Show $ \int_ {-\infty} ^{\infty} (\arctan(x+a)) {{1}\over {\sqrt{2 \pi T}}} e^{-x^2 /2T} dx$ can assume any value in $(-\pi/2, \pi/2)$

Question

I would like to show

$$ \int_ {-\infty} ^{\infty} (\arctan(x+a)) {{1}\over {\sqrt{2 \pi T}}} e^{-x^2 /2T} dx$$ can assume any value in $(-\pi/2, \pi/2)$ where $T>0$ is fixed and $a$ may be any real number.

In the limit $T \rightarrow 0$, I see the integral has value $\arctan a$ which can assume any value in the above interval. I have tried to use properties of convolution but it gets me nowhere.

Any hint is appreciated.

Answer 1

Let your integral be $F(a)$. First note that by dominated convergence, $\lim_{a\to a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $\lim_{a\to \pm \infty} F(a)= \pm \pi/2$.

Use IVT to conclude.