Evaluating $\int e^{-(t-k)^2} dt$, where $k$ is a constant I have done the following, I just wanted to know if I have made any blunders. If so please suggest accordingly. It follows in the line of evaluating $\int_{-\infty}^{\infty}e^{-x^2}dx$
$$I = \int_{-\infty}^{\infty}e^{-(t-k)^2} \tag{1}dt$$
$$I^2 = \int_{-\infty}^{\infty}e^{-(t-k)^2}dt.\int_{-\infty}^{\infty}e^{-(t-k)^2}\tag{2}dt$$
By change of variable:
$$I^2 = \int_{-\infty}^{\infty}e^{-(t-k)^2}dt.\int_{-\infty}^{\infty}e^{-(s-k)^2}ds\tag{3}$$
Let,
$$t-k = a \implies dt = da$$ and 
$$s-k = b \implies ds = db$$
Hence,
$$I^2 = \int_{-\infty}^{\infty}e^{-a^2}da.\int_{-\infty}^{\infty}e^{-b^2}db\tag{4}$$
$$I^2 = \int_{-\infty}^{\infty}e^{-(a^2+b^2)}dadb\tag{5}$$
converting to polar form: (i.e. taking $r^2 = a^2 + b^2$ where $a=rcos(\theta)$ and $b=rsin(\theta)$)
$$I^2 = \int_{0}^{2\pi}\int_{0}^{\infty}re^{-r^2}drd\theta\tag{6}$$
Now, let $u=r^2$ then, $du = 2rdr$. Therefore,
$$I^2 = \int_{0}^{2\pi}d\theta\int_{0}^{\infty}re^{-u}\frac{du}{2r}\tag{7}$$
$$I^2 = \pi\int_{0}^{\infty}e^{-u}du$$
$$\implies I=\sqrt{\pi}\tag{8}$$
Therefore:
$$I = \int_{-\infty}^{\infty}e^{(t-k)^2}dt = \sqrt{\pi}\tag{9}$$
But I am not understanding how to account for the substitutions $(t-k)=a$ and $(s-k)=b$
Thanks in advance. I appreciate for taking your time out to patiently read through the entire thing.
 A: You have accounted accordingly for the substitutions in your calculations already. There is a (positive) linear dependence in your substitution, so you will get a Jacobian that equals 1 ($dt=da$, and $ds=db$), and you will have the same integration limit. To see this, consider again:
$$I_2 = \int_{-\infty}^{\infty} e^{-(t-k)^2}dt\cdot\int_{-\infty}^{\infty} e^{-(s-k)^2}ds.$$
In the left hand integral, make the substitution $t-k = a$ and in the right integral $s-k = b$. This means that 
\begin{aligned}
t & = k + a \Rightarrow \frac{dt}{da}=1 \Rightarrow dt=da \\
s & = k + b \Rightarrow \frac{ds}{db}=1 \Rightarrow ds=db,
\end{aligned}
and
\begin{aligned}
a(t) & = t - k \Rightarrow \lim_{t\to\pm\infty} a(t) \to \pm \infty \\
b(t) & = s - k \Rightarrow \lim_{s\to\pm\infty} b(s) \to \pm \infty.
\end{aligned}
The rest follows just as you have already calculated.
Update: This variable substitution corresponds to a translation. The integrand is, as @user86418 states, translation invariant. Therefore, the final result should not depend on $k$.
