Improper Integrals

Introduction 5.3.1  
It is a very natural question to ask if the area under the graph of

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{1}{x^2}$$ (5.3.1)

from $1$ to $\infty$ is finite or not. Looking at the graph doesn't realy help all that much:

200pt

Clearly $\frac{1}{x^2}$ goes to 0 as $x$ gets larger, but we kind of have a situation where we have to measure the area of a surface that is infinitely long and infinitely thin, so which one wins? Sometimes the area will turn out to be infinite, and sometimes it will turn out to be finite. In this section we will develop a way to decide in which case we are and what the final answer to this question is.

Definition 5.3.2  

• The improper integral of a function over an infinite interval $[a,+\infty[$ is defined as

  $$\int_a^{+\infty} f(x)dx = \lim_{n \to +\infty} \int_a^n f(x)dx$$ (5.3.2)

  
  
  if the series on the right hand side exists and converges. Otherwise, the improper integral is undefined. This case is also referred to as divergent.
• Of course, we have the same reasoning for improper integrals over the infinite interval $]-\infty,a]$:

  $$\int_{-\infty}^a f(x)dx = \lim_{n \to \infty}\int_{-n}^a f(x)dx$$ (5.3.3)

  
  

• For integrals over the entire real axis, we then use the following:

  $$\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^a f(x)dx + \int_a^{+\infty} f(x)dx$$ (5.3.4)

  
  
  for any $a$.

Examples 5.3.3  

• Our original example (5.3.1) turns out to be a convergent improper integral:

  $$\int_1^{+\infty} \frac{dx}{x^2} = \lim_{n \to \infty}\int_1^{+\in... ...o \infty}\left[-\frac{1}{x}\right]_1^n = 1 - \lim_{n \to \infty}\frac{1}{n} = 1$$ (5.3.5)

  
  

• On the other hand, $f(x)=\frac{1}{\sqrt{x}}$ integrated over $[1,+\infty[$ will turn out to diverge:

  $$\int_1^{+\infty} \frac{dx}{\sqrt{x}} = \lim_{n \to \infty}\int_1^... ...\lim_{n \to \infty}[2\sqrt{x}]_1^n = \lim_{n \to \infty}2\sqrt{n} - 2 = +\infty$$ (5.3.6)

  
  

• It is natural to wonder where the boundary lies between convergent and divergent. For the functions $f(x)=\frac{1}{x^p}$ this question can be answered very precisely:

  $$\int_1^{+\infty} \frac{dx}{x^p} = \begin{cases}\frac{1}{p-1} & \text{if $p>1$} \\ +\infty & \text{if $p \leq 1$} \end{cases}$$ (5.3.7)

  
  
  The details are left to the industrious reader. No other method is used than for the calculation of both previous improper integrals.

Definition 5.3.4  

• Assume that we have a function $f$ and a point $m$ such that $f(m)=\pm\infty$. Then the improper integral of $f$ over the interval $[a,m]$ is defined as

  $$\int_a^m f(x)dx = \lim_{t \to m^-} \int_a^t f(x)dx$$ (5.3.8)

  
  
  if the series on the right hand side exists and converges. Otherwise the improper integral over $[a,m]$ is undefined. This case is often referred to as divergent.
• The same reasoning holds if our troublemaker $m$ is at the beginning of the interval:

  $$\int_m^b f(x)dx = \lim_{t \to m^+} \int_t^b f(x)dx$$ (5.3.9)

  
  
  Again, the improper integral is well defined if the series on the right hand side exists and converges. Otherwise the improper integral over $[m,b]$ is undefined.
• As is to be expected, taking the integral over an interval $[a,b]$ where there is a point $m$ such that $f(m)=\pm\infty$ is defined as:

  $$\int_a^b f(x)dx = \int_a^m f(x)dx + \int_m^b f(x)dx$$ (5.3.10)

  
  
  This integral is only defined if both integrals on the right hand side exist.

Examples 5.3.5  

• Our original example (5.3.1) turns out to be a divergent over $[0,1]$:

  $$\int_0^{1} \frac{dx}{x^2} = \lim_{t \to 0^+} \int_t^{1} \frac{dx}... ...0^+} \left[-\frac{1}{x}\right]_t^1 = \lim_{t \to 0^+} \frac{1}{t} - 1 = +\infty$$ (5.3.11)

  
  

• On the other hand, $f(x)=\frac{1}{\sqrt{x}}$ integrated over $[1,+\infty[$ will turn out to converge:

  $$\int_0^{1} \frac{dx}{\sqrt{x}} = \lim_{t \to 0^+} \int_t^{1} \fra... ...lim_{t \to 0^+} \left[2\sqrt{x}\right]_t^1 = 1 - \lim_{t \to 0^+} 2\sqrt{t} = 1$$ (5.3.12)

  
  

• This seems like the mirror image of (5.3.3). So, one wonders, if this is always the case for functions of the form $f(x)=\frac{1}{x^p}$. And indeed:

  $$\int_0^1 \frac{dx}{x^p} = \begin{cases}\frac{1}{p-1} & \text{if $p < 1$} \\ +\infty & \text{if $p \geq 1$} \end{cases}$$ (5.3.13)

  
  
  Again, the details are left to the industrious reader.

Remark 5.3.6  
It is of course not the case that only rational functions have improper integrals. In fact, improper integral show up whenever an asymptote shows up. Look e.g.:

$$\int_0^{\frac{\pi}{2}} \tan(x) dx = \int_0^{\frac{\pi}{2}} \frac{\sin(x)}{\cos(x)} dx \cr$$ (5.3.14)