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I am trying to derive the Fresnel Integrals $$\int_{-\infty}^\infty \sin(x^2)\rm dx ~~~~and ~~~~~\int_{-\infty}^\infty \cos(x^2)\rm dx$$ through the Gaussian Integral via $I=\int_{-\infty}^\infty e^{ix^2}\rm dx$.

I got to $$I^2=\int_{0}^{2\pi}\int_{0}^\infty re^{ir^2}\rm dr$$

After integrating, I am left with the expression$$I^2=\int_{0}^{2\pi}\frac{i}{2}(\lim_{r\to\infty}-e^{ir^2}+1)\rm d\theta$$

I don't believe that this approaches a defined value but if I use $\lim_{r\to\infty}-e^{ir^2}=0$, I arrive at the correct answer of $\Re(I)=\sqrt{\frac{\pi}{2}}$ and $\Im(I)=\sqrt{\frac{\pi}{2}}$

So my question is why does using $0$ for that expression work? Or am I incorrect in using that value.

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    $\begingroup$ This recent question is strongly related. See in particular the answer by Guy Fsone $\endgroup$ Nov 21, 2017 at 22:11
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    $\begingroup$ Your approach works when replacing $I$ by $\lim_{a \to 0^+} I_a =\lim_{a \to 0^+}\int_0^\infty e^{(-a+i) x^2}dx$. Regularizing integrals in such a way is very common in complex and Fourier analysis. $\endgroup$
    – reuns
    Nov 21, 2017 at 22:28
  • $\begingroup$ @reuns I see, so it requires an additional limit of the function $I(a)$? $\endgroup$
    – aleden
    Nov 21, 2017 at 22:31
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    $\begingroup$ $\lim_{r \to \infty} e^{i r^2}$ diverges (the original integral is only conditionally convergent so the circular change of variable, changing the order of summation, isn't allowed) but $\lim_{a \to 0^+} \lim_{r \to \infty} e^{(-a+i) r^2} = 0$ $\endgroup$
    – reuns
    Nov 21, 2017 at 22:34

1 Answer 1

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Observes that the Gaussian $ e^{ix^2}$ is $$ e^{ix^2}=\cos(x^2)+i\sin(x^2)$$

I applying polar coordinates to $e^{ix^2}$ does not provide a convergent integral. But rather it is more wise to think about Feymann's tricks by considering the the following( perturbed) Gaussian $$\color{blue}{e^{(t+i)x^2} = e^{tx^2}\cos(x^2)+ie^{tx^2}\sin(x^2)}$$ then apply polar coordinates and let $t\to0$ aftermath to get the result. this is exactly what I did in my answer here (see all details below)

Proof: Indeed, we proceed as follow: Let, $$I=\int_0^\infty \cos(x^2) dx =\frac{1}{2}\int^\infty_{-\infty}\cos(x^2)dx~~~ and ~~~~ J=\int_0^\infty \sin(x^2) dx=\frac{1}{2}\int^\infty_{-\infty} \sin(x^2) \,dx $$ We set, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the present of the integrand factor $e^{-tx^2}$.

Using Fubini we have,

\begin{split} I_t^2-J^2_t&=& \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) \\&-& \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\ &=& \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\cos(x^2+y^2)dxdy\\ &=&\int_0^{\frac\pi2}\int_0^\infty re^{-tr^2}\cos r^2 drd\theta\\&=&\frac\pi4 Re\left( \int_0^\infty \left[\frac{1}{i-t}e^{(i-t)r^2}\right]' dr\right)\\ &=&\color{blue}{\frac\pi4\frac{t}{1+t^2}\to 0~~as ~~~t\to 0} \end{split}

However using similar technique in above prove one can easily arrives at the following $$\color{blue}{I_tJ_t = \frac\pi8\frac{1}{t^2+1}}$$ from which one get the following explicit value of $$\color{red}{I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac{\pi}{8}}$$

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