Evaluate this improper integral as a function of $t$ Evaluate $\displaystyle f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(x-t)^2/2}\ dx$
My attempt:
I know that $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-x^2/2}\ dx=1$$
So, to evaluate $\displaystyle f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(x-t)^2/2}\ dx$, 
I guess we could let $x-t=y$.
Then $dx-dt=dy$
$dx=dy+dt$
What would the new limits of integration be? Am I on the right track? How do I proceed?
 A: The variable $t$ has absolutely no effect on anything.
What you are doing is finding the area under a bell curve.
Changing $t$ shifts this bell curve to the left or right, doesn't change the area, doesn't change the integral.
Can be seen by making the substitution $u=x-t$.
And so for all values of $t$, this integral has value $1$.
Sooooo finally, $f(t)=1$
A: As $x$ goes from $-\infty$ to $+\infty$, the quantity $t$ does not change. In other words, $t$ is a constant.
If $x-t = y,$ then $dx=dy.$  (Just as, if $x-8=y,$ then $dx=dy.$)
As $x$ goes from $-\infty$ to $+\infty$ while $t$ remains fixed, then $y$ also goes from $-\infty$ to $+\infty,$ so those are your bounds of integration.
A: The error function is defined as
$$\text{erf(x)}=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt $$
with $\text{erf}(\mp\infty)=\mp 1$. Therefore,
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-x^2/2}\ dx=\frac{1}{2\sqrt{2}}\frac{2}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-x^2/2}\ dx=\frac{1}{\sqrt{2}}$$
And finally,
$$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(x-t)^2/2}\ dx=\frac{1}{\sqrt{2}}$$
because the limits are infinite so that it transforms into the same thing. I have verified these results numerically.
