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When we integrate $$I = \int_{-\infty}^\infty \sin(x^2) \ \mathrm{d}x,$$ is it justifiable to use use the fact that $\sin(x^2)$ = $-\mathrm{Im}\left(e^{-ix^2}\right)$? The reason I ask is because when evaluating this integral we obtain $$I = \int_{-\infty}^\infty e^{-ix^2} \ \mathrm{d}x. $$ Here is where my issue resides. By the generalized Gaussian integral we have $$\int_{-\infty}^\infty e^{-ax^2} \ \mathrm{d}x = \sqrt{\frac{\pi}{a}}.$$ But I believe this only holds $\forall a\in \mathbb{R}\setminus \{0\}.$ Correct me if I am wrong but the use of Euler's identity is not justified, solely because we are dealing with the complex number, $i$. However, in the computation of the latter we obtain $$I = \sqrt{\frac{\pi}{2}}$$ which is the correct result. Is this just based off of luck?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Did you mean $-\mathrm{Im}\left(e^{-ix^\color{red}2}\right)$? $\endgroup$ – J. W. Tanner Apr 23 at 23:19
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    $\begingroup$ Yes, I did. I just fixed it. $\endgroup$ – user753116 Apr 23 at 23:23
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I guess you missed a factor 2 somewhere, as $\sqrt{\pi/8}$ is the value of the Fresnel integral between $0$ and $+\infty$. Anyway, your intuition is correct, and actually you can indeed take the imaginary part by continuity of $\textrm{Im}$.

In addition the value of the Gaussian integral is also known (and it is what you hinted indeed) for complex offsets, but the proof requires to integrate over a well chosen contour. What I would advise you in not to think of it as a black box result but instead to try to follow the proof step by step and see what is the difference with the real case and how you can fix it.

As it is very classical, I point you a reference on the same topic a few years ago: Some way to integrate $\sin(x^2)$?

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  • $\begingroup$ So the generalized Gaussian integral is also true for all a in C (including every real and imaginary number)? $\endgroup$ – user753116 Apr 23 at 23:39
  • $\begingroup$ Yes indeed, and this is an elegant demonstration :) $\endgroup$ – user70925 May 18 at 23:53
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    $\begingroup$ very interesting; thanks! $\endgroup$ – user753116 May 19 at 0:59

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