MATH 2133 - SPRING 2001 - REVIEW FOR POST EXAM III MATERIAL

Three topics were covered after the third exam: partial fractions, tables of integrals, and improper integrals.

PARTIAL FRACTIONS

In this section we consider ${\displaystyle\int \frac{p(x)}{q(x)}\,dx}$, where p(x) and q(x) are polynomials.

1.

The highest power of x in a polynomial is called the degree of the polynomial. For example, x³+3x²-2x+5 has degree 3. If the degree of p(x) is greater than or equal to the degree of q(x), then we first use long division to write ${\displaystyle\frac{p(x)}{q(x)}=s(x)+\frac{r(x)}{q(x)}}$, where s(x) and r(x) are polynomials and the degree of r(x) is less than the degree of q(x). For example ${\displaystyle\frac{x^3+3x^2-2x+5}{x^2+1}=x+3+\frac{-3x+2}{x^2+1}}$. We know how to integrate polynomials, so we have reduced the problem to the case when the degree of the numerator is less than that of the denominator. So from now on we will assume that the degree of p(x) is less than the degree of q(x).

2.

In this situation it turns out that ${\displaystyle\frac{p(x)}{q(x)}}$ can always be written as a sum of special things called partial fractions. There are four kinds of partial fractions:

(a)

${\displaystyle\frac{A}{ax+b}}$, where $a\neq 0$

(b)

${\displaystyle\frac{A}{(ax+b)^n}}$, where $a\neq 0$ and n>1

(c)

${\displaystyle\frac{Bx+C}{ax^2+bx+c}}$, where b²-4ac<0

(d)

${\displaystyle\frac{Bx+C}{(ax^2+bx+c)^n}}$, where b²-4ac<0 and n>1

The condition b²-4ac<0 is equivalent to ax²+bx+c being ``irreducible'' in the sense that it cannot be factored into a product $(\alpha x+\beta)(\gamma x+\delta)$ where $\alpha$, $\beta$, $\gamma$, and $\delta$ are all real numbers. For example, x²+5x+6 has b²-4ac=25-24=1>0 and factors as (x+2)(x+3), and x²+6x+9 has b²-4ac=36-36=0 and factors as (x+2)². On the other hand x²+2x+5 has b²-4ac=4-25=-21<0 and does not factor as a product of this sort; it is irreducible. So x²+2x+5 is allowed as the denominator of a partial fraction of the third or fourth types listed above. x²+5x+6 and x²+6x+9 are not. There are two issues now to consider: How do we express ${\displaystyle\frac{p(x)}{q(x)}}$ as a sum of partial fractions? How do we integrate the partial fractions?

3.

There is a hard theorem in algebra which says that every polynomial can be factored into a product of linear factors ax+b and irreducible quadratic factors ax²+bx+c. Suppose this has been done for q(x). We gather like factors together into distinct powers. For example we might have q(x)=(x+2)x³(x²+2x+5)(x²+4)². For each factor of the form (ax+b)^m we take the sum of m partial fractions

$$\frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+\cdots+\frac{A_n}{(ax+b)^m}.$$

For each factor of the form (ax²+bx+c)^m we take the sum of m partial fractions

$$\frac{B_1x+C}{ax^2+bx+c}+\frac{B_2x+C_2}{(ax^2+bx+c)^2}+\cdots+ \frac{B_nx+C_n}{(ax^2+bx+c)^m}.$$

Let's look at this for ${\displaystyle\frac{p(x)}{q(x)}= \frac{x^4-3x^2+2}{(x+2)x^3(x^2+2x+5)(x^2+4)^2}}$. Note that since p(x) has degree 4 and q(x) has degree 10 we do not have to divide first. Note also that instead of using subscripts, it is customary to just go through the letters of the alphabet as far as needed.

$$\frac{x^4-3x^2+2}{(x+2)x^3(x^2+2x+5)(x^2+4)^2}= \frac{A}{x+2... ...frac{Ex+F}{x^2+2x+5}+\frac{Gx+H}{x^2+4}+\frac{Ix+J}{(x^2+4)^2}.$$

4.

