Let's consider the function in $(eq. 1)$, i'm not solving explicitely that integral, but since it is the convolution of some other functions, it's easier to express $f(x)$ as the inverse of its fourier transform. I'm wondering if the existence of $(eq. 2)$ is sufficient to state that, or i need certain convergence conditions.. In particular, i know that the fourier inversion theorem holds for schwartz functions, or at least for square integrable function. Furthermore, i know that a prototype of schwartz functions is something as $(eq. 3)$, where $p_n(x)$ is any $n$-degrees polynomial. And this is the case for the integrand function in $(eq. 1)$!
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MY QUESTION: is $f(x)$ still a schwartz function? If yes, why? If not, does the fourier inversion theorem still hold for certain conditions? Which ones?
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EDIT 1: computing the integral in $(eq. 1)$, i got a function that is again a schwartz function.. so i guess the answer would be "yes".
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$$f(x) = \int_{-\infty}^{+\infty} y(y-x)e^{-\frac{y^2}{2}}e^{-\frac{(y-x)^2}{2}}dy = -\frac{1}{4}\sqrt{\pi}(x^2-2)e^{-\frac{x^2}{4}}$$
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EDIT 2: i should report an extract of on an interesting article i found (please, look at page 22). (http://www.mat.unimi.it/users/salvatori/ha-aa1718.pdf)
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$$f(x) = \int_{-\infty}^{+\infty} y(y-x)e^{-\frac{y^2}{2}}e^{-\frac{(y-x)^2}{2}}dy \qquad (1)$$
$$\hat{f}(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} f(x)e^{-ikx}dx \qquad (2)$$
$$f(x) = p_n(x)e^{-\alpha x^2} \longrightarrow x \in \mathbb{R}, \text{ } a \in \mathbb{R^+} \qquad (3)$$