I want to calculate the fouriertransform of $x^2 e^{-\lambda x}$, so i need to evaluate $$ \int_{-\infty}^{\infty} x^2 e^{-x(\lambda + i\xi)} \mathrm{d}x $$ i wanted to do integration by parts two times to get rid of the $x^2$ and then integrate, but after i applied integration by parts once i get $$ \int_{-\infty}^{\infty} x^2 e^{-x(\lambda + i\xi)} \mathrm{d}x = -(\lambda + i\xi)^{-1} \left( [ x^2 e^{-x(\lambda + i\xi)} ]_{-\infty}^{\infty} \int_{-\infty}^{\infty} 2x e^{-x(\lambda + i\xi)} \mathrm{d}x \right) $$ and here the first term $[ x^2 e^{-x(\lambda + i\xi)} ]_{-\infty}^{\infty}$ goes to infinity so the integral does not converge? is this right?
Fourier transformation of $x^2 e^{-\lambda x}$
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analysis
fourier-analysis
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0What are the conditions that must be met in order for the Fourier Transform to exist, and does your function meet the requirements? Alternatively, is the function given to you as $x^2\exp(-\lambda x)\mathbf 1_{(0,\infty)}$ and you have extrapolated it to the entire real line? – 2012-06-23
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0Does the Fourier transform exist for your function? – 2012-06-23
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0In the distributional sense, using differentiation under the integral sign you'll get sth like $\delta''(\epsilon + i \lambda)$. – 2012-06-23
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1A nasty convergence problem looms large on $(-\infty, 0]$. – 2012-06-23
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0@Mercy as I mentioned it only exists in distributional sense. For normal functions the integral obviously diverges. – 2012-06-23
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0$x^2e^{-\lambda x}$ is not a distribution on ${\mathbb R}$ so you can't take its Fourier transform. The integral you wrote there is divergent. – 2012-06-23