Working backward from Area Under Curve for $y=\frac{1}{x^c}$ to calculate $"c"$?

Question

I am working on a synthesizer project and have reached a point that I am stumped on.

I am in this part trying to work from a basic curve of $y=\frac{1}{x^c}$ (where $x≥1$): Graph of $y=\frac{1}{x^c}$

As I understand, area under the curve between $x=1$ and $x=100,000,000$ (ie. more than I need for a rough approximation) would be: Formula of Integral (Symbolab Integral calculator link)

So essentially,

$$AUC = \frac{100,000,000^{(-c+1)} - 1}{-c+1}$$

Can this equation be rearranged to roughly approximate for "c="?

I would like to be able to give an "AUC" level between an arbitrary range, and from each AUC level given, calculate approximately "c". This $"c"$ will then be put back into the original $y=\frac{1}{x^c}$ equation. The shape of the $y=\frac{1}{x^c}$ curve will then dictate a part of the synth's sound.

The point is to be able to adjust the $y=\frac{1}{x^c}$ curve with a knob based on "AUC" rather than directly based on $"c"$, as "AUC" makes more sense in this case.

Is this possible to do even approximately in this way?

Thanks a lot for any help.

Answer 1

Clearly, your reduced expression is invalid, if, $c = 1$.

And also, if, you are interested in $c = 1$, then, $\int_{1}^{M} \frac{1}{x} = \ln M$ ($M$ is arbitrarily large number you have defined.) Therefore, if, your $AUC \approx \ln M$ then $c \approx 1$.

Now suppose, $M^{-c + 1} \gg 1$. Then, $(1 - c) \ln M \gg 0$, or, $c \ll 1$ (because, $\ln M > 0$). Which reduces the approximation to, $AUC \approx M$. Therefore, if your $AUC \approx M$ then, $c \ll 1$

Again suppose, $M^{-c + 1} \ll 1$. Then, $(1 - c) \ln M \ll 0$, or, $c \gg 1$ (because, $\ln M > 0$). Which reduces the approximation to, $AUC \approx \frac{1}{c}$. Or, $c \approx \frac{1}{AUC}$.

Answer 2

Let $1-c=x$, $AUC=y$, $k=10^8$, then $$x+\frac1{y}=\frac{k^x}{y}$$ $$k^{1/y}(x+\frac1y)=\frac{k^{x+1/y}}{y}$$

Let $z=x+\frac1y$, $$-(\ln k)ue^{-u\ln k}=\frac{k^{-1/y}}{y}(-\ln k)$$ $$-u\ln k=W(-\frac{k^{-1/y}}{y}\ln k)$$

Then $x$ can be solved easily.

You should first calculate the argument inside the W function, then see if it is large or small. There are good approximations for the W function at large or small arguments. For more information, see the Wikipedia page of Lambert W function.