Prove that $\frac{t}{t^2-1}$ is a tempered distribution I want to compute the Fourier transform of $\frac{t}{t^2-1}$, and in order to do so I need to prove in which space is the function. Clearly the function is not $L^1(\mathbb{R})$ neither $L^2(\mathbb{R})$ because the function isn't continuos in $\pm 1$. How do I prove that the function is a tempered distribution ? My exercise book states that it is without proving it, and according to the statement that T is a tempered distribution if holds the inequality:
$$ |f(t)| \leqslant K(1 + |t|^p) \qquad K >0, \ p \in \mathbb{N}_0$$
Or if T is a derivative of some order of the function that holds the inequality, for me the function is not a tempered distribution, so where is my error ? Thanks
 A: The ''function'' as it is has no right to be called a distribution, since it has non-summable singularities in the form $1/y$ at $x=\pm1$: therefore I'm going to assume that your teacher understood it as Cauchy principal value:
$$
u(x) = PV \frac{x}{x^2-1}.
$$
According to your book, a function $f\in L^1_{\text{loc}}(\mathbb R)$ is a tempered distribution in $\mathscr S'(\mathbb R)$ if it is of at most polynomial growth which means
$$
\left|f(x)\right|\le K(1+\left|x \right|^p) 
$$
for some positive constant $K$ and a positive power $p$. 
Now, consider 
$$
f(x) = \frac{1}{2} \log\left|x^2-1\right|.
$$
This is a function in $L^1_{\text{loc}}(\mathbb R)$, since the logarithm has integrable singularities at $x=\pm1$. Using the (distributional) identity
$$
\frac{d}{dx}\log|y| = PV\left( \frac{1}{y}\right),
$$
one gets precisely
$$
f'(x) = PV\frac{x}{x^2-1} = u(x).
$$
Consider the decomposition
$$
f(x) = \chi_{[-2,2]}(x)f(x) + \left( 1-\chi_{[-2,2]}(x)\right) f(x)\equiv g(x) + h(x),
$$
where $\chi$ is the characteristic function. Clearly, $h(x)$ is a function of at most polynomial growth, e.g.
$$
\left| h(x) \right| \le 1+x^2.
$$
Also, $g$ is in $\mathscr S'(\mathbb R)$ since it is in $L^1(\mathbb R)$.
Since the tempered distributions form a vector space and since derivatives of distributions in $\mathscr S'(\mathbb R)$ are still tempered,
$$
u(x)\in\mathscr S'(\mathbb R).
$$
