MATH 152 Chapter 8.9

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Improper Integrals

2 types:

1. $\int _{a}^{b}{f\left(x\right)\,dx}$ — number $x$ on interval $[a,b]$ is not in the domain of $f$
2. $\int _{a}^{b}{f\left(x\right)\,dx}$ — $a=\infty$ and/or $b=\infty$

NOTE: If the limit is ±∞ or DNE, we say the integral diverges.

Example 1

By convention, integrals can be taken over finite intervals

${\begin{aligned}\int _{0}^{\infty }{{\frac {1}{(x+1)^{2}(x+2)}}\,dx}&=\lim _{a\to \infty }\int _{0}^{a}{{\frac {1}{(x+1)^{2}(x+2)}}\,dx}\\&=\lim _{a\to \infty }\int _{0}^{a}{{\frac {-1}{x+1}}+{\frac {1}{(x+1)^{2}}}+{\frac {1}{x+2}}\,dx}\\&=\lim _{a\to \infty }{\bigg (}-\ln {|x+1|}-{\frac {1}{x+1}}+\ln {|x+2|}{\bigg )}_{0}^{a}\\&=\lim _{a\to \infty }-\ln {|a+1|}-{\frac {1}{a+1}}+\ln {|a+2|}+1-\ln {2}\\&=\lim _{a\to \infty }-\ln {{\bigg (}{\frac {a+2}{a+1}}{\bigg )}}+1-\ln {2}\\&=1-\ln {2}\end{aligned}}$

The integral converges to $1-\ln {2}$

Example 2

${\begin{aligned}\int _{1}^{e}{{\frac {1}{x{\sqrt {\ln {x}}}}}\,dx}&=\lim _{a\to 1^{+}}\int _{a}^{e}{{\frac {1}{x{\sqrt {\ln {x}}}}}\,dx}\\&=\lim _{a\to 1^{+}}\int _{a}^{e}{{\frac {1}{\sqrt {u}}}\,dx}&({\text{Let }}u=\ln {x}\;\therefore \;du={\frac {dx}{x}})\\&=\lim _{a\to 1^{+}}\int _{a}^{e}{u^{-1/2}\,dx}\\&=\lim _{a\to 1^{+}}2(\ln {x})^{1/2}{\bigg |}_{a}^{e}&({\text{Substitute }}u=\ln {x}{\text{ back into equation}})\\&=\lim _{a\to 1^{+}}2{\sqrt {\ln {e}}}-2{\sqrt {\ln {a}}}\\&=2&(\ln {1^{+}}{\text{ approaches }}0)\end{aligned}}$

The integral converges to 2

Example 3

Find values $p$ where integral converges (b > 0)

${\begin{aligned}\int _{b}^{\infty }{{\frac {1}{x^{p}}}\,dx}&=\lim _{a\to \infty }\int _{b}^{a}{x^{-p}\,dx}\\&=\lim _{a\to \infty }{\begin{cases}-{\frac {1}{(p-1)x^{p-1}}}{\big |}_{b}^{a}&(p>1;converges)\\\ln {x}{\big |}_{b}^{a}&(p=1;diverges)\\{\frac {1}{p+1}}x^{p+1}{\big |}_{b}^{a}&(p<1;diverges)\end{cases}}\\\therefore \;\int _{b}^{\infty }{{\frac {1}{x^{p}}}\,dx}&{\text{ converges iff }}p>1\end{aligned}}$

Example 4

Find values $p$ where $\int _{0}^{1}{x^{p}\ln {x}\ dx}$ converges and solve.

${\begin{aligned}\int _{0}^{1}x^{p}\ln x\ dx=&{\frac {1}{p+1}}\left[x^{p+1}\ln x-{\frac {1}{p+1}}x^{p+1}\right]_{0}^{1}&p\neq -1\\=&{\frac {1}{p+1}}\lim _{a\to 0^{+}}\left[{\frac {\ln x-{\frac {1}{p+1}}}{\frac {1}{x^{p+1}}}}\right]_{a}^{1}&({\tfrac {\infty }{\infty }}\ {\mbox{LHospital}})\\=&{\frac {1}{p+1}}\lim _{a\to 0^{+}}\left[{\frac {\frac {1}{x}}{\frac {-\left(p+1\right)}{x^{p+2}}}}\right]_{a}^{1}\\=&{\frac {1}{p+1}}\lim _{a\to 0^{+}}\left[{\frac {1}{x}}\ {\frac {x^{p+2}}{-\left(p+1\right)}}\right]_{a}^{1}\\=&-{\frac {1}{\left(p+1\right)^{2}}}\lim _{a\to 0^{+}}\left[x^{p+1}\right]_{a}^{1}\\=&-{\frac {1}{\left(p+1\right)^{2}}}\lim _{a\to 0^{+}}1-a^{p+1}\\=&-{\frac {1}{\left(p+1\right)^{2}}}&\left(p>-1\right)\end{aligned}}$

Wednesday, October 6, 2010

Note: Split up type 2 integrals over asymptotic regions

$\int _{-2}^{1}{{\frac {1}{x^{2}}}\,dx}=\int _{-2}^{0}{{\frac {1}{x^{2}}}\,dx}+\int _{0}^{1}{{\frac {1}{x^{2}}}\,dx}$

Comparison Theorem

Suppose $f$ and $g$ are continuous and $0<=f(x)<=g(x)$ for $x>N$

If $\int _{N}^{\infty }{g(x)\,dx}$ converges, then $\int _{N}^{\infty }{f(x)\,dx}$ also converges

If $\int _{N}^{\infty }{f(x)\,dx}$ diverges, then $\int _{N}^{\infty }{g(x)\,dx}$ also diverges

If $\int _{N}^{\infty }{g(x)\,dx}$ diverges or $\int _{N}^{\infty }{f(x)\,dx}$ converges, we know nothing about the other function</math>

Example 5

Method 1

${\begin{aligned}\int _{3}^{\infty }{{\frac {x+1}{x^{2}-4}}\,dx}&=\lim _{a\to \infty }\int _{3}^{a}{{\frac {3/4}{x-2}}+{\frac {1/4}{x+2}}\,dx}\\&=\lim _{a\to \infty }{\frac {3}{4}}\ln {|x-2|}+{\frac {1}{4}}\ln {|x+2|}{\bigg |}_{3}^{a}\\&=\lim _{a\to \infty }{\frac {3}{4}}\ln {|a-2|}+{\frac {1}{4}}\ln {|a+2|}-{\frac {3}{4}}\ln {1}-{\frac {1}{4}}\ln {5}\\&=\lim _{a\to \infty }\ln {\bigg (}(a-2)^{3/4}-(a+2)^{1/4}{\bigg )}-{\frac {1}{4}}\ln {5}&=\ln {\infty }-{\frac {1}{4}}\ln {5}\;\therefore \;{\text{it diverges}}\end{aligned}}$

Method 2

Compare function with dominating terms: $\int _{3}^{\infty }{{\frac {x+1}{x^{2}-4}}\,dx}\to \int _{3}^{\infty }{{\frac {x}{x^{2}}}\,dx}\to \int _{3}^{\infty }{{\frac {1}{x^{(1)}}}\,dx}$

The simplified function diverges, therefore the original function diverges:

Proof: ${\frac {x+1}{x^{2}-4}}>{\frac {1}{x}}\quad x\in [3,\infty )$

$\therefore \int _{3}^{\infty }{{\frac {x+1}{x^{2}-4}}\,dx}{\text{ diverges by comparison with }}\int _{3}^{\infty }{{\frac {1}{x}}\,dx}$