Evaluate this improper integral as a function of $t$

Question

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?

Answer 1

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, $f(t)=1$

Answer 2

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.

Answer 3

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.