Fourier transform of $t^2$ discrepancy I encountered a discrepancy when taking the fourier transform of $t^2$ that I don't understand. I would expect $\mathcal{F}[t^2]$,
$$\mathcal{F}[t^2]=\int_{-\infty}^\infty t^2 e^{-i \omega t}dt$$
 to diverge because $t^2$ isn't absolutely integrable. Typing this integral into Wolfram Alpha confirms that it does diverge.

However, typing "Fourier Transform of $t^2$" into Wolfrom Alpha produces the following result:
$$\mathcal{F}[t^2]=-\sqrt{2\pi}\delta''(\omega)$$

What is the reason for this discrepancy and what does this answer represent physically? I was taught that if a function isn't absolutely integrable, it was not possible to take its Fourier transform, but this seems to suggest otherwise.

EDIT:
I think I wasn't clear on my exact question. I'm more interested in the implications of this integral and why you can take the Fourier transform of $t^2$ even though it isn't absolutely integrable. However, that brings up another question - why does Wolfram Alpha tell me that the integral diverges when it is actually equal to the second derivative of the delta function?
We can build the theory of the Fourier transform up in several steps. The integral in the definition converges by an immediate estimate if the function is in $L^1 (\mathbb{R})$. Since $L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$ (the set of functions which are both absolutely integrable and square-integrable) is dense in $L^2 (\mathbb{R})$ (i.e. every function in $L^2$ can be approximated of functions in the former space which converges in the $L^2$ norm), and the Fourier transform is a continuous linear operator from $L^1 (\mathbb{R})$ to $L^2 (\mathbb{R})$, we can define the Fourier transform for f in $L^2 (\mathbb{R})$ by approximating $f$ with a sequence of functions $f_n \in L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$, and defining the Fourier transform of $f$ to be the $L^2$-limit of the Fourier transforms of $f_n$. This idea of taking a function defined on a dense subset and extending it by continuity to a larger set is very useful. 

Of course, your example isn't in $L^2 (\mathbb{R})$ either. It turns out we can extend the Fourier transform to an even broader class of functions (actually, to a class that includes things that aren't functions in the traditional sense at all!). This broad class is the set of tempered distributions. We do this by first noting that the Fourier integral is well-defined on the Schwartz class, which is roughly speaking the set of functions which decay to zero at infinity faster than any polynomial (hence these functions are all in $L^1 (\mathbb{R})$. We give the Schwartz class a particular (locally convex) topology, which (in particular) determines which sequences in this space are convergent and which functions defined on the Schwartz class are continuous. 

The set of tempered distributions is the dual space to the Schwartz class, which means it is the set of all continuous linear functions which map from the Schwartz class to the real (or complex numbers). An example is the Dirac delta, which satisfies $\delta(f) = f(0)$ for $f$ in the Schwartz class. Distributions are objects which act on functions. Operations on distributions may therefore be characterized by how they affect the way the distribution acts on functions. It turns out that for any tempered distribution $\phi$, we can define the Fourier transform $F \phi$ as the tempered distribution defined by 
\begin{equation*}
F \phi (g) = \phi (Fg) \quad \text{ for all } g \text{ in the Schwartz class}.
\end{equation*}
So the Fourier transform of a tempered distribution is the distribution which first applies the Fourier transform to the test function, and then applies the original distribution to the result. 

Roughly speaking, all functions with polynomial growth at infinity have corresponding tempered distributions, and hence we can define their Fourier transform. Given such a function $f$, we define the tempered distribution corresponding to $f$ by 
\begin{equation*}
\Lambda_f (g) = \int_{-\infty}^\infty f(x) g(x) \, dx \text{ for any } g \text{ in the Schwartz class}.
\end{equation*}
In particular, the function in your example has a corresponding tempered distribution, and so we can define the Fourier transform of that distribution. Notice that the Fourier transform involves the Dirac delta, and hence is not a traditional function, but instead is defined in terms of how it acts on other functions, as we might expect from our definition of the Fourier transform on the Schwartz class. 

edit: Responding to your question in another comment, it is correct that we cannot take the Fourier transform of $e^{-t}$, since $e^{-t}$ is not a tempered distribution (it "grows too fast"). There are several ways to prove this. Suppose that $f(t) = e^{-t}$ was a tempered distribution, and take for granted that the rules of derivatives and Fourier transforms work for tempered distributions more or less like you expect they should (they do). Then $e^{-t}$ would (as a distribution) solve the differential equation 
\begin{equation*}
f' + f = 0. 
\end{equation*}
Taking the Fourier transform, we obtain 
\begin{equation*}
(2 \pi i \xi + + 1) \hat{f} (\xi) = 0 \text{ for every } \xi \in \mathbb{R}
\end{equation*}
where $\hat{f}$ is the Fourier transform of $f$. The only way for this equation to be satisfied is if $\hat{f}$ is identically zero. But (it turns out that) the Fourier transform on the set of tempered distributions is an isomorphism, and in particular is invertible, so $\hat{f} = 0$ implies $f = 0$, since the Fourier transform of $0$ is clearly zero. This is a contradiction, so $f$ cannot be a tempered distribution, so we can't define its Fourier transform.
You simply use a trick to bypass the question. 

$$\int_{-\infty}^{+\infty}t^2 e^{-i\omega t}\ dt = -\int_{-\infty}^{+\infty} \frac{d^2}{d\omega^2} e^{-i\omega t} = -\frac{d^2}{d\omega^2} \int_{-\infty}^{+\infty} e^{-i\omega t} =  -\frac{d^2}{d\omega^2} \sqrt{2\pi}\delta(\omega) = -\sqrt{2\pi}\delta''(\omega)$$