Find $\int_{-\infty}^{\infty} e^{\frac{1}{2}(x+i \cdot s)^2}\mathrm{d}x$

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How would one go about integrating the following integral:

$$\int_{-\infty}^{\infty} e^{\frac{1}{2}(x+i \cdot s)^2}\mathrm{d}x$$

My intuition tells me this integral diverges since we have

$$e^{\frac{1}{2}(x+i \cdot s)^2} = e^{\frac{1}{2}(x^2-s^2)}\cdot (\cos(2xs) + i \cdot \sin(2xs))$$

This looks to me like a spiral with a radius that increases to infinity at the endpoints of the interval. However, I have trouble coming up with an argument other than the fact that $e^{\frac{1}{2}(x^2-s^2)}$ does not converge and neither does $(\cos(2xs) + i \cdot \sin(2xs))$.

Best regards, kasp9201.

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$\int_{\infty} ^{\infty} |e^{\frac 1 2 (x-is)^{2}}|\, dx=\int_{\infty} ^{\infty} e^{\frac 1 2 x^{2}} e^{-s^{2}x/2}\, dx=\infty$ as seen by making the substitution $y=x-\frac {s^{2}} 2$.