Is $x \mapsto \int_{-\infty}^{-a+x}\exp(-\frac{t^2}{2})\, dt + \int^{\infty}_{a+x}\exp(-\frac{t^2}{2})\, dt$ minimized at $x=0$? 
As the title suggest: I wonder if the integral
$$\int_{\mathbb R \setminus (-a,a) } \exp(-\frac{t^2}{2})\, dt$$
increases when whe shift the area of integration by any number $x$?

Intuitively one could look at the following picture and consider that a right shift of the area of integration amounts to adding to the integral a part where the area under the curve is rapidly increasing and subtracting a part where the area isn't increasing as fast.

There is also the following case:

Of courese if we shift past $t=0 $, that is we let $x>a $, we get an increase in the integral as then certainly the area under the curve on $(-a +x, a+ x) $ is less than the area under $(-a , a) $.

How would you go about prove this?

 A: The general answer depends on the value of $a$. To find the minimum points, we will first need to consider $f'(x)$ which is easy enough to find since $f(x)$ is defined by an integral. The reader can verify that
$$f'(x)=e^{-\frac{1}{2}(-a+x)^2}-e^{-\frac{1}{2} (a+x)^2}$$
Now it's plain to see that if $x=0$ then the derivative is zero. Therefore, $f'(x)$ has a stationary point at $x=0$. However, the stationary point could be a minimum, maximum or a inflection point.
To confirm that $x=0$ is a minimum point, we will need the sign of $f''(0)$. Finding $f''(x)$ requires differentiating exponentials which in general works as follows:
$$\frac{\partial e^{g(x)}}{\partial x}=e^{g(x)}g'(x)$$
In our case, $g(x)=-\frac{1}{2} (\pm a+x)^2$. Using this, the reader should find that
$$f''(x)=e^{-\frac{1}{2}(a-x)^2}(a-x)+e^{-\frac{1}{2}(a+x)^2}\ (a+x)$$
In general, a function $h(x)$ has a minimum point when when $h'(0)=0$ and $h''(0)>0$, and a maximum point when $h'(0)=0$ and $h''(0)<0$. Further consideration is required when $h'(0)=0$ and $h''(0)=0$.
In our case, $f'(0)=0$ and $f''(0)=2 a e^{-\frac{a^2}{2}}$. Therefore, $f''(0)>0$ when $a>0$ and $f''(0)<0$ when $a<0$. At $x=0$, the function $f(x)$ has a minimum when $a>0$ and maximum when $a<0$. In the case where $a=0$, we know from the definition that $f(x)=1$ for all $x$. At $a=0$ the entire function is a constant.
