Calculate $\rho *f(x)$ where $f(x) = x$ and $\rho$ is a standard symmetric mollifier on $R$

Question

As title, I want to calculate as an exercise $\rho *f(x)$ where $f(x) = x$ and $\rho$ is a standard symmetric mollifier on $R$.

I know that $\rho *f(x) = \int_R \rho(x-y)f(y)dy$

I have the following definition for $\rho$:

$\rho$:= $e^{-{1 \over {} 1-|x|^2}}$ for $|x|<1$, $0$ otherwise.

I've tried to solve it by parts, but I came up with an unintegrable integral. What is the best advice for me to continue and also complete the exercise?

Answer 1

Partial Answer

$$(\rho \ast f)(x)=\int_{-1}^1 e^{-{1 \over {} 1-y^2}}(x-y)dy$$ $$=x\int_{-1}^1 e^{-{1 \over {} 1-y^2}}dy-\int_{-1}^1 ye^{-{1 \over {} 1-y^2}}dy$$

The second integral is zero because the integrand is an odd function.

Also the first integral is a constant. I am not sure if you can calculate it explicitly.