009B Sample Final 2, Problem 7

Evaluate the following integrals or show that they are divergent:

(a)  ${\displaystyle \int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}$

(b)  ${\displaystyle \int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx}$

Foundations:
1. How could you write   ${\displaystyle \int _{0}^{\infty }f(x)~dx}$ so that you can integrate?

You can write   ${\displaystyle \int _{0}^{\infty }f(x)~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}f(x)~dx.}$

2. How could you write   ${\displaystyle \int _{0}^{1}{\frac {1}{x}}~dx?}$

The problem is that  ${\displaystyle {\frac {1}{x}}}$  is not continuous at  ${\displaystyle x=0.}$

So, you can write  ${\displaystyle \int _{0}^{1}{\frac {1}{x}}~dx=\lim _{a\rightarrow 0}\int _{a}^{1}{\frac {1}{x}}~dx.}$

Solution:

(a)

Step 1:
First, we write
${\displaystyle \int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx=\lim _{a\rightarrow \infty }\int _{1}^{a}{\frac {\ln x}{x^{4}}}~dx.}$
Now, we use integration by parts.
Let  ${\displaystyle u=\ln x}$  and  ${\displaystyle dv={\frac {1}{x^{4}}}dx.}$
Then,  ${\displaystyle du={\frac {1}{x}}dx}$  and  ${\displaystyle v={\frac {1}{-3x^{3}}}.}$
Using integration by parts, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln x}{-3x^{3}}}{\bigg |}_{1}^{a}+\int _{1}^{a}{\frac {1}{3x^{4}}}~dx}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln x}{-3x^{3}}}-{\frac {1}{9x^{3}}}{\bigg |}_{1}^{a}.}\end{array}}}$
Step 2:
Now, using L'Hopital's Rule, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln a}{-3a^{3}}}-{\frac {1}{9a^{3}}}-{\bigg (}{\frac {\ln 1}{-3}}-{\frac {1}{9}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln(a)}{-3a^{3}}}+0+0+{\frac {1}{9}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln(x)}{-3x^{3}}}+{\frac {1}{9}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{-9x^{2}}}+{\frac {1}{9}}}\\&&\\&=&\displaystyle {{\frac {1}{9}}.}\end{array}}}$

(b)

Step 1:
First, we write
${\displaystyle \int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx=\lim _{a\rightarrow 0^{+}}\int _{a}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx.}$
Now, we use integration by parts.
Let  ${\displaystyle u=3\ln x}$  and  ${\displaystyle dv={\frac {1}{\sqrt {x}}}dx.}$
Then,  ${\displaystyle du={\frac {3}{x}}dx}$  and  ${\displaystyle v=2{\sqrt {x}}.}$
Using integration by parts, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx}&=&\displaystyle {\lim _{a\rightarrow 0^{+}}(3\ln x)(2{\sqrt {x}}){\bigg |}_{a}^{1}-\int _{a}^{1}{\frac {6}{\sqrt {x}}}~dx}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 0^{+}}6{\sqrt {x}}\ln(x)-12{\sqrt {x}}{\bigg |}_{a}^{1}.}\end{array}}}$
Step 2:
Now, using L'Hopital's Rule, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx}&=&\displaystyle {\lim _{a\rightarrow 0^{+}}(6{\sqrt {1}}\ln(1)-12{\sqrt {1}})-(6{\sqrt {a}}\ln(a)-12{\sqrt {a}})}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 0^{+}}-12-6{\sqrt {a}}\ln(a)+12{\sqrt {a}}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 0^{+}}-12-6{\sqrt {a}}\ln(a)+0}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0^{+}}-12-6{\sqrt {x}}\ln(x)}\\&&\\&=&\displaystyle {-12-\lim _{x\rightarrow 0^{+}}{\frac {6\ln(x)}{\frac {1}{\sqrt {x}}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {-12-\lim _{x\rightarrow 0^{+}}{\frac {\frac {6}{x}}{-{\frac {1}{2x^{3/2}}}}}}\\&&\\&=&\displaystyle {-12+\lim _{x\rightarrow 0^{+}}12{\sqrt {x}}}\\&&\\&=&\displaystyle {-12.}\end{array}}}$

(a)    ${\displaystyle {\frac {1}{9}}}$
(b)    ${\displaystyle -12}$