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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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
1 Answers
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Note that $x^2e^{-\lambda x}=\frac{d^2}{d\lambda^2}e^{-\lambda x}$ and so it must exist the integral $ \int_{-\infty}^\infty e^{-\lambda x}e^{i\xi x}dx. $ But, depending on the sign of $\lambda$, this will diverge from the positive or the negative half-line.