009B Sample Final 2, Problem 7 Detailed Solution

Evaluate the following integrals or show that they are divergent:

(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 \int_1^\infty \frac{\ln x}{x^4}~dx}

(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 \int_0^1 \frac{3\ln x}{\sqrt{x}}~dx}

---

Background Information:
1. How could you 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^{\infty} f(x)~dx} so that you can integrate?
You can 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^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.}
2. How could you 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{1}{x}~dx?}
The problem is that  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}{x}}   is not continuous at  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 x=0.}
So, you can 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{1}{x}~dx=\lim_{a\rightarrow 0} \int_{a}^1 \frac{1}{x}~dx.}

Solution:

(a)

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_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 (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=\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}{x^4}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{1}{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=\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
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 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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=3\ln x}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv={\frac {1}{\sqrt {x}}}dx.}
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {3}{x}}dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{9}}}
(b)    Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -12}

Return to Sample Exam