# Create 'smooth breakpoint function' by using integral?

Experts,

I am a biologist and thus my natural strength is not math, yet I´m quite okay with statistics. Now I am facing the problem that I have to find an unusual (?) mathematical solution for a function with certain properties.

In biology, there is something called the 'Species-Area Relationship' (SAR) which describes the increase of species numbers with an increase in area investigated. Usually, a power functions is best describing this. Now there is the special case of very small areas where vales are close to zero for a number of area sizes.

One can use piecewise regressions such as

y = f1 (x) = c + (x ≤ T) z1 x + (x > T) [(z1 – z2) T + z2 x],

y = number of species
x = Area
T = Breakpoint
z1, z2 = slopes of breakpoint functions on LHS and RHS of T


Sadly, the breakpoint function has an unrealistic "break" which is quite rough. Hence, I am trying to find a smooth version of this breakpoint function that connects the two breakpoint functions on the left and right handside with a smooth transition function.

Trying a logistic function it becomes already quite close. However, it is unrealistically bound between 0 and 1.

Now I am looking for an integral of this logistic function

z = f4’ (x) = z1 + (z2 - z1) (1 /(1 +exp (–k x)))


Then I only had to add a constanct 'c' to lift it beyond 1...and here is where I have to stop since my brain starts hurting.

Could anyone help me out and try to find the integral of the second function? Or give a hint how to achieve a smooth transition in the BP-function?

many thanks already for all grey matter turned into words,

best,

Jens

PS: I am not sure if this post would better fit to 'crossvalidated' list on stackexchange?

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I'm not quite sure that I understand the question, but if you're simply looking for $f_4(x)$ with $f_4'(x)$ given by your expression, you can find it by using $$\int \frac{1}{1+e^{-kx}} dx = \int \frac{e^{kx}}{e^{kx}+1}dx = \frac{1}{k} \int \frac{(e^{kx}+1)'}{e^{kx}+1} dx = \frac{1}{k} \ln(e^{kx}+1) + C.$$