009B Sample Final 2, Problem 7

Evaluate the following integrals or show that they are divergent:

(a)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\infty }f(x)~dx} so that you can integrate?
You can write   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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  $u=\ln x$  and  $dv={\frac {1}{x^{4}}}dx.$
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {1}{x}}dx}   and  $v={\frac {1}{-3x^{3}}}.$
Using integration by parts, we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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
${\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {\ln x}{x^{4}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln a}{-3a^{3}}}-{\frac {1}{9a^{3}}}-{\bigg (}{\frac {\ln 1}{-3}}-{\frac {1}{9}}{\bigg )}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {\ln(a)}{-3a^{3}}}+0+0+{\frac {1}{9}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln(x)}{-3x^{3}}}+{\frac {1}{9}}}\\&&\\&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{-9x^{2}}}+{\frac {1}{9}}}\\&&\\&=&\displaystyle {{\frac {1}{9}}.}\end{array}}$

(b)

Step 1:
First, we write
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 \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  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 u=3\ln x}   and  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 dv=\frac{1}{\sqrt{x}}dx.}
Then,  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 du=\frac{3}{x}dx}   and  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 v=2\sqrt{x}.}
Using integration by parts, we get
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 \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
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 \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)    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 \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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