1. ## Improper integrals

Hi , I'm Mahler,
Someone can help me? with these integrals.

Need to know if I did it well.

The first one is :

$\displaystyle \int_{2}^{4}\frac{1}{x\sqrt{x^{2}-4}}dx$

$\displaystyle f(x)=\frac{1}{x\sqrt{x^{2}-4}}=\frac{1}{x(2-x)^{1/2}(2+x)^{1/2}}$

When $\displaystyle x\rightarrow2 , (2-x)^{1/2}\rightarrow0$
$\displaystyle x\rightarrow2 , (2+x)^{1/2}\rightarrow2$

Then,$\displaystyle f(x)\sim\frac{1}{4(2-x)^{1/2}}=\frac{A}{(2-x)^{m}} ,A=1/4\neq0,\neq+\infty$
$\displaystyle m=1/2<1$ , the the integral converges.

...the second is $\displaystyle \int_{0}^{\infty}e^{-x^{2}}dx$ I think it should apply comparsion test, $\displaystyle \frac{1}{e^{x}}$ converges , ($\displaystyle \left|1/e^{x^{2}}\right|\leq1/e^{x}$) , $\displaystyle \int_{0}^{\infty}e^{-x^{2}}dx$ also converges .

finally .. $\displaystyle \int_{2}^{\infty}\frac{1}{\sqrt{x}ln\left|x\right| }dx$ I think should compare with this function $\displaystyle \frac{1}{\sqrt{x}}$ . but I'm not sure

thanks for everything.

bye

2. For the first one , substitute $\displaystyle x = \frac{1}{t}$ we have

$\displaystyle \int_2^4 \frac{dx}{x\sqrt{x^2-4}}$

$\displaystyle = \int_{0.25}^{0.5} \frac{dt}{\sqrt{1- (2t)^2 }}$

$\displaystyle = \frac{1}{2} \left[ \sin^{-1}(2t) \right]_{0.25}^{0.5}$

$\displaystyle = \frac{\pi}{6}$

For the second one , It is unnecessary to write $\displaystyle |e^{-x^2}|$ because $\displaystyle e^{-x^2} > 0 , x\in \mathbb{R}$

Also , your inequality is false for $\displaystyle 0 < x < 1$ because

$\displaystyle x^2 < x$ for $\displaystyle 0<x<1$

However , we still use this to prove it convergent :

$\displaystyle \int_{0}^{\infty}e^{-x^{2}}dx = (\int_0^1 + \int_1^{\infty} ) e^{-x^2} ~dx$

$\displaystyle = \int_0^1 e^{-x^2}~dx + \int_1^{\infty} e^{-x^2} ~dx$

$\displaystyle < \int_0^1 e^{-x^2}~dx + \int_1^{\infty} e^{-x}~dx$

Also we have $\displaystyle e^{-x^2} < 1$ for $\displaystyle x > 0$

we conclude that

$\displaystyle I < (1) - [ e^{-\infty} - e^{-1} ] = 1 + e^{-1}$

thus it is convergent .

For the last one , we have $\displaystyle \frac{1}{\sqrt{x}} > \frac{1}{x}$ , for $\displaystyle x >2$ ( later you will know that why i compare with this )

$\displaystyle \int_2^{\infty} \frac{dx}{\sqrt{x}\ln(x)} > \int_2^{\infty} \frac{dx}{x\ln(x)}$

sub $\displaystyle \ln(x) = t$ we obtain

$\displaystyle \int_2^{\infty} \frac{dx}{x\ln(x)} = \lim_{a\to\infty} \ln[\ln(x)]_2^a = \infty$ ......

3. thanks so much simplependulum , now I have another doubt ,

to prove if this integral is convergent $\displaystyle I={\displaystyle \int_{-\infty}^{\infty}\cos(}\pi x)dx$ I do this:

$\displaystyle I=\left(\int_{-\infty}^{0}+\int_{0}^{\infty}\right)\cos(\pi x)dx$ (thanks for this notation too, I didn't it know)

$\displaystyle {\displaystyle -1\leq\cos(x)\leq1\Rightarrow}-1\leq\cos(\pi x)\leq1,$ like $\displaystyle {\displaystyle \int_{-\infty}^{\infty}}dx$ or ( $\displaystyle {\displaystyle \int_{0}^{\infty}}dx$ )diverges , $\displaystyle {\displaystyle \int_{-\infty}^{\infty}\cos(}\pi x)dx$ or also $\displaystyle {\displaystyle \int_{0}^{\infty}\cos(}\pi x)dx$ diverges, my reasoning is correct??.

Generally ,this is correct? if $\displaystyle \displaystyle\int_{a}^{b}f$ and $\displaystyle \displaystyle\int_{a}^{b}g$ are two improper integrals , $\displaystyle {\displaystyle f(x)\leq g(x)}$, $\displaystyle \displaystyle\int_{a}^{b}g$ diverges, implies $\displaystyle \int_{a}^{b}f$ also diverges??

When $\displaystyle \displaystyle\int_{a}^{b}g$ converges I know that if it is true ( implies $\displaystyle \int_{a}^{b}f$ converges ), but I'm not sure if is true when it diverges

bye and thanks

4. $\displaystyle I=\left(\int_{-\infty}^{0}+\int_{0}^{\infty}\right)\cos(\pi x)dx$

haha actually , it is not a formal writing , i was just lazy to write longer ...

For this problem , you can find that

$\displaystyle \lim_{x\to\infty} f(x)~ \neq 0$

so the integral diverges . I think it is not suitable to compare with other integral . In your case , $\displaystyle \int_0^{\infty}\cos(\pi x)~dx < [ divergent ~integral ]$ doesn't imply the integral also diverges .

However , in other case if you can prove that $\displaystyle I > [ divergent ~integral ]$ , then $\displaystyle I$ diverges . (I assume the two integrals we compare are always positive . )