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Math Help - Integration Query

  1. #1
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    Integration Query

    Say we want to find  \displaystyle \int \frac{1}{x^2+1}\;{dx}. Normally, we let \displaystyle x = \tan{t}, or realise that the integrand is the derivative of \displaystyle \tan^{-1}{x}, or do something along similar lines, but suppose we ignore those. Instead, we do the following:

    Write \displaystyle \int\frac{1}{x^2+1}\;{dx} = \int\frac{1}{(x-i)(x+i)}\;{dx} and let:

    \displaystyle t = \frac{x-i}{x+i} \Rightarrow \frac{dt}{dx} = \frac{2i}{(x+i)^2} \Rightarrow dx = \frac{(x+i)^2}{2i}\;{dt}

    \displaystyle \therefore ~ \int\frac{1}{x^2+1}\;{dx} = \int\frac{1}{(x-i)(x+i)}\;{dx} = \int\frac{(x+i)^2}{2i(x+i)(x-i)}\;{dt}

    \displaystyle = \frac{1}{2i}\int\frac{x+i}{x-i}\;{dt} = \frac{1}{2i}\int\frac{1}{t}\;{dt} = \frac{1}{2i}\ln{t}+k = \frac{1}{2i}\ln\left(\frac{x-i}{x+i} \right)+k.

    More generally, letting \displaystyle t = \frac{x-ai}{x+ai} gives:

    \displaystyle  \int\frac{1}{x^2+a^2}\;{dx} = \int\frac{1}{(x-ai)(x+ai)}\;{dx} = \int\frac{(x+ai)^2}{2ai(x+ai)(x-ai)}\;{dt}

    \displaystyle = \frac{1}{2ai}\int\frac{x+ai}{x-ai}\;{dt} = \frac{1}{2i}\int\frac{1}{t}\;{dt} = \frac{1}{2i}\ln{t}+k = \frac{1}{2ai}\ln\left(\frac{x-ai}{x+ai} \right)+k.

    My Question (1): why are these identities not used and instead the arctangent ones are? I even suspected that I was missing something, but differentiating quickly confirmed that they are indeed correct.

    We know that \displaystyle \int_{0}^{\infty}\frac{1}{x^2+1}\;{dx} = \frac{\pi}{2}

    But using the relation found above gives:

    \displaystyle \int_{0}^{\infty}\frac{1}{x^2+1}\;{dx} = \frac{1}{2i}\int_{-1}^{1}\frac{1}{t}\;{dt}, which doesn't converge, since the logarithmic function is not defined on [-1, 1].

    Question (2): why is this? I probably need to pick up a complex analysