f(\xi)\ d\xi]=\frac{1}{i\lambda}\hat{f}(\lambda) $$
I tried this: \begin{align*} F[\int_{-\infty}^x f(\xi)\ d\xi] &= \int_{-\infty}^{+\infty} \int_{-\infty}^x f(\xi)\ d\xi\ e^{-i\lambda x}\ dx\\ &= \int_{-\infty}^{+\infty}\ d\xi \int_{\xi}^{+\infty} f(\xi)e^{-i\lambda x}\ dx\\ &= \int_{-\infty}^{+\infty}\ d\xi \int_{0}^{+\infty} f(\xi)e^{-i\lambda y}e^{-i\lambda \xi}\ dy\\ &= \hat{f}(\lambda) \int_{0}^{+\infty} e^{-i\lambda y}\ dy\\ &= \hat{f}(\lambda) \int_{0}^{+\infty}[ \cos(\lambda y)-i\sin(\lambda y)]\ dy \end{align*} Why $\int_{0}^{+\infty} e^{-i\lambda y}\ dy = \frac{1}{i\lambda}$? It seems to me $\lim_{y \to +\infty} e^{-i\lambda y}$ does not exist. And neither $\cos(\lambda y)$ nor $\sin(\lambda y)$ is integrable on $[0,+\infty)$. Thank you for any help!
Your idea is good, but you forget to check the hypothesis under which you can switch integration order, i.e. you applied Fubini theorem when the hypothesis don't hold. To be specific, the problem arises when you try to switch integrals to obtain: $$\int_{-\infty}^{+\infty}\int_{-\infty}^xf(\xi)e^{-i\lambda x}d\xi dx = \int_{-\infty}^{+\infty}\int_{\xi}^{+\infty}f(\xi)e^{-i\lambda x}dx d\xi.$$ In fact, if $\|f\|_1\neq0$, this double integral isn't absolutely convergent, because: $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}|\chi_{(-\infty,x]}(\xi)f(\xi)|d\xi dx=\int_{-\infty}^{+\infty} \|f\|_{L^1((-\infty,x])} dx = +\infty,$$ and so we are not under the hypothesis of Fubini theorem.
However, we can adjust your argument to get the correct result introducing a convergence factor (i.e. to use a summation method). A simple one is:
$$x\mapsto e^{-\varepsilon |x|}.$$
First, let's state explicitly the hypothesis: assume that $f\in L^1(\mathbb{R})$ and $x\mapsto \int_{-\infty}^x f(\xi)d\xi \in L^1(\mathbb{R})$.
Here the idea: by Lebesgue dominated convergence theorem: $$\int_{-\infty}^{+\infty}\left(\int_{-\infty}^xf(\xi)d\xi\right) e^{-i\lambda x} dx = \lim_{\varepsilon\rightarrow0^+}\int_{-\infty}^{+\infty}\left(\int_{-\infty}^xf(\xi)d\xi \right) e^{-i\lambda x} e^{-\varepsilon|x|} dx$$ and now, for $\varepsilon>0$: $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}|\chi_{(-\infty,x]}(\xi)f(\xi)e^{-\varepsilon x}|d\xi dx<+\infty,$$ so we can use Fubini theorem to switch integration order.
