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Math Help - Integral

  1. #1
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    Integral

    Evaluate \int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx

    But without applying \int\frac1{x^2+1}\,dx=\arctan x+c.
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  2. #2
    GAMMA Mathematics
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    Let x = tan(t)
    x^2 + 1 = tan^2(t) + 1
    x^2 + 1 = sec^2(t)

    dx = sec^2(t)dt
    dt = \frac{dx}{sec^2(t)}

    \int \frac{1}{x^2 + 1} dx
    =\int dt
    =t + c
    =arctan(x) + c
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  3. #3
    Eater of Worlds
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    This is a good case for a little CA since yu can't use arctan.

    \frac{1}{x^{2}+1}=\frac{1}{(x+i)(x-i)}

    The function has a simple pole at i, and the residue is \frac{1}{2i}. Therefore, hence, and whence:

    \int\frac{1}{z^{2}+1}dz=2{\pi}i(\frac{1}{2i})={\pi  }
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by liyi View Post
    Evaluate \int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx

    But without applying \int\frac1{x^2+1}\,dx=\arctan x+c.
    Hmph. I'm not going to do it, but I wonder if you could manage to apply the definition of an integral to this? You know
    \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{(n \Delta x)^2 + 1} \cdot \Delta x

    The two problems I see are that \Delta x depends on the integration limits, essentially forcing us to calculate \frac{\infty + \infty}{n}, and the resulting series summation that the problem leaves could be a real bear to solve. But these problems might not be insurmountable.

    -Dan
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  5. #5
    Math Engineering Student
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    Nice problem!!!

    Quote Originally Posted by liyi View Post
    Evaluate \int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx
    Start by showin' that \int_0^\infty {e^{ - ux} \sin u\,du} = \frac{1}<br />
{{x^2 + 1}}.

    We're going to create a double integral. (Of course, the only reason that I'd ever create a double integral in the first place is so I could reverse the integration order.) This yields

    \int_{ - \infty }^{ + \infty } {\frac{1}<br />
{{1 + x^2 }}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,du} \,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,dx} \,du} .

    So \int_{ - \infty }^{ + \infty } {\frac{1}<br />
{{1 + x^2 }}\,dx} = 2\int_0^\infty {\frac{{\sin u}}<br />
{u}\,du} .

    The right integral can be proven applyin' double integration, but I'll use another method.

    Define f(\varphi ) = \int_0^\infty {\frac{{e^{ - \varphi u} \sin u}}<br />
{u}\,du} . (Note that we require f(0).)

    Now f'(\varphi ) = - \int_0^\infty {e^{ - \varphi u} \sin u\,du} = - \frac{1}<br />
{{\varphi ^2 + 1}}.

    Integrate for \varphi\in(0,\infty),

    f(\infty ) - f(0) = - \frac{\pi }<br />
{2} \implies f(0) = \frac{\pi }<br />
{2} + f(\infty ).

    Since f(\infty)=0, we happily get f(0) = \frac{\pi }<br />
{2}.

    --

    Yes, I used the fact that \int\frac1{x^2+1}\,dx=\arctan x+k, but the target was to compute the Dirichlet Integral
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  6. #6
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    Maybe this will help.

    Quote Originally Posted by topsquark
    Hmph. I'm not going to do it, but I wonder if you could manage to apply the definition of an integral to this? You know
    I cannot image that you can do it. Because integration is defined for bounded sets (or intervals as in this case).
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