Differentiable probability distribution on $\mathbb{R}$ with the minimim variance

Question

As the title says, I want to find a way of determining the function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continuous and differentiable and has the least possible variance. That is, the function which minimises the functional \begin{aligned} \text{Var}[f]&= \langle x^2 \rangle -\langle x\rangle^2=\int_{-\infty}^{\infty}x^2f(x)\text{d}x -\left(\int_{-\infty}^{\infty}xf(x)\text{d}x\right)^2 \\ &=\int_{-\infty}^{\infty}\left\{x^2f(x)-xf(x) \int_{-\infty}^{\infty}yf(y)\text{d}y\right\}\text{d}x. \end{aligned} I know that a way to do this is to use the Euler-Lagrange equation $$ \frac{\partial \mathcal{L}}{\partial f}= \frac{\text{d}}{\text{d}x}\frac{\partial\mathcal{L}}{\partial f'} $$ where $$ \mathcal{L}= x^2f(x)-xf(x) \int_{-\infty}^{\infty}yf(y)\text{d}y $$ but I have no idea how to differentiate the integral with respect to $f$. I would be very grateful for some help.