009B Sample Final 2, Problem 7

Evaluate the following integrals or show that they are divergent:

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

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

Foundations:
1. How could you write   $\int _{0}^{\infty }f(x)~dx$ so that you can integrate?
You can write   $\int _{0}^{\infty }f(x)~dx=\lim _{a\rightarrow \infty }\int _{0}^{a}f(x)~dx.$
2. How could you write   $\int _{0}^{1}{\frac {1}{x}}~dx?$
The problem is that  ${\frac {1}{x}}$  is not continuous at  $x=0.$
So, you can write  $\int _{0}^{1}{\frac {1}{x}}~dx=\lim _{a\rightarrow 0}\int _{a}^{1}{\frac {1}{x}}~dx.$

Solution:

(a)

Step 1:
First, we write
$\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx=\lim _{a\rightarrow \infty }\int _{1}^{a}{\frac {\ln x}{x^{4}}}~dx.$
Now, we use integration by parts.
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\ln x}   and  $dv={\frac {1}{x^{4}}}dx.$
Then,  $du={\frac {1}{x}}dx$  and  $v={\frac {1}{-3x^{3}}}.$
Using integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln x}{-3x^{3}}}{\bigg |}_{1}^{a}+\int _{1}^{a}{\frac {1}{3x^{4}}}~dx}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln x}{-3x^{3}}}-{\frac {1}{9x^{3}}}{\bigg |}_{1}^{a}.}\end{array}}$

Step 2:
Now, using L'Hopital's Rule, we get
${\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln a}{-3a^{3}}}-{\frac {1}{9a^{3}}}-{\bigg (}{\frac {\ln 1}{-3}}-{\frac {1}{9}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln(a)}{-3a^{3}}}+0+0+{\frac {1}{9}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln(x)}{-3x^{3}}}+{\frac {1}{9}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{-9x^{2}}}+{\frac {1}{9}}}\\&&\\&=&\displaystyle {{\frac {1}{9}}.}\end{array}}$

(b)

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

Step 2:
Now, using L'Hopital's Rule, we get
${\begin{array}{rcl}\displaystyle {\int _{0}^{1}{\frac {3\ln x}{\sqrt {x}}}~dx}&=&\displaystyle {\lim _{a\rightarrow 0}(6{\sqrt {1}}\ln(1)-12{\sqrt {1}})-(6{\sqrt {a}}\ln(a)-12{\sqrt {a}})}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 0}-12-6{\sqrt {a}}\ln(a)+12{\sqrt {a}}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow 0}-12-6{\sqrt {a}}\ln(a)+0}\\&&\\&=&\displaystyle {\lim _{x\rightarrow 0}-12-6{\sqrt {x}}\ln(x)}\\&&\\&=&\displaystyle {-12-\lim _{x\rightarrow 0}{\frac {6\ln(x)}{\frac {1}{\sqrt {x}}}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {-12-\lim _{x\rightarrow 0}{\frac {\frac {6}{x}}{-{\frac {1}{2x^{3/2}}}}}}\\&&\\&=&\displaystyle {-12+\lim _{x\rightarrow 0}12{\sqrt {x}}}\\&&\\&=&\displaystyle {-12.}\end{array}}$

Final Answer:
(a)    ${\frac {1}{9}}$
(b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -12}

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