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Analytic continuation and Regularization simplified.


Well-known member
MHB Math Helper
Jan 17, 2013
Hello MHB members

In this set of lectures we are going to explore the nice idea of analytic continuation and regularization of divergent series and integrals. Don't get panic ,the idea is so simple that you are actually using it without knowing. I'll try to make the tutorials as simple as possible so people with basic knowledge of complex variables will understand it. Basic knowledge of calculus will actually suffice. Even a high school student will be able to understand the concepts once introduced to calculus. My treatment of these concepts will not be rigorous actually it is for the sake of fun and learning something new and astonishing. There will not be lots of proofs of theorems or evaluations of complicated results but I will rather emphasize on the intuition behind these interesting ideas.


  • Basice knowledge of calculus of single variable (differentation and integration)
  • Basic knowledge of complex variables.
  • Basic knowledge of convergence tests of series and integrals.
  • Willing to learn.


Well-known member
MHB Math Helper
Jan 17, 2013
The intuition :

During the history of mathematics, mathematician have tried to extend their knowledge of things around them. They used Geometry and symbols to describe the objects around them. The simplest number system was Natural numbers. They used it to count real-live objects. As humans started to work heavily on agriculture, they were forced to use fractions as a simple extension of the old system. If we define fractions as a ratio of two natural numbers then it is an extension of the Natural numbers by putting the denominator equal to $1$. This is actually a simple example of extending a domain where the bigger domain contains the smaller domain. Remember that the structure of the smaller domain is conserved in the larger domain.
The idea of analytic continuation is as simple as extending functions of real variables to be defined on the complex domain $\mathbb{C}$. We clarify the idea of analytic continuation by putting some examples that are not neccessarily the application of analytic continuation but they help us understand the content in the next sections.

You have in a way or another used a similar idea to analytic continuation.Consider the following function
$$f(x)=\frac{x^2-1}{x-1}$$ then $f(1)$ has no meaning or does it ?
Actually by some algebraic manipulation -simply factorization- we get $f(x)=x+1$ so $f(1)=2$. Suppose that we didn't know that $x^2-1=(x-1)(x+1)$ then $f(1)=2$ has no sense for us. Simply we extended the function to have the value $f=2$ at $x=1$, not interesting enough?

Let us take another example, consider the function $f(x) = x \sin\left(\frac{1}{x} \right)$ then the limit $\lim_{ x \to 0}\, x \sin\left( \frac{1}{x}\right)=0$ which can be proved by the squeeze theorem,right ?

Now define the new function $g$ as

$$g(x) =
f(x) & \text{if } x \neq 0 \\
0 & \text{if } x = 0

Now, the function $f$ is continuous for all $x \neq 0$ what about the function $g$ ?
What we have actually done is that we have extended $f$ to a continuous function $g$ where $f\equiv g$ for $x \neq 0$. This is justified by $g(0)= \lim_{x \to 0}f(x)=0$ so $g$ is continuous on $\mathbb{R}$ while $f$ is continuous on $\mathbb{R}^*=\mathbb{R} /\{0\}$ but $g$ agrees with $f$ on $\mathbb{R}^*$.

Consider the function $f(n)=n!$ where $n \in \mathbb{N}$. You have surely recognized the factorial function as $n! = n(n-1) \cdots 3 \times 2\times 1 $. The mere existence of $n!$ is in combinatorics. Suppose that you have 5 books then in how many ways you can order these books in 5 cells ?
Then by simple logic , you put the first book in one of the cells so you have 5 choices then once one cell is reserved you have only 4 remaining cells so the second book has 4 empty cells to populate, hence it has 4 choices and so on. So we have $5\times 4 \times 3 \times 2=120$ different ways to order the books in the cells.
Now , having that definition in mind what is the factorial of $0$ or $f(0)=0!$?
How many ways to order $0$ books in $0$ cells ?!. Well, that has no meaning at least using our previous definition.

Define $f(n)= nf(n-1) \,\,\,n \in \mathbb{N} $ putting $n=1$ we have $f(1)=f(0)$ but what is $f(1)$ ? by our combinatorial definition it is $1!=1$ since there is only $1$ way to order one book in one cell , "just but it in the cell!". Fair enough ?
So we have $0! = 1$ which for some reason didn't make any sense in the previous paragraph!
Hence we define the extended factorial as $n!=n (n-1)!$ .
Well, believe it or not the factorial can be extended to factional numbers and even negative and complex numbers , which to be explained in the next sections.


\(\displaystyle \mathbb{N}=\{1,2,3, \cdots \}\)
\(\displaystyle - 1 \leq \sin \left( \frac{1}{x}\right) \leq 1 \,\,\, ,\forall \,\, x \neq 0\)