I’d like to present my dodgy proof of the Fresnel Integrals, which I wrote before I knew anything about complex analysis. It takes some… liberties; yet, it still managed to produce the right value for both integrals, so I thought it might be worth sharing. Perhaps, with some input, I could change it into something more rigorous? I also wanted to leave a fun teaser at the end.

The Proof

So, the goal is to evaluate both $\int_0^\infty \cos{\left(x^2\right)}dx$ and $\int_0^\infty \sin{\left(x^2\right)}dx$, the Fresnel Integrals. Start by using Euler’s identity: $$e^{ix} = \cos{x} + i\sin{x}$$

Substituting $x$ for $-x^2$, this leaves: $$e^{-ix^2} = \cos{\left(x^2\right)} - i\sin{\left(x^2\right)}$$ $$\implies \int_0^\infty e^{-ix^2}dx = \int_0^\infty\cos{\left(x^2\right)}\,dx - i\int_0^\infty\sin{\left(x^2\right)}\,dx$$

Notice how the left-hand side looks almost exactly like the Gaussian integral? Using that observation, I used the substitution $ix^2 = u^2 \implies u =\pm\sqrt{i} \cdot x$. Since this integral is taken within the interval $[0, \infty)$, we can take only the positive branch from the square root. When $x = 0, u = 0$; as $x\to\infty, u\to\infty$. $dx = \frac{du}{\sqrt{i}} = -i\sqrt{i} \cdot du$.

$$I = -i\sqrt{i}\int_0^\infty e^{-u^2}du$$

$\sqrt{i} = \frac1{\sqrt{2}} + i\frac1{\sqrt{2}}$ (taking only the positive branch), $\implies -i\sqrt{i} = \frac1{\sqrt{2}} - i\frac1{\sqrt{2}}$. The new integral $\int_0^\infty e^{-u^2}du = \frac{\sqrt{\pi}}{2}$. All of this leaves: $$I = \left(\frac1{\sqrt{2}} - i\frac1{\sqrt{2}}\right)\frac{\sqrt{\pi}}{2}$$ $$ = \sqrt{\frac{\pi}8} - i\sqrt{\frac{\pi}8} $$

With this, all that’s left to do is compare the real and imaginary parts of $I$:

$$\Re{\left(I\right)} = \sqrt{\frac{\pi}{8}} = \int_0^\infty\cos{\left(x^2\right)}dx$$

$$\Im{\left(I\right)} = -\sqrt{\frac{\pi}{8}} = -\int_0^\infty\sin{\left(x^2\right)}dx$$

$$\therefore \int_0^\infty\cos{\left(x^2\right)}dx = \int_0^\infty\sin{\left(x^2\right)} = \sqrt{\frac{\pi}{8}}$$

The Problem

The most blaring issue here is my assumption that $\lim_\limits{x\to\infty} \sqrt{i} \cdot x = \infty$. This is nonsense when you look at it this idea of the limit on the complex plane.

The complex function $f(x) = \sqrt{i} \cdot x$ can be expressed as $f(x) = \frac{x}{\sqrt2} + i\frac{x}{\sqrt2}$. To help visualize this, I plotted this on a graphing calculator, which gives the following:

enter image description here

As you can see, as this x approaches infinity, this function does approach an infinity; the function approaches the corner of the real-imaginary plane $\infty + i\infty$. However, this is not the same as $\infty$, which represents approaching the edge of the horizontal real number line. To put this more rigorously, this picture shows that $\lim_\limits{x\to\infty}\sqrt{i}\cdot x \neq \lim_\limits{x\to\infty}x$.

But why did this proof still work? It clearly returned the right value for both integrals. Is this step really as nonsensical as it seems? And if not, what kind of implications could that carry?

The Teaser — The Icing on the Cake

For positive n, $$\int_0^\infty e^{-ax^n}dx = a^{-\frac1{n}}\Gamma{\left(1+\frac1{n}\right)}$$

You could derive this yourself with the substitution $u = ax^n$. But all this thinking about limits to complex infinities has me thinking: why not plug in $a = i$? Is that valid? This would yield:

$$\int_0^\infty e^{-ix^n}dx = i^{-\frac1{n}}\Gamma{\left(1+\frac1{n}\right)}$$

Using the same logic as before, this would not make sense, since as $x\to\infty$ in the substitution above, $u\to -i\infty$, which is not the same as $u$ approaching normal infinity. But it managed to work before—could it work now? With $n = 2$—our original case—it does seem to return the same value as before.

  • $\begingroup$ You use Cauchy which connects the two integrals (one on the reals and one on the complex path you get after the naive change variables - technically you take the integrals from zero to endpoint of modulus $R$ and connect those by a circle arc say and show that the integral on that goes to zero as $R$ goes to infinity) $\endgroup$
    – Conrad
    May 1, 2022 at 14:11
  • $\begingroup$ That makes sense, I’ve seen similar approaches when looking into other ways of proving this. But I’m curious as to why my approach works anyways. Did I just get lucky? Or is there something deeper to it? $\endgroup$
    – Mailbox
    May 1, 2022 at 14:16
  • 1
    $\begingroup$ $e^{-z^2}$ is governed at infinity by the real part of $z^2$ and in this case it happens that goes to zero in the domain we consider which is the first octant, except on the boundary $x=y$ but the integrand goes to zero fast enough on the circle arc connecting the reals with $x=y$ at level $R$ so the integral still goes to zero; so it all has to do with estimates that here are good enough - this is a common method and works precisely when you have estimates that lead to the integral on the connecting path being zero in the limit $\endgroup$
    – Conrad
    May 1, 2022 at 14:29


You must log in to answer this question.

Browse other questions tagged .