Calculus – Proof by Transformation: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

analysiscalculuscomplex-analysisfresnel-integralsintegration

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?

Best Answer

<Here is what I found

Employing the change of variables $2u =x^2$ We get $$I=\int_0^\infty \cos(x^2) dx =\frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx$$ $$ J=\int_0^\infty \sin(x^2) dx=\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx $$

Summary: We will prove that $J\ge 0$ and $I\ge 0$ so that, proving that $I=J$ is equivalent to $$ \color{blue}{0= (I+J)(I-J)=I^2 -J^2 =\lim_{t \to 0}I_t^2-J^2_t}$$ Where, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the present of the integrand factor $e^{-tx^2}$.

However, By 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}

To end the proof: Let us show that $I> 0$ and $J> 0$. Performing an integration by part we obtain $$J = \frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx=\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\color{red}{>0}$$ Given that $\color{red}{\sin 2x= 2\sin x\cos x =(\sin^2x)'}$. Similarly we have, $$I = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\= \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx\color{red}{>0}$$

Conclusion: $~~~I^2-J^2 =0$, $I>0$ and $J>0$ impliy $I=J$. Note that we did not attempt to compute neither the value of $~~I$ nor $J$.

Extra-to-the answer 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 $$\color{red}{I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac{\pi}{8}}$$

See also here for more on (The Fresnel Integrals Revisited)

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