Here the details for the case $\lambda\neq 0$: $$\mathcal{F}\left(x\mapsto\int_{-\infty}^xf(\xi)d\xi\right)(\lambda) = \int_{-\infty}^{+\infty}\int_{-\infty}^xf(\xi)d\xi e^{-i\lambda x} dx = \lim_{\varepsilon\rightarrow 0^+} \int_{-\infty}^{+\infty}\int_{-\infty}^xf(\xi)d\xi e^{-i\lambda x} e^{-\varepsilon |x|}dx \\ = \lim_{\varepsilon\rightarrow 0^+} \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \chi_{(-\infty,x]}(\xi) f(\xi) e^{-i\lambda x} e^{-\varepsilon |x|}d\xi dx \\ = \lim_{\varepsilon\rightarrow 0^+} \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \chi_{(-\infty,x]}(\xi) f(\xi) e^{-i\lambda x} e^{-\varepsilon |x|}dx d\xi \\ = \lim_{\varepsilon\rightarrow 0^+} \int_{-\infty}^{+\infty} f(\xi) \int_{-\infty}^{+\infty} \chi_{[\xi, +\infty)}(x) e^{-i\lambda x} e^{-\varepsilon |x|}dx d\xi = (*). $$ Now, having to deal with an absolute value, let's compute the integral in $dx$ first in $[0,+\infty)$ and then in $(-\infty,0)$. We have: $$\int_{-\infty}^{+\infty} f(\xi) \int_{0}^{+\infty} \chi_{[\xi, +\infty)}(x) e^{-i\lambda x} e^{-\varepsilon |x|}dx d\xi \\ = \int_{-\infty}^{+\infty} f(\xi) \int_{0}^{+\infty} \chi_{[\xi, +\infty)}(x) e^{-(\varepsilon+i\lambda) x} dx d\xi \\ = \int_{-\infty}^{0} f(\xi) \int_{0}^{+\infty} \chi_{[\xi, +\infty)}(x) e^{-(\varepsilon+i\lambda) x} dx d\xi + \int_{0}^{+\infty} f(\xi) \int_{0}^{+\infty} \chi_{[\xi, +\infty)}(x) e^{-(\varepsilon+i\lambda) x} dx d\xi \\ = \int_{-\infty}^{0} f(\xi) \int_{0}^{+\infty} e^{-(\varepsilon+i\lambda) x} dx d\xi + \int_{0}^{+\infty} f(\xi) \int_{\xi}^{+\infty}e^{-(\varepsilon+i\lambda) x} dx d\xi \\ = \frac{1}{\varepsilon+i\lambda} \int_{-\infty}^{0} f(\xi) d\xi + \frac{1}{\varepsilon+i\lambda} \int_{0}^{+\infty} f(\xi)e^{-i\lambda \xi} e^{-\varepsilon\xi} d\xi.$$ On the other hand: $$\int_{-\infty}^{+\infty} f(\xi) \int_{-\infty}^{0} \chi_{[\xi, +\infty)}(x) e^{-i\lambda x} e^{-\varepsilon |x|}dx d\xi \\ = \int_{-\infty}^{+\infty} f(\xi) \int_{-\infty}^{0} \chi_{[\xi, +\infty)}(x) e^{(\varepsilon-i\lambda) x} dx d\xi \\ = \int_{-\infty}^{0} f(\xi) \int_{\xi}^{0} e^{(\varepsilon-i\lambda) x} dx d\xi = \int_{-\infty}^{0} f(\xi) \frac{1-e^{(\varepsilon-i\lambda)\xi}}{\varepsilon-i\lambda} d\xi \\ = \frac{1}{\varepsilon-i\lambda} \int_{-\infty}^{0} f(\xi) d\xi - \frac{1}{\varepsilon-i\lambda} \int_{-\infty}^{0} f(\xi) e^{-i\lambda\xi} e^{\varepsilon\xi} d\xi.$$ So, again using Lebesgue dominated convergence theorem, we get: $$(*) = \lim_{\varepsilon\rightarrow 0^+} \left(\frac{1}{\varepsilon+i\lambda} \int_{-\infty}^{0} f(\xi) d\xi + \frac{1}{\varepsilon+i\lambda} \int_{0}^{+\infty} f(\xi)e^{-i\lambda \xi} e^{-\varepsilon\xi} d\xi + \frac{1}{\varepsilon-i\lambda} \int_{-\infty}^{0} f(\xi) d\xi - \frac{1}{\varepsilon-i\lambda} \int_{-\infty}^{0} f(\xi) e^{-i\lambda\xi} e^{\varepsilon\xi} d\xi\right) \\ = \frac{1}{i\lambda} \int_{-\infty}^{0} f(\xi) d\xi + \frac{1}{i\lambda} \int_{0}^{+\infty} f(\xi)e^{-i\lambda \xi} d\xi + \frac{1}{-i\lambda} \int_{-\infty}^{0} f(\xi) d\xi - \frac{1}{-i\lambda} \int_{-\infty}^{0} f(\xi) e^{-i\lambda\xi} d\xi \\ = \frac{1}{i\lambda} \int_{-\infty}^{+\infty} f(\xi) e^{-i\lambda\xi} d\xi = \frac{1}{i\lambda} \mathcal{F}(f)(\lambda). $$
A final remark: this isn't the only way to prove the result, in fact, you can start backward from the derivative and use integration by parts to get there. However, this way to prove the result is a nice illustration of the power of summation methods.