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

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.

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 $$\lim_{a\to \pm \infty} F(a)= \pm \pi/2$$.