Integrate translated gaussian function by change of variable

[cozycoz]

I want to integrate this:$$\int_0^∞ re^{-\frac{1}{2σ^2} (r-iσ^2q)^2} \, dr.$$
If I change the variable r into t with this relation:$$r-iσ^2q=t,$$
then the integral becomes$$\int_{-iσ^2q}^∞ (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt$$
so it seems I cannot use the famous gaussian integral formula. But I got the correct result for the imaginary part of this problem with just ignoring the lower limit(actually I just forgot to rearrange upper and lower limits but anyway):$$iσ^2q\int_{0}^{∞} e^{-\frac{1}{2σ^2} t^2}\,dt= \frac{iσ^2q}{2} \int_{-∞}^∞ e^{-\frac{1}{2σ^2}t^2} \, dt =\frac{iσ^2q}{2}\sqrt{2σ^2π}$$ So why are they the same: the integration over the interval {$[-iσ^2q, 0]+[0, ∞]$} and $[0, ∞]$?

 

[haruspex]

then the integral becomes$$\int_{-iσ^2q}^∞ (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt$$

No, it becomes $$\int_{-iσ^2q}^{∞-iσ^2q} (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt$$, where the integral is understood to be along the straight line $t=x-iσ^2q$.

 

[cozycoz]

No, it becomes $$\int_{-iσ^2q}^{∞-iσ^2q} (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt$$, where the integral is understood to be along the straight line $t=x-iσ^2q$.

Yes, that's what my mentor noted on my homework, but I don't get it..
Could you tellme how you think about my explanation below?
Maybe I could write it as

$$\int_{-iσ^2q}^{∞-iσ^2q}e^{-\frac{1}{2σ^2} t^2} \, dt$$
$$=<span style="font-size: 16px">\frac{1}{2} \int_{-∞-iσ^2q}^{∞-iσ^2q} e^{-\frac{1}{2σ^2}t^2} \, dt </span>$$
$$≅\frac{1}{2} \int_{-∞}^{∞} e^{-\frac{1}{2σ^2}t^2} \, dt.$$

Sorry I don't know what the heck is wrong with the second code...Hope you recognize the limits:(

 

[haruspex]

Maybe I could write it as

I do not understand how you got that form.
Perhaps you should consider integrals around different contours. What do you know about integrals around poles in the complex plane?