TITLE: Proof only by transformation that : $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $
QUESTION [38 upvotes]: This was a  question in our exam and I did not know which change of variables or trick to apply
How to show  by inspection ( change of variables or whatever trick  ) that
$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx \tag{I} $$
Computing the values of these integrals are known as routine. Further from their values, the equality holds. But can we show equality beforehand?

Note: I am not asking for computation since it can be found here
and we have as well that,
$$ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx =\sqrt{\frac{\pi}{8}}$$
and the result can be recover here, Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?.

Is there any trick to prove the equality in (I) without computing the exact values of these integrals beforehand?

REPLY [2 votes]: Note that
$\int_0^\infty(\mathcal{L} g)(x)f(x)\,dx = \int_0^\infty g(x)(\mathcal{L}f)(x)\,dx$
with $f(x) =\sin x -\cos x$, $g(x)=\frac1{\sqrt x}$ and their transformations $\mathcal{L} f=\frac{x-1}{x^2+1}$, $
 \mathcal{L} g=\sqrt{\frac \pi x}$
leads to
$$\sqrt\pi\int_0^\infty\frac{ \sin x -\cos x}{\sqrt x}dx=
\int_0^\infty\frac{ x-1}{\sqrt x(x^2+1)}dx
 \overset{x\to \frac1x}= \int_0^\infty\frac{1- x}{\sqrt x(x^2+1)}dx=0
$$
where the vanishing LHS with $x\to x^2$ yields $\int_0^\infty \sin x^2 dx= \int_0^\infty \cos x^2dx$.