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what exactly means if an integral exists just in the distributional sense, for example the fourier-transform of $x^2 e^{-\lambda x}$ or of $H(R-|x|)$ where $R > 0$ and $H$ is the Heaviside-Function?

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I know next to nothing about distribution theory, but I would guess that it means that the Fourier transform might not exist as a function, but it does as a distribution. –  tomasz Jun 24 '12 at 16:42

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While the derivative in the distributional sense always exist, the same is not true for the Fourier transform or arbitrary (convolution) integrals. The Fourier transform in the distributional sense can be defined for tempered distributions. The space of tempered distributions $\mathcal{S}'$ is the dual space of the Schwartz space $\mathcal{S}$ (which is the function space of functions all of whose derivatives are rapidly decreasing). The Fourier transform is an automorphism on the Schwartz space, which enables to define the Fourier transform for its dual space by transposition, i.e. the Fourier transform of the tempered distribution $T\in \mathcal{S}'$ is defined via $$<\mathcal{F}(T),\phi>:=<T,\mathcal{F}(\phi)> \forall \phi\in\mathcal{S}$$

Regarding your question, $H(R-|x|)$ is a tempered distribution (or can be identified with a ...), but $x^2e^{-\lambda x}$ is not a tempered distribution. Perhaps you meant $x^2e^{-i\lambda x}$, which would be a tempered distribution. I propose this because your question might be motivated by the inverse Laplace transformation, where the $i$ is omitted.

Things get more tricky when it comes to multiplication (or convolution) of tempered distributions. These operations are not always defined. Even so it's easy to decide for two given tempered distributions whether their product (or convolution) is defined, writing down a set of sufficient conditions general enough to cover all cases which can occur in practice is nearly impossible.

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