Simplify ratio of integrals $\frac{\int f(x-t) t e^{-t^2/2} dt}{\int f(x-t)e^{-t^2/2} dt}$

Question

I am trying to simplify the following expression:

\begin{align*} \frac{\int_{-\infty}^\infty f(x-t) t e^{-t^2/2} dt}{\int_{-\infty}^\infty f(x-t)e^{-t^2/2} dt} \end{align*}

by getting it in terms of just one integral. For example, is the following possible

\begin{align*} \frac{\int_{-\infty}^\infty f(x-t) t e^{-t^2/2} dt}{\int_{-\infty}^\infty f(x-t)e^{-t^2/2} dt}=\int_{-\infty}^\infty y(x-t)g(t) dt \end{align*}

for some function $y(t)$ and $g(t)$. Can we find $y,g$ in term of $f$ and $e^{-t^2/2}$?

Few comment:

1. observe that both integrals are convolutions
2. we can assume that both integrals exist and are finite
3. Can assume that $f(t)>0$ and $f \in L^1$

Would like to hear any suggestion or thoughts you can give me. Thank you

Answer 1

$$ \frac{\int{f(x-t) t e^{\frac{-t^2}{2}} dt}}{\int{f(x-t)e^{\frac{-t^2}{2}} dt}} $$

Let's consider this integral first:

$$ \int{f(x-t)e^{\frac{-t^2}{2}} dt} $$

Using integration by parts, we get

$$ \int{f(x-t)e^{\frac{-t^2}{2}} dt} = F(x - t)e^{\frac{-t^2}{2}} + \int{F(x - t)te^{\frac{-t^2}{2}} dt} $$

where $F(x - t)$ is the anti-derivative of $f(x - t)$

Next, let's consider the top integral.

$$ \int{f(x-t) t e^{\frac{-t^2}{2}} dt} = f(x - t)e^{\frac{-t^2}{2}} + \int{f'(x - t)e^{\frac{-t^2}{2}} dt} $$

Depending on your function $f$, the following fraction may be easier to deal with:

$$ \frac{\int{f(x-t) t e^{\frac{-t^2}{2}} dt}}{\int{f(x-t)e^{\frac{-t^2}{2}} dt}} = \frac{f(x - t)e^{\frac{-t^2}{2}} + \int{f'(x - t)e^{\frac{-t^2}{2}} dt}} {F(x - t)e^{\frac{-t^2}{2}} + \int{F(x - t)te^{\frac{-t^2}{2}} dt}} $$

Answer 2

You might use: $$ \int_{-\infty}^{\infty} f(x-t) t e^{-t^2/2} dt = -\int_{-\infty}^{\infty} f(x-t) \frac{d}{dt} e^{-t^2/2} dt $$ Upon integration by parts then: $$ -\int_{-\infty}^{\infty} f(x-t) \frac{d}{dt} e^{-t^2/2} dt = \int_{-\infty}^{\infty} \frac{d}{dt} f(x-t) e^{t^2/2} dt \\ = -\int_{-\infty}^{\infty} \frac{d}{dx} f(x-t) e^{-t^2/2} dt \\ = -\frac{d}{dx} \int_{-\infty}^{\infty} f(x-t) e^{-t^2/2} dt $$ Letting $ \widetilde{f}(x) = \int_{-\infty}^{\infty} f(x-t) e^{-t^2/2} dt $, the original quantity is: $$ -\frac{d}{dx} \log( \widetilde{f}(x) ) $$