Oh! Matryoshka!

Given polynomial f(x) = a_0 + a_1 x+a_2 x^2 + ... + a_{n-1}x^{n-1}+a_n x^n, we wish to evaluate integral

\int \frac{f(x)}{(x-a)^p}\;dx, \quad p \in N^+\quad\quad\quad(1)

When p = 1,

\int \frac{f(x)}{x-a} \;dx= \int \frac{f(x)-f(a)+f(a)}{x-a}\;dx

= \int \frac{f(x)-f(a)}{x-a}\;dx + \int \frac{f(a)}{x-a}\;dx

=\int \frac{f(x)-f(a)}{x-a}\;dx + f(a)\cdot \log(x-a).

Since

f(x) = a_0 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + a_n x^n

and

f(a) = a_0 + a_1 a + a_2 a^2 + ... + a_{n-1}a^{n-1} + a_n a^n

It follows that

f(x)-f(a) = a_1(x-a) + a_2(x^2-a^2) + ... + a_{n-1}(x^{n-1}-a^{n-1}) + a_n (x^n-a^n).

That is

f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x^k-a^k)

By the fact (see “Every dog has its day“) that

x^k-a^k =(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1},

we have

f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}=(x-a)\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})

or,

\frac{f(x)-f(a)}{x-a}=  \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\quad\quad\quad(2)

Hence,

\int\frac{f(x)}{x-a}\;dx = \int \sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\;dx + f(a)\log(x-a)

=\sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}\int x^{k-i}a^{i-1}\; dx)+ f(a)\log(x-a)

i.e.,

\int \frac{f(x)}{x-a} = \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a)

Let us now consider the case when p>1:

\int \frac{f(x)}{(x-a)^p}\; dx

=\int \frac{f(x)-f(a)+f(a)}{(x-a)^p}\;dx

=\int  \frac{f(x)-f(a)}{(x-a)^p} + \frac{f(a)}{(x-a)^p}\;dx

=\int \frac{f(x)-f(a)}{(x-a)}\cdot\frac{1}{(x-a)^{p-1}} + \frac{f(a)}{(x-a)^p}\;dx

= \int \frac{f(x)-f(a)}{x-a}\cdot\frac{1}{(x-a)^{p-1}}\;dx + \int\frac{f(a)}{(x-a)^p}\; dx

\overset{(2)}{=}\int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}

where

g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}), a polynomial of order n-1.

What emerges from the two cases of p is a recursive algorithm for evaluating (1):

Given polynomial f(x) = \sum\limits_{k=0}^{n} a_k x^k,

\int \frac{f(x)}{(x-a)^p} \;dx, \; p \in N^+= \begin{cases}p=1: \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a) \\p>1:  \int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}, \\ \quad\quad\quad g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}). \end{cases}


Exercise-1 Optimize the above recursive algorithm (hint: examine how it handles the case when f(x)=0)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s