Determine a function $f(x)$ such that $ \int_{-\infty}^{+\infty}\left(3-4 y^2\right) e^{-y^2} f(x-y) d y=e^{-x^2}, \quad x \in \mathbb{R} $

Question

Determine a function $f(x)$ such that $$ \int_{-\infty}^{+\infty}\left(3-4 y^2\right) e^{-y^2} f(x-y) d y=e^{-x^2}, \quad x \in \mathbb{R} $$

I know that we are using Fourier transform but could someone please explain that how could we get this as stated in the solution. And also why the solution switched the y into x?

After Fourier transformation, the equation takes the form $$ \mathcal{F}\left[\left(3-4 x^2\right) e^{-x^2}\right](\xi) \hat{f}(\xi)=\mathcal{F}\left[e^{-x^2}\right](\xi) $$

Thanks in advance for your answer!

Answer 1

Define the Fourier transform, $F(k)$, of $f(x)$ as

$$F(k)=\int_{-\infty}^\infty f(x)e^{ikx}\,dx$$

Then, using the convolution theorem we find that

$$\begin{align} F(k)&=\frac{\int_{-\infty}^\infty e^{-x^2}e^{ikx}\,dx}{\int_{-\infty}^\infty (3-4x^2)e^{-x^2}e^{ikx}\,dx}\\\\ &=\frac{\sqrt\pi e^{-k^2/4}}{\sqrt\pi (k^2+1) e^{-k^2/4}}\\\\ &=\frac1{k^2+1} \end{align}$$

Taking the inverse Fourier transform, yields

$$\begin{align} f(x)&=\frac1{2\pi}\int_{-\infty}^\infty F(k)e^{-ikx}\,dk\\\\ &=\frac1{2\pi}\int_{-\infty}^\infty \frac{e^{-ikx}}{k^2+1}\,dk\\\\ &=\frac12 e^{-|x|} \end{align}$$

And we are done!