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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 $k\in\mathbb{C}$, 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 $R\rightarrow \infty$ seem not easy to handle either.

Andrews
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Ruairi
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  • Hi Collin - Welcome to MSE - in order for the community to be able to assist you, you need to provide all working you have done so far. If you are looking for a starting point, please ask - but this site is not a 'homework for free' site. –  Feb 25 '19 at 03:46
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    Hint: This is all you need to solve:

    $$ \int_{-\infty}^\infty e^{-x^2}:dx = \sqrt{\pi}$$

     –  Feb 25 '19 at 03:48
  • Assuming the jump into the complex domain is valid. I always do. – marty cohen Feb 25 '19 at 05:39

3 Answers3

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

Claude Leibovici
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  • Hi! I think it's because I'm not familiar with complex integral; so how can you guys play so freely on the $\mathbb{C}$ plane, without worrying much? The very reason I ask this is because I don't know why your second equation holds for k being a complex number. I tried to rotate this integral path by $\pi/4$ on the complex plane, but the two arcs at $|x|\rightarrow \infty$ seem not trying to vanish. – Ruairi Feb 25 '19 at 06:01
  • @Collin, The reasoning behind this 'extrapolation' is that $f$ defines a holomorphic function on the right half-plane $H={k:\operatorname{Re}(k) > 0}$ and a continuous function on $\overline{H}\setminus{0} = {k : \operatorname{Re}(k) \geq 0, k \neq 0 }$. Now, in view of identity theorem, if you can identify another holomorphic function $g$ on $H$ such that $f = g$ on $(0,\infty)$, then $f = g$ on all of $H$ as well. We can check that this argument works with $g(k)=\sqrt{\pi/k}$, and so, $f = g$ on $H$ by identity theorem and on all of $\overline{H}\setminus{0}$ by continuity. – Sangchul Lee Feb 25 '19 at 06:24
  • @SangchulLee Hi! So this is the material covered by complex analysis? I have encountered many times similar problems when doing complex integral. As a physicist, complex analysis is only usually covered by Math-physics course within a chapter. – Ruairi Feb 25 '19 at 19:57
  • See my answer that purposely avoids this problem. – marty cohen Feb 26 '19 at 00:03
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Trying to avoid complex funniness.

$\begin{array}\\ \int_{-\infty}^\infty e^{-ix^2}dx &=\int_{-\infty}^\infty (\cos(x^2)-i\sin(x^2))dx\\ &=2\int_{0}^\infty (\cos(x^2)-i\sin(x^2))dx\\ &=2\int_{0}^\infty \cos(x^2)dx-2i\int_{0}^\infty\sin(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.

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

csch2
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