Improper integrals (Comparison Test)

[Fernando Revilla]

I quote a question from Yahoo! Answers

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.

 

[Fernando Revilla]

$(1)$ $I_1=\displaystyle\int_0^{\pi/2}\frac{dx}{\sin^px}$ 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 $(3)$ ...

 

[renyikouniao]

$(1)$ $I_1=\displaystyle\int_0^{\pi/2}\frac{dx}{\sin^px}$ 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?

 

[Prove It]

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 \(\displaystyle \displaystyle \begin{align*} \int_0^1{\frac{1}{1 - x^2}\,dx} \end{align*}\) is convergent.

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

Since it should be obvious that this will go to \(\displaystyle \displaystyle \begin{align*} \infty \end{align*}\), the integral is divergent.

 

[Fernando Revilla]

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.

thank you,
can you help me with 2 and 3?

Please, show some work for 3).