Jtdylktuy

How to calculate this elementary complex integral? This is what we would encounter if we are studying the Green's function for Schroedinger's equation.
$$int_{-infty}^infty e^{-ix^2}d x=?$$
However, I think there should be someone that posted similar question on Math SE, though I don't know how to search by equation.

Thank you very much if you can help me out! And I would be grateful if you can give more than one approach

P.S.: The equation $int _{-infty}^{infty}e^{-kt^2}d sqrt{k}t=sqrt{pi}$ surely comes to my mind, but I don't know why it holds for $kinmathbb{C}$, because for me, the above integral is over real line, however, the question here is like integral on $y=e^{i pi/4}x$ ( So I think it's the problem with my complex integral knowledge.)

I tried to rotate this integral path by $pi/4$, but the two arcs at $Rrightarrow infty$ seem not easy to handle either.

Trying to avoid complex funniness.

$begin{array}\
int_{-infty}^infty e^{-ix^2}dx
&=int_{-infty}^infty (cos(x^2)-isin(x^2))dx\
&=2int_{0}^infty (cos(x^2)-isin(x^2))dx\
&=2int_{0}^infty cos(x^2)dx-2iint_{0}^inftysin(x^2))dx\
end{array}
$

and these are the
Fresnel integrals
$C(x)$ and $S(x)$
both of which approach
$dfrac{sqrt{pi}}{8}
$
as $x to infty$.

Therefore the result is
$(1-i)sqrt{frac{pi}{2}}
$
as Claude Leibovici
got.

Hint
$$int e^{-k x^2},dx=frac{sqrt{pi } }{2 sqrt{k}},text{erf}left(sqrt{k} xright)$$
$$f(k)=int_{-infty}^infty e^{-k x^2},dx=frac{sqrt{pi }}{sqrt{k}}$$
$$f(i)=frac{sqrt{pi }}{sqrt{i}}=(1-i) sqrt{frac{pi }{2}}$$

Hint:$$int_{-infty}^infty e^{-kx^2}dx=int_{-infty}^infty e^{-left(xsqrt kright)^2}dx$$
Use the $u$-substution $u=xsqrt k$ and this transforms the integral into the form given in DavidG's suggestion. Can you take it from here?