Integral of exponential of complex expression $\int_{-\infty}^{\infty}\exp\left(-a\left[\left(y+ib/2a\right)^2-i^2b^2/4a^2\right]\right)dy$

I have the following expression

$$R\int_{-\infty}^{\infty}exp \bigg(-a\bigg[\bigg(y+ib/2a\bigg)^2-i^2b^2/4a^2\bigg]\bigg)dy$$ Where $R$ is real numbers and $i$ denotes complex numbers.

Which should result in the following $$=exp\bigg(-b^2/4a\bigg)\sqrt{\pi/a}$$

I am not sure how to get to that result however, any help would be highly appreciated :)

• sorry this was soled by Gauss integral as seen below :-) Commented Aug 4, 2018 at 8:39

Hint: $$\int_{-\infty}^{\infty}\exp \bigg(-a\bigg[\bigg(y+ib/2a\bigg)^2-i^2b^2/4a^2\bigg]\bigg)dy=\exp\bigg(-b^2/4a\bigg)\ \int_{-\infty}^{\infty}\exp \bigg(-a\bigg[\bigg(y+ib/2a\bigg)^2\bigg]\bigg)dy$$ now let $\sqrt{a}\bigg(y+ib/2a\bigg)=u$ and after substitution use gamma function. Note that $e^{-ay^2}$ is even.
• thanks good hint :-). I found another way, by splitting the function up as you did and letting $\bigg(y+ib/2a\bigg)=u$, and then use the Gauss integral, i.e. $\int_{-\infty}^{\infty}e^{-a \cdot(x+b)^2}=\sqrt{\pi/a}$. Commented Jul 15, 2018 at 20:16