Is there a (relatively simple) function bounding the two Fresnel integrals?

Question

I just discovered the two Fresnel integrals and it really seems like they are bounded by two functions (maybe exponentials?). Here is an image explaining what I mean:

As you can see, I just drew the green and yellow functions by hand, these are the functions I'm interested in. As far as I can tell there are a few constraints that define these two functions. Since those constraints are very similar, I'll only provide them for the green function, which I will call $f(x)$:

1. $f(x)$ has to be strictly convex ($f''(x)>0$)
2. $f(x) \geq S(x)\,\forall x > 0$
3. $f(x)=S(x)\Rightarrow f'(x)=S'(x)$
4. $\int_0^\infty f(x)-S(x)\,\mathrm{d}x$ should be minimized

These 4 constraints (and their counterparts for the yellow function) should in theory define the functions I'm searching well. However, I haven't yet found a method to actually determine these functions.

Can anyone help me?

Answer 1

Although of arguable simplicity, there are two such functions, detailed further in An Atlas of Functions (Chapter 39: The Fresnel Integrals).

Associated with the Fresnel integrals are the auxiliary Fresnel cosine and sine integrals (sometimes designated $f$ and $g$), which the book calls Fres and Gres respectively.

These functions can be defined in several ways, the simplest probably being as follows: $$ F(x)=(\frac{1}{2}-S(x))\cos(x^2)-(\frac{1}{2}-C(x))\sin(x^2)\\ G(x)=(\frac{1}{2}-S(x))\sin(x^2)+(\frac{1}{2}-C(x))\cos(x^2) $$ where $S(x)$ and $C(x)$ are the sine and cosine Fresnel integrals.

When graphed, the auxiliary functions look like this: (Image from book)

The upper bounds of the Fresnel integrals is $\frac{1}{2}+F(x)+G(x)$, and the lower $\frac{1}{2}-F(x)-G(x)$. These are only valid when $x\gt0$ (though in practice, this range actually extends a small amount back).

I've created an interactive graph of these functions here. Note the definitions of the Fresnel integrals; the $\sqrt{\frac{2}{\pi}}$ scaling factor is necessary for the above auxiliary definitions to work.

As to your requirements:

1. Since the bounds functions extend into negative values of $x$, where they oscillate, the functions as a whole are clearly not either purely concave or convex. On the domain $(0, \infty)$ however, I think they are.
   The above definitions of the Fresnel integrals converge to $\frac{1}{2}$. In the auxiliary functions, these are subtracted from $\frac{1}{2}$ to converge to $0$ overall.
   Each auxiliary is composed of two opposing 'blocks' ($(\frac{1}{2}-S(x))\cos(x^2)$ and $(\frac{1}{2}-C(x))\sin(x^2)$ in $F(x)$) that cancel out each other's oscillations. As the oscillations present in the Fresnel integrals decrease in amplitude as $x$ approaches $\infty$, you are left with essentially the difference between the 'center' of each Fresnel integral and $\frac{1}{2}$, which will always (I think) decrease towards $0$, at least in $G(x)$. $F(x)$ actually becomes briefly concave near $0$, but this issue seems to vanish when combined with $G(x)$.
   You could calculate the second derivative yourself to try and prove this if you wanted.
2. I cannot provide a proof for this either, but as per the book, the Fresnel integrals exist purely in the envelope created by the two above bounds functions.
3. As far as I know, the bounds functions never actually touch either of the Fresnel integrals, though (as usual) I cannot prove this. I do not know of any other bounds functions which do.
   That being said, since the derivative of either of the Fresnel integrals decreases after a maxima, necessarily to a greater degree than the upper bounds (as it oscillates), there should always be a point on either of the bounds functions (applying the same logic to the lower bound) near an extrema point which satisfies this requirement.
4. I'm not sure if the above functions satisfy this, sorry. I'd guess probably not, as a theoretically perfect function would essentially trace a straight line between the maxima (or as close as possible, rather).
   With some slight curvature at the tangent points to keep requirement 3, and some infinitely small curvature on the 'straight' sections to keep 1, there probably exists some 'better' function, though I have no idea what that might be.

Apologies for not being able to rigorously answer any of your questions, or even show how you obtain these functions, but hopefully this is of some use to you.

Oldham, Keith B.; Myland, Jan; Spanier, Jerome, An atlas of functions. With Equator, the atlas function calculator. With CD-ROM, New York, NY: Springer (ISBN 978-0-387-48806-6/hbk). xi, 748 p. (2008). ZBL1167.65001.