1
$\begingroup$

I'm supposed to evaluate the following integral by decomposing the integrand into a sum of partial fractions. Note, this question isn't about evaluating.

$$\int\frac{6x+6}{(x^2+1)(x-1)^3}dx$$

From my understanding the numerator should be equated to the following form:

$$\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}$$

I know how to evaluate for constants $A$ through $E$, but I'm still unsure as to how the above form was equated to the integrand. If someone could help clarify that'd be great.

$\endgroup$
0
2
$\begingroup$

By setting $f(x)=\frac{6x+6}{(x^2+1)(x-1)^3}$ we have: $$ E = \lim_{x\to 1}(x-1)^3 f(x) = \lim_{x\to 1}\frac{6x+6}{x^2+1} = 6 \tag{E} $$ $$ D = \lim_{x\to 1}\frac{d}{dx}(x-1)^3 f(x) = -\lim_{x\to 1}\frac{6(x^2+2x-1)}{(x^2+1)^2}=-3\tag{D} $$ $$ C = \frac{1}{2}\lim_{x\to 1}\frac{d^2}{dx^2}(x-1)^3 f(x) = 6\lim_{x\to 1}\frac{(x-1)(1+4x+x^2)}{(1+x^2)^3}=0 \tag{C} $$ and $$ f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3} = \frac{3}{1+x^2}\tag{A,B}$$ so $(A,B,C,D,E)=\color{red}{(0,3,0,-3,6)}$. This method clearly shows the strong relation between partial fraction decomposition and the residue theorem. The reason for such a decomposition to hold is that $f(x)$ is a meromorphic function with a triple pole at $x=1$ and simple poles at $x=\pm i$. If we remove from $f(x)$ the "polar part" given by the triple pole at $x=1$, i.e. if we consider $$g(x)=f(x)-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$ we know in advance that such a meromorphic function is regular in a neighbourhood of $x=1$ and has two simple poles at $x=\pm i$. By removing from $g(x)$ the "polar part" associated with such poles, i.e. by subtracting from $g(x)$ something of the form $\frac{Ax+B}{x^2+1}$, we get that $$p(x)=f(x)-\frac{Ax+B}{x^2+1}-\frac{C}{x-1}-\frac{D}{(x-1)^2}-\frac{E}{(x-1)^3}$$ is an entire function and a rational function, i.e. a polynomial. By inspecting the removed part and the degrees of the polynomials whose ratio defines $f(x)$, we may also easily get what the degree of $p(x)$ has to be, and deduce $$f(x)=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}+\frac{E}{(x-1)^3}.$$ $p(x)\equiv 0$ also follows from computing $\lim_{x\to +\infty}f(x)$ and $\lim_{x\to +\infty}$ of the removed part.

$\endgroup$
2
  • $\begingroup$ @OP's probably not asking how to compute these but rather why can we be sure that the integrand decomposes like that $\endgroup$ – John Doe Oct 12 '16 at 19:32
  • $\begingroup$ @j___d: all right, updating. $\endgroup$ – Jack D'Aurizio Oct 12 '16 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.