complex analysis - Curve with one negative and one positive sine-like impulse

Is it possible to create an algebraic function that is smooth and continuous (i.e., a function in the form $f(x)$ using algebraic functions, with no curly braces that stipulate different behaviour for different domains of $x$, and not piecewise-continuous, smoothly continuous even in the limit) that has the following properties:

LIST EDITED FOR CLARITY:

For positive integer $a:

• In the range $x$ equals $-\infty$ to $-b$, the function yields $0$


• In the range $x$ equals $-b$ to $-a$, the function yields a single negative 'pulse' of some form with only one local minimum (an example would be $-\cos\bigl(\frac{1}{b-a}\pi (x+\frac{b-a}{2})\bigr)-1$.)



• In the range $x$ equals $-a$ to $a$, the function yields $0$


• In the range $x$ equals $a$ to $b$, the function yields a single positive 'pulse' of some form with only one local maximum (an example would be $\cos\bigl(\frac{1}{b-a}\pi (x+\frac{b-a}{2})\bigr)+1$.)


• In the range $x$ equals $b$ to $\infty$, the function yields $0$

END OF EDIT

The result would be a curve with vaguely sinusoidal behaviour: a single (i.e., non-oscillating) negative pulse between $x=(-b,a)$ and a positive pulse between $x=(a,b)$, and for other values of $x$, the curve either oscillates or flat-lines, but integrates to $0$ across the ranges specified.

I suspect this requires some form of Fourier analysis, and I am sure that the solution will require complex analysis, which is fine.

I understand (to some degree!) complex numbers. But Fourier analysis is something I simply don't know how to do. I just want to find such a curve so I can play with it :-)

$\color{red}{\text{Edits:}}$

An example that doesn't quite work is $\DeclareMathOperator{\sinc}{sinc}(\pi (x-n))-\DeclareMathOperator{\sinc}{sinc}(\pi (x+n))$. This creates a curve that could easily be adjusted to create an area of $1$ between points $a$ and $b$ (and the negative equivalent), but it's hard to hit the points $a$ and $b$ precisely (needs to cross $y=0$ at precisely those points), and the integral of the other ranges are not zero.

If the answer is that it's impossible, that's useful info too.

Note that the function should be smoothly continuous, at least in the limit ($\DeclareMathOperator{\sinc}{sinc}$ is an example).

If behaviour is different in the complex plane, that's also fine. So long as it's continuous, and the real part has the behaviour above.

Answer

By substitution $x = At + B$, you can shift and compress/expand the location of the bumps in this function to be at any two points you like. The function itself seems as if it captures what you're looking for:

$$
y = \frac{\arctan(t)}{1 + t^2}
$$

Here's a plot via desmos.com:

====

Oops. Now that I look at your question more clearly, I see that I've produced something more interesting than what you asked for. I'm gonna leave it there anyhow.

There's a small problem because you say that the function should be "odd", but I think that you mean that it should be symmetric (in an "even" sense) around the point $\frac{a+b}{2}$; otherwise most of what you write is inconsistent in the case $a = -1, b = 1$.

The way I derived my example was to take the derivative of $arctan^2(t)$ (and then drop a factor of $2$ to simplify). Starting with arctan-like stuff is generally a good way to work things like this.

But in this case, it's even easier: take a look at the function $y = \frac{sin(x)}{x}$, where you have to define $y = 1$ when $x = 0$; this (or a scaled version of it) is commonly given then name $sinc$, and it has an everywhere convergent power series without any "cases" in it, namely
$$
sinc(x) = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \ldots
$$
This function, or better,
$$
f(x) = \frac{sin(\pi x)}{\pi x}
$$
has a bump at the origin, and minima at either side of the origin, around $x = \pm 1.43$. Let's call that $x = \pm c$. Then you can arrange, by multiplying by a small constant, to make $\int_{-c}{c} kf(x) dx = 1$. The problem is that the integral over the "tails" won't be $0$. But it'll be pretty close to zero, and you can make a small adjustment to make it work: you just add in a little bit of $\frac{1}{1+x^2}$. So what I'm claiming is that if you look at
$$

u(x) = A\left( \frac{sin(\pi x)}{\pi x} + \frac{B}{1+x^2} \right),
$$
you'll have, for $A$ somewhere near $2$, and $B$ around $0.2$, a function that has the properties you want. Using the intermediate value theorem, you can prove that there are values for $A$ and $B$ that achieve your goals (assuming that $a$ and $b$ are approximately $\pm 1.5$). Once you achieve this, you can scale vertically and horizontally to make it work for $a = -1, b = +1$, and then scale and translate to make it work for any $a$ and $b$.