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:
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.
