# Fourier transformation of $x^2 e^{-\lambda x}$

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?

-
What 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? – Dilip Sarwate Jun 23 '12 at 13:27
Does the Fourier transform exist for your function? – Mercy King Jun 23 '12 at 13:29
In the distributional sense, using differentiation under the integral sign you'll get sth like $\delta''(\epsilon + i \lambda)$. – qoqosz Jun 23 '12 at 13:47
A nasty convergence problem looms large on $(-\infty, 0]$. – ncmathsadist Jun 23 '12 at 14:04
@Mercy as I mentioned it only exists in distributional sense. For normal functions the integral obviously diverges. – qoqosz Jun 23 '12 at 14:54

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.

-

$$\int_{-\infty}^{\infty}e^{- 2 \pi i\xi x} \mathrm{d}x=\delta(\xi)$$

Where $δ(ξ)$ is the Dirac delta function

$$-4\pi^2 \int_{-\infty}^{\infty} x^2e^{- 2 \pi i\xi x} \mathrm{d}x=\delta''(\xi)$$

$$\int_{-\infty}^{\infty} x^2e^{- 2 \pi i\xi x} \mathrm{d}x=-\frac{\delta''(\xi)}{4\pi^2}$$

$$\int_{-\infty}^{\infty} x^2e^{- 2 \pi i\xi x} \mathrm{d}x=-\frac{\delta''(\xi)}{4\pi^2}$$

$\xi ---> \xi + \frac{ λ }{2 \pi i}$

$$\int_{-\infty}^{\infty} x^2 e ^ {-λx}e^{- 2 \pi i\xi x} \mathrm{d}x=-\frac{\delta''(\xi + \frac{ λ }{2 \pi i} )}{4\pi^2}$$ (Fourier transform table)

-
$x^2e^{-\lambda x}$ is not a distribution on ${\mathbb R}$. What you are doing isn't valid... the shift rule for Fourier transforms only applies to shifts in ${\mathbb R}^n$, not complex shifts, which can result in expressions that are not well-defined, such as your last line there. The left-hand integral is divergent. – Zarrax Jun 23 '12 at 14:59