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Math Help - Improper integrals

  1. #1
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    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


    <br />
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 <br />
    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
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  2. #2
    Super Member
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    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<x<1

    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 ......
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  3. #3
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    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
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  4. #4
    Super Member
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    <br />
I=\left(\int_{-\infty}^{0}+\int_{0}^{\infty}\right)\cos(\pi x)dx<br />

    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 . )
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