In the process of proving that the fourier transform $L^1(\mathbb R) \rightarrow \mathcal C_0(\mathbb R)$ is not surjective, I am asked to show that given an odd integrable function $f$ on $\mathbb R$, the integral $$\int_0^{+\infty} \frac{\hat{f}(t)}{t} dt$$ is convergent, with the convention $\hat{f}(t) = \int_{-\infty}^{+\infty}f(x)\exp(-ixt)dx$.
Because $f$ is odd, we may also write $\hat{f}(t) = -2i\int_{0}^{+\infty}f(x)\sin(xt)dx$.
In order to prove the convergence, we consider a positive real number $A$ and we would like to write the following
$$\int_0^{A} \frac{\hat{f}(t)}{t} dt = -2i\int_0^{A} \int_{0}^{+\infty} \frac{f(x)\sin(xt)}{t} dxdt = -2i\int_0^{+\infty} \int_{0}^{A} \frac{f(x)\sin(xt)}{t} dtdx$$
My problem is that I can't justify the last equality properly.
I would like to use Fubini-Lebesgue theorem in order to exchange both integrals, but I can't prove that the function inside is integrable in both $t$ and $x$ on the domain $[0,+\infty]\times[0,A]$.
Could someone please explain why it is possible to exchange the integrals ? I thank you very much in advance.
As already mentioned (see previously below) you can't apply Fubini directly to $\int_0^A\hat f(t)/t\,dt$, because the hypotheses need not be satisfied.
Edit: The answer is yes or no, depending on what sort of "convergence" we want for the integral.
Yes: Sure enough, looking at $\int_\delta^A$ leads very easily to at least a partial result:
The proof is straightforward. For $0<\delta<A$ define $$K_{\delta,A}(t)=\frac1t\chi_{[-A,-\delta]\cup[[\delta,A]}(t).$$
Since $f$ and $K_{\delta,A}$ are both integrable it's immediate from Fubini that $$\int_\delta^A\frac{\hat f(t)}{t}\,dt=\frac12\int\hat f K_{\delta,A}=\frac12\int f\widehat{K_{\delta,A}}.$$It's easy to see that $$\lim_{\delta\to0,A\to\infty}\widehat{K_{\delta,A}}(x) =m(x)=\text{sgn}(x)\int_0^\infty\frac{\sin(t)}t\,dt,$$and the fact that $\int_0^A\frac{\sin(t)}t$ is bounded shows, by a little change of variable, that $$\left|\widehat{K_{\delta,A}}\right|\le c \quad(0<\delta<A, x\in\Bbb R),$$ so the existence of the limit follows from Dominated Convergence.
No: Otoh there exists an odd integrable $f$ such that the integral does not converge absolutely.
First note that if $\mu=\delta_1-\delta_{-1}$ then (ignoring irrelevant constants) $\hat\mu(t)=\sin(t)$, so $\int_0^\infty\frac{|\hat\mu(t)|}{t}=\infty$.
Now say $f_n$ is a sequence of odd integrable functions with $f_n \to\mu$ in an appropriate weak* topology. Then $||f_n||_1$ is bounded, but $\widehat{f_n}\to\hat \mu$ pointwise, so Fatou's lemma implies that $\int_0^\infty|\widehat{f_n}(t)|/t\to\infty$. The Closed Graph Theorem now shows that the integral is infinite for some odd integrable $f$.
Previously: The detail you ask about in the argument you started can't be fixed: Define $$I(A)=\int_0^A\frac{|\sin(t)|}{t}\,dt.$$ Then $$\int_0^A\frac{|\sin(xt)|}{t}\,dt=I(xA),$$hence $$\int_0^{+\infty} \int_{0}^{A} \frac{|f(x)\sin(xt)|}{t} dtdx=\int_0^\infty|f(x)|I(xA)\,dx,$$which need not be finite, since $I$ is unbounded.
So you can't apply Fubini, at leat not in such a straighforward way.. Maybe you get somewhere starting with $\int_\delta^A$ instead of $\int_0^A$, or maybe the result is simply false.