The next step is to find the coefficients A, B, etc. What you do is to add together all the terms on the right hand side, putting everything over a common denominator. This common denominator will be equal to q(x). (Remember that you multiply each numerator on the right by the parts of q(x) that it does not have in its own denominator.) Since the two denominators are both q(x) the two numerators must be equal, so we have that p(x) is equal to some polynomial with the A, B, etc. in it. The easiest way to find these coefficients is to plug values in for x to both sides (usually choosing values of x which make things vanish), thereby getting a system of equations in the coefficients, which we then solve. Let's look at this for a simpler example than the one above.

$$\frac{x^2+x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x... ...}+\frac{Cx^2}{x^2(x-1)}= \frac{Ax(x-1)+B(x-1)+Cx^2}{x^2(x-1)}.$$

So we have x²+x+1=Ax(x-1)+B(x-1)+Cx². Setting x=0 we get 1=-B, so B=-1. Setting x=1 we get 3=C. There are no other values of x which make things vanish so we just choose a convenient value, say x=2; this gives 7=2A+B+4C. Plugging in B=-1 and C=3 we get 7=2A-1+12, so 7=2A+11, -4=2A, -2=A. Thus we have

$$\frac{x^2+x+1}{x^2(x-1)}=-\frac{2}{x}-\frac{1}{x^2}+\frac{3}{x-1}.$$

5.

To integrate the partial fraction ${\displaystyle\frac{A}{ax+b}}$ we set u=ax+b; this turns the integral into a logarithmic integral.

6.

To integrate the partial fraction ${\displaystyle\frac{A}{(ax+b)^n}}$ where n>1 we again set u=ax+b, but this time we note that this gives a general power rule integral.

7.

Now consider the partial fraction ${\displaystyle\frac{Bx+C}{ax^2+bx+c}}$. We first rewrite this in the form ${\displaystyle\alpha\frac{2ax+b}{ax^2+bx+c}+ \beta\frac{1}{ax^2+bx+c}}$. Note that if we add these two expressions together we get a numerator of $\alpha(2ax+b)+\beta=2\alpha ax+(\alpha b+\beta)$; this is equal to Bx+C, so we have $2\alpha a=B$, giving ${\displaystyle\alpha=\frac{B}{2a}}$. then we can compute $\beta=C-\alpha b$. For example,

$$\frac{6x+7}{x^2+4x+5}=\alpha\frac{2x+4}{x^2+4x+5}+\beta\frac{1}{x^2+4x+5}= \frac{2\alpha x+4\alpha+\beta}{x^2+4x+5}.$$

So $2\alpha x+4\alpha+\beta=6x+7$, hence $2\alpha=6$, and thus $\alpha=3$, and $4\alpha+\beta=7$, so $12+\beta=7$, and thus $\beta=-5$. So we have

$$\frac{6x+7}{x^2+4x+5}=3\frac{2x+4}{x^2+4x+5}-5\frac{1}{x^2+4x+5}$$

${\displaystyle\int\frac{2ax+b}{ax^2+bx+c}\,dx}$ is integrated by setting u=ax²+bx+c; it is a logarithmic integral. ${\displaystyle\int\frac{dx}{ax^2+bx+c}}$ is integrated by completing the square; this turns it into an Arctan integral. In our example we have

$$\int\frac{6x+7}{x^2+4x+5}\,dx= 3\int\frac{2x+4}{x^2+4x+5}\,d... ...{x^2+4x+5}= 3\ln\vert x^2+4x+5\vert-5\,\mathrm{Arctan}(x+2)+C.$$

8.

We now consider the partial fraction ${\displaystyle\frac{Bx+C}{(ax^2+bx+c)^n}}$, where n>1. Again the idea is to write it as ${\displaystyle\alpha \frac{2ax+b}{(ax^2+bx+c)^n} + \beta\frac{1}{(ax^2+bx+c)^n}}$. This is done in the same way as above. ${\displaystyle\int\frac{2ax+b}{(ax^2+bx+c)^n}\,dx}$ is handled by the substitution u=ax²+bx+c; this time it gives a general power rule integral. ${\displaystyle\int\frac{1}{(ax^2+bx+c)^n}\,dx}$ is more troublesome. As before we complete the square and make a substitution to express it in terms of ${\displaystyle\int \frac{du}{(k^2+u^2)^n}}$. (The letter k is used here instead of a because we are already using a for something else.) This is a tough integral. Using a tricky integration by parts one can obtain a formula which expresses it in terms of ${\displaystyle\int \frac{du}{(k^2+u^2)^{(n-1)}}}$, so one can inductively work down to an integral one can do. However, the best way to do this is probably to use instead the substitution $u=k\tan\theta$; this will ultimately turn it into ${\displaystyle k^{1-2n}\int\cos^{2n-2}\theta\,d\theta}$, which is an integral we know how to do. On the final you will not need to do integrals of this type unless they occur in a problem that uses tables of integrals.

TABLES OF INTEGRALS

There are many standard forms for integrals that we have not seen in this course. Such things are collected into tables. One then tries to recognize that a given integral fits the form and to do the correct substitutions needed. For example one might be given ${\displaystyle\int\frac{dx}{(x^2+4x+13)^2}}$. In looking through a table one might see the formula

$$\int\frac{du}{(a^2+u^2)^2}=\frac{u}{2a^2(a^2+u^2)}+ \frac{1}{2a^2}\int\frac{du}{a^2+u^2}.$$

Note that with our given integral we can complete the square to get x²+4x+13=(x+2)²+3², which suggests u=x+2 and a=3. Then du=dx and we have

$$\int\frac{du}{(3^2+u^2)^2}=\frac{u}{18(3^2+u^2)}+ \frac{1}{1... ...frac{1}{3}\right) \mathrm{Arctan}\left(\frac{u}{3} \right)+C=$$

$$\frac{x+2}{18(x^2+4x+13)}+\frac{1}{54}\mathrm{Arctan}\left(\frac{x+2}{3}\right)+C.$$

IMPROPER INTEGRALS

An improper integral is a definite integral ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx}$ in which either $\alpha=-\infty$, $\beta=\infty$, or f(x) is not continuous, or some combination of these conditions. All of these conditions are handled by taking limits. If the relevant limits exist, the integral is said to converge; otherwise it diverges.

1.

${\displaystyle\int_{\alpha}^{\infty}f(x)\,dx=\lim_{b\rightarrow\infty} \int_{\alpha}^bf(x)\,dx}$ For example ${\displaystyle\int_1^{\infty}\frac{dx}{x^2}=\lim_{b\rightarrow\infty} \int_1^b... ...y}\frac{-1}{x}\right\vert _1^b= \lim_{b\rightarrow\infty}\frac{-1}{b}+1=0+1=1}$. On the other hand ${\displaystyle\int_1^{\infty}\frac{dx}{x}=\lim_{b\rightarrow\infty} \int_1^b\f... ...\left.\right\vert _1^b= \lim_{b\rightarrow\infty}\ln{b}-\ln 1=\infty-0=\infty}$, so this integral does not converge, i.e. it diverges. (Note: An integral may diverge without being infinite; see the book for an example.)

2.

${\displaystyle\int_{-\infty}^{\beta}f(x)\,dx=\lim_{a\rightarrow -\infty}\int_a^{\beta}f(x)\,dx}$.

3.

${\displaystyle\int_{-\infty}^{\infty}f(x)\,dx= \int_{-\infty}^{0}f(x)\,dx+\int_{0}^{\infty}f(x)\,dx}$.

4.

Suppose ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx}$ is given, and that f(x) is not continuous at $\beta$. Then we define ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx= \lim_{b\rightarrow \beta}\int_{\alpha}^bf(x)\,dx}$. For example, ${\displaystyle\int_0^1\frac{dx}{(1-x)^{1/2}}= \lim_{b\rightarrow 1}\int_0^b\fr... ...w 1}-2(1-x)^{1/2}\left.\right\vert _0^b= \lim_{b\rightarrow 1}-2(1-b)+2=0+2=2}$

5.

If f(x) is not continuous at $\alpha$ we define ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx=\lim_{a\rightarrow \alpha}\int_a^{\beta}f(x)\,dx}$.

6.

If f(x) is not continuous at either endpoint then we choose c between $\alpha$ and $\beta$ and define ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx= \int_{\alpha}^{c}f(x)\,dx+\int_{c}^{\beta}f(x)\,dx}$.

7.

If f(x) is not continuous at some point $\gamma$ between $\alpha$ and $\beta$, then we define ${\displaystyle\int_{\alpha}^{\beta}f(x)\,dx= \int_{\alpha}^{\gamma}f(x)\,dx+\int_{\gamma}^{\beta}f(x)\,dx}$.