# A question involving Fourier transform

I need a hand with this question:

I have to find a function $g$ verifying the following:

$$xe^{\frac{-x^2}{2}}=\int_{-\infty}^{x}g(t)e^{t-x} \mathrm{d}t,\quad \text{for }x\in\mathbb{R}$$

I know this somehow involves Fourier transform, but I don't know how to solve it.

Thanks a lot for any help.

• Why do you think Fourier transform is involved? Commented Mar 8, 2014 at 14:15
• I know its involved because the problem comes from some class notes about Fourier transform. Commented Mar 8, 2014 at 14:27

The RHS is $$\int\limits_{-\infty}^x g(t)e^{t-x} \mathrm{d}t \\ = \int\limits_{-\infty}^x g(t)e^{-(x-t)}\mathcal{H}(x-t) \mathrm{d}t \\ = \int\limits_{-\infty}^\infty g(t)e^{-(x-t)}\mathcal{H}(x-t) \mathrm{d}t \ \ \ (\text{Because}\ \mathcal{H}(x-t) = 0 \ \forall \ t > x)\\ = g(x) * (e^{-x}\mathcal{H}(x))$$ Now apply Fourier transform to the given equation $$\mathcal{F}(xe^{\frac{-x^2}{2}}) = \mathcal{F}(g(x)) \mathcal{F}(e^{-x}\mathcal{H}(x)) \\ \mathcal{F}(xe^{\frac{-x^2}{2}}) = \mathcal{F}(g(x)) \frac{1}{(1+i\omega)} \\ \mathcal{F}(g(x)) = \mathcal{F}(xe^{\frac{-x^2}{2}}) (1+i\omega) \\ \mathcal{F}(g(x)) = \mathcal{F}(xe^{\frac{-x^2}{2}}) +i\omega \mathcal{F}(xe^{\frac{-x^2}{2}}) \\ \mathcal{F}(g(x)) = \mathcal{F}(xe^{\frac{-x^2}{2}}) + \mathcal{F}\left(\frac{\mathrm{d}}{\mathrm{d}x}xe^{\frac{-x^2}{2}}\right) \\ \mathcal{F}(g(x)) = \mathcal{F}(xe^{\frac{-x^2}{2}}) + \mathcal{F}\left(\left(1-x^2\right)e^{\frac{-x^2}{2}}\right) \\ \mathcal{F}(g(x)) = \mathcal{F}\left(\left(x + 1-x^2\right)e^{\frac{-x^2}{2}}\right) \\ g(x) = (1+x-x^2)e^{\frac{-x^2}{2}}$$ Simple Answer
Differentiate the given expression wrt $x$. $$(1-x^2)e^{\frac{-x^2}{2}} = e^{-x} g(x) e^x - e^{-x} \int\limits_{-\infty}^x g(t) e^t\mathrm{d}t \\ (1-x^2)e^{\frac{-x^2}{2}} = g(x) - \int\limits_{-\infty}^x g(t) e^{t-x}\mathrm{d}t \\ (1-x^2)e^{\frac{-x^2}{2}} = g(x) - xe^{\frac{-x^2}{2}} \\ g(x) = (1+x-x^2)e^{\frac{-x^2}{2}}$$
• We work with this definition of Fourier transform: $T(f)(\xi)=\int_{\mathbb{R}}f(x)e^{-i\xi x}\;dx$, Commented Mar 8, 2014 at 14:34
If derivativs of the sides are taken, on the right side one gets $g(x)-\int_{-\infty}^xg(t)e^{t-x}dt$ [used differentiation under integral sign, see here]. Now note the subtracted integral here is $xe^{-x^2/2}$ from your initial equation.