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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 :)

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  • $\begingroup$ sorry this was soled by Gauss integral as seen below :-) $\endgroup$
    – user469216
    Commented Aug 4, 2018 at 8:39

1 Answer 1

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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.

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  • $\begingroup$ 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. $\int_{-\infty}^{\infty}e^{-a \cdot(x+b)^2}=\sqrt{\pi/a}$. $\endgroup$
    – user469216
    Commented Jul 15, 2018 at 20:16
  • $\begingroup$ Good point . . . $\endgroup$
    – Nosrati
    Commented Jul 15, 2018 at 20:54

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