# Can we use the real part of $e^{ix}$ in the evaluation of the Fresnel integral?

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?

• Welcome to Mathematics Stack Exchange. Did you mean $-\mathrm{Im}\left(e^{-ix^\color{red}2}\right)$? – J. W. Tanner Apr 23 at 23:19
• Yes, I did. I just fixed it. – user753116 Apr 23 at 23:23

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