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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.

Priyatham
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Mark_Hoffman
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2 Answers2

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Using Fourier Transform
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 $x$. $$ (1-x^2)e^{\frac{-x^2}{2}} = e^{-x} g(x) e^x - e^{-x} \int\limits_{-\infty}^x g(t) e^t\mathrm{d}t \\ (1-x^2)e^{\frac{-x^2}{2}} = g(x) - \int\limits_{-\infty}^x g(t) e^{t-x}\mathrm{d}t \\ (1-x^2)e^{\frac{-x^2}{2}} = g(x) - xe^{\frac{-x^2}{2}} \\ g(x) = (1+x-x^2)e^{\frac{-x^2}{2}} $$

Priyatham
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  • Thanks for the answer, Priyatham. But are the hypothesis of differentiation under integral sign verifyed? And do you have any idea on how could this be solved using Fourier transform techiniques? – Mark_Hoffman Mar 08 '14 at 14:29
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    Yes, the hypothesis is satisfied. And will you tell me how FT is defined in your notes(many conventions exist)? – Priyatham Mar 08 '14 at 14:32
  • We work with this definition of Fourier transform: $T(f)(\xi)=\int_{\mathbb{R}}f(x)e^{-i\xi x};dx$, – Mark_Hoffman Mar 08 '14 at 14:34
  • I am not getting anywhere with FT. Are you sure that the upper limit of integration is x and not infinity. – Priyatham Mar 08 '14 at 14:55
  • I also thought that. In the notes is written this way. With $\infty$, would it make more sense I think. Looks like some convolution – Mark_Hoffman Mar 08 '14 at 14:58
  • If we take $\infty$ instead of $x$, taking Fourier tranforms, we get that $T(xe^{\frac{-x^2}{2}})=T(e^{-x}\ast g)=T(e^{-x})\cdot T(g)$. Using the fact that $T(xe^{-x})(\xi)=\sqrt{2\pi}i(-\xi) e^{\frac{-\xi^2}{2}}$ and $T(e^{-x})(\xi)=\frac{1}{1+i\xi}$ we arrive to $T(g)(\xi)=\sqrt{2\pi}i(-\xi)e^\frac{-\xi^2}{2}(1+i\xi)$. With Fourier inversion, we would get $g$, but i'm having problems with it... – Mark_Hoffman Mar 08 '14 at 15:38
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    I updated my answer with a method using FT. – Priyatham Mar 08 '14 at 16:15
  • Very good answer, Priyatham. Thanks a lot! – Mark_Hoffman Mar 08 '14 at 16:20
  • let us continue this discussion in chat – Priyatham Mar 08 '14 at 16:30
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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.

coffeemath
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