MHB Improper integrals (Comparison Test)

AI Thread Summary
The discussion focuses on determining the convergence of three improper integrals using the Comparison Test. For the first integral, it is established that it converges if p<1, as both parts of the integral are shown to be convergent. The second integral diverges due to the behavior near x=1, where it approaches infinity, confirmed by the Comparison Test with a known divergent integral. The third integral's convergence is not addressed in detail, but participants express a need for assistance in solving it. Overall, the thread emphasizes the application of the Comparison Test for analyzing improper integrals.
Fernando Revilla
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Use the comparison test to find out whether or not the following improper integral exist(converge)?
1)integral(upper bound:pi lower bound:0)1/((sinx)^p) dx,p<1
2)integral(upper bound:1 lower bound:0) 1/(1-x^2) dx
3)integral(upper bound:infinity lower bound:2) 1/(1-x^2) dx
I have given a link to the topic there so the OP can see my response.
 
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$(1)$ $I_1=\displaystyle\int_0^{\pi/2}\frac{dx}{\sin^px}$ is improper at $x=0$ and $I_2=\displaystyle\int_{\pi/2}^{\pi}\frac{dx}{\sin^px}$ is improper at $x=\pi$. Then, $I=\displaystyle\int_0^{\pi}\frac{dx}{\sin^px}$ is convergent iff $I_1$ and $I_2$ are both convergent. If $p<1$, $h(t)=t^p$ is decreasing on $(0,1]$ so if $x\in(0,1]$: $$0<\sin x<x\Rightarrow 0<x^p<\sin^p x\Rightarrow 0<\frac{1}{\sin^p x}<\frac{1}{x^p}$$ According to a well known property $\displaystyle\int_0^{1}\frac{dx}{x^p} $ is convergent, so $\displaystyle\int_0^{1}\frac{dx}{\sin^px}$ is also convergent. But $$I_1=\displaystyle\int_0^{1}\frac{dx}{\sin^px}+ \displaystyle\int_1^{\pi/2}\frac{dx}{\sin^px}$$ and $f(x)=1/\sin^p x$ is continuous on $[1,\pi/2]$, so $I_1$ is convergent. With the substitution $t=\pi-x$ we easily verify $I_2=I_1$ that is, $I$ is convergent if $p<1.$

P.S. If someone wants to solve $(2)$ and $(3)$ ...
 
Fernando Revilla said:
$(1)$ $I_1=\displaystyle\int_0^{\pi/2}\frac{dx}{\sin^px}$ is improper at $x=0$ and $I_2=\displaystyle\int_{\pi/2}^{\pi}\frac{dx}{\sin^px}$ is improper at $x=\pi$. Then, $I=\displaystyle\int_0^{\pi}\frac{dx}{\sin^px}$ is convergent iff $I_1$ and $I_2$ are both convergent. If $p<1$, $h(t)=t^p$ is decreasing on $(0,1]$ so if $x\in(0,1]$: $$0<\sin x<x\Rightarrow 0<x^p<\sin^p x\Rightarrow 0<\frac{1}{\sin^p x}<\frac{1}{x^p}$$ According to a well known property $\displaystyle\int_0^{1}\frac{dx}{x^p} $ is convergent, so $\displaystyle\int_0^{1}\frac{dx}{\sin^px}$ is also convergent. But $$I_1=\displaystyle\int_0^{1}\frac{dx}{\sin^px}+ \displaystyle\int_1^{\pi/2}\frac{dx}{\sin^px}$$ and $f(x)=1/\sin^p x$ is continuous on $[1,\pi/2]$, so $I_1$ is convergent. With the substitution $t=\pi-x$ we easily verify $I_2=I_1$ that is, $I$ is convergent if $p<1.$

P.S. If someone wants to solve $(2)$ and $(3)$ ...
thank you,
can you help me with 2 and 3?
 
Fernando Revilla said:
I quote a question from Yahoo! Answers


I have given a link to the topic there so the OP can see my response.

For 2) you wish to determine if [math]\displaystyle \begin{align*} \int_0^1{\frac{1}{1 - x^2}\,dx} \end{align*}[/math] is convergent.

[math]\displaystyle \begin{align*} \int_0^1{\frac{1}{1 - x^2}\,dx} &= \int_0^1{\frac{1}{(1 - x)(1 + x)}\,dx} \\ &= \int_0^1{\frac{1}{2(1 - x)} + \frac{1}{2(1 + x)}\,dx} \\ &= \lim_{\epsilon \to 1^+} \int_0^\epsilon{ \frac{1}{2(1 - x)} + \frac{1}{2(1 + x)}\,dx } \\ &= \lim_{\epsilon \to 1^+} \left[ -\frac{1}{2}\ln{|1 - x|} + \frac{1}{2}\ln{|1 + x|} \right]_0^\epsilon \end{align*}[/math]

Since it should be obvious that this will go to [math]\displaystyle \begin{align*} \infty \end{align*}[/math], the integral is divergent.
 
For all $x\in [0,1)$: $$0\leq 1-x^2=(1+x)(1-x)\leq 2(1-x)\Rightarrow \frac{1}{2(1-x)}\leq \frac{1}{1-x^2}$$ But $\displaystyle\int_0^1\frac{dx}{2(1-x)}=\left[-\frac{\color{red}1}{\color{red}2}\log (1-x)\right]_0^{1}=+\infty$ so, by the Comparison Test $\displaystyle\int_0^1\frac{dx}{1-x^2}$ is divergent.

renyikouniao said:
thank you,
can you help me with 2 and 3?

Please, show some work for 3).
 
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