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I have an integral of type:

$$\int_{-\infty}^{\infty} e^{iax}\cdot e^{-(bx^2 + cx)}~ dx$$

And I have no clue on how to integrate that properly. What I've tried so far is writing everything above one single e-function and use

$$\int e^{a(x)} dx = \bigg(\frac{d}{dx} a(x) \bigg)^{-1}e^{a(x)} + C$$

But this hasn't brought me any far ....

Do you have some trick for me I didn't see?

Thanks for helping!

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  • $\begingroup$ Do $a,b,c \in \mathbb{R}$? $\endgroup$
    – Rammus
    Nov 15 '14 at 14:32
  • $\begingroup$ Yes. Sorry, I forgot about that. $\endgroup$ Nov 15 '14 at 14:35
  • $\begingroup$ Did you try term by term integration ? $\endgroup$
    – Airbag
    Nov 15 '14 at 14:35
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Hint

Let us start rewriting $$I=\int e^{iax}\cdot e^{-(bx^2 + cx)}dx=\int e^{-\big(bx^2 + (c-ia)x\big)}dx$$ Now, completing the square, $$bx^2 + (c-ia)x=b\Big(\big(x+\frac{c-ia}{2b}\big)^2-\big(\frac{c-ia}{2b}\big)^2\Big)=b\big(x+\frac{c-ia}{2b}\big)^2+\frac{(a+i c)^2}{4 b}$$ So, $$I=e^\frac{(a+i c)^2}{4 b} \int e^{-b\big(x+\frac{c-ia}{2b}\big)^2}dx$$ Now, remember that $$\int e^{-\alpha y^2}dy=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{\alpha } y\right)}{2 \sqrt{\alpha }}$$

I am sure that you can take from here.

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First note that $e^{iax}e^{-(bx^2 +cx)}=e^{ix(a+ic)}e^{-bx^2}$. Taking $a+ic=z$, $\int_{-\infty}^\infty e^{izx}e^{-bx^2}dx $ can be written in terms of the fourier transform of the function $e^{-bx^2}$.

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