# B Integrating to infinity issue

1. May 29, 2017

### Saracen Rue

Say we have two functions; $f\left(x\right)=\frac{1}{e^x-1}$ and $g\left(x\right)=\ln \left(\frac{1}{x}+1\right)$. Let us find the limit of both functions as x approaches infinity;
$\lim_{x \rightarrow \infty} {f(x)} = \frac{1}{e^\infty-1} = \frac{1}{\infty} = 0$ Therefore as $x \rightarrow \infty$, $f(x) \rightarrow 0$.
$\lim_{x \rightarrow \infty} {g(x)} = \ln \left(\frac{1}{\infty}+1\right) = \ln \left(0+1\right) = 0$ Therefore as $x \rightarrow \infty$, $g(x) \rightarrow 0$.

Now, when we integrate $f(x)$ over the domain $[1, \infty)$, it works perfectly fine and has a result of
0.45868. This makes sense because the area enclosed under the graph is constantly getting smaller due to $f(x) \rightarrow 0$.

However, when I try to integrate $g(x)$ over the same domain my calculator wont give me an answer. $g(x)$ approaches 0 just as $f(x)$ does, meaning that the enclosed area should increasingly get smaller also until it reaches the point where the area incriminates being added become negligible. So why can't I get a numerical answer while integrating $g(x)$?

2. May 30, 2017

### Orodruin

Staff Emeritus
It is not sufficient that the function goes to zero. It must go to zero sufficiently fast.

3. May 30, 2017

### Saracen Rue

Could you please elaborate on what you mean? How could the function not being 'sufficiently fast' change the fact that it approaches zero as x approaches infinity? Integrating gives the area under the graph, and we know that as x approaches infinity the function approaches 0. This means there has to be a point in which the 'height' of the function above the x-axis becomes negligible (that is equal to zero) and therefore the area under the the graph must stop having incriminates added onto it.

4. May 30, 2017

### PeroK

A function tending to 0 is not the same as a function becoming 0. Your functions never actually reach 0, hence there are always"increments".

Look up the series $\Sigma 1/n$ for a clue as to what is going on here.

5. May 30, 2017

### Orodruin

Staff Emeritus
No it doesn't. This argument is not rigorous. Even if the function value decreases towards zero, it does not mean that the area will. Consider the piecewise constant function that takes the value $1/2^n$ between $x = 2^n$ and $x = 2^{n+1}$. Clearly, this function converges to zero as $x \to \infty$. However, integrating the function from $2^n$ to $2^{n+1}$ gives $2^n/2^n = 1$. Therefore, the integral from 1 to $2^n$ is equal to $n$ and therefore the integral diverges when taking the upper limit to infinity. When you compute an area, it is not only a matter of "height", you also need to consider the width of the region and your region is infinite.

6. May 30, 2017

### Saracen Rue

I'm sorry for being so dense everyone, I haven't even heard of things like piecewise constants. I guess I'm just beyond my current level of understanding here.

There is one thing I stumbled across though - the derivative function of $g\left(x\right)=\ln \left(\frac{1}{x}+1\right)$ is $g'(x)=\frac{-1}{x(x+1)}$. The squared factor on the denominator implies that; as x approaches infinity, the gradient function approaches 0 at a faster rate than function $g(x)$ approaches 0. Could you then extrapolate from this and state that $g(x)$ approaches a non-zero constant value before approaching 0, and thus the area under the graph will be infinite?

7. May 30, 2017

### Orodruin

Staff Emeritus
No.

Think of it this way, if the integrant approached a constant non-zero value $c_0$ and the integral upper boundary was set to $x = N$, then the integral would grow as $\sim c_0 N$ as $N \to \infty$. However, if the integrand goes to zero, the integral will increase slower. However, there are many possible functions that grow slower than $N$ as $N \to \infty$ but that still tend to infinity, $\log(N)$ being one such example.