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 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 (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_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 (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_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:

Step 2:

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)

Return to Sample Exam