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 :

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

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

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

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

$m=1/2<1$ , the the integral converges.

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

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

thanks for everything.

bye

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

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

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

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

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

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

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

$x^2 < x$ for $0

However , we still use this to prove it convergent :

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

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

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

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

we conclude that

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

thus it is convergent .

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

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

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

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

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

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

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

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

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

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

bye and thanks

4. $
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

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

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

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