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Math Help - Diverge or Converge

  1. #1
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    Diverge or Converge

    Determine whether \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx converges or diverges. If it converges, evaluate the integral.
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  2. #2
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    Quote Originally Posted by FLTR View Post
    Determine whether \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx converges or diverges. If it converges, evaluate the integral.
    Hello,FLTR,

    this is a standard integral which you probably know:

    \int \frac{1}{1+x^2}dx=\arctan(x)+C

    As you may know the arctan-function has the limit pi/2 if x approaches infinity:

    \lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2} and
    \lim_{x\rightarrow -\infty} \arctan(x)=-\frac{\pi}{2}

    So you get: \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}-\left( -\frac{\pi}{2} \right) = \pi

    EB

    Unfortunately the preview doesn't work properly. So I send you this post without checking. Please be careful, there might be some ugly mistakes!
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  3. #3
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    Quote Originally Posted by earboth View Post
    Hello,FLTR,

    this is a standard integral which you probably know:

    \int \frac{1}{1+x^2}dx=\arctan(x)+C

    As you may know the arctan-function has the limit pi/2 if x approaches infinity:

    \lim_{x\rightarrow \infty} \arctan(x)=\frac{\pi}{2} and
    \lim_{x\rightarrow -\infty} \arctan(x)=-\frac{\pi}{2}

    So you get: \int_{-\infty}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}-\left( -\frac{\pi}{2} \right) = \pi

    EB

    Unfortunately the preview doesn't work properly. So I send you this post without checking. Please be careful, there might be some ugly mistakes!
    I don't see anything wrong with the Math and my calculator, at least, agrees with you.

    -Dan
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  4. #4
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    Since,
    f(x)=\frac{1}{1+x^2}
    Is continous on the entire number line, we well-define

    \int_{-\infty}^{\infty}
    At any point,
    Choosing x=0

    \int_{-\infty}^0+\int_0^{\infty}
    Both the improper integrals converge to what Earboth said.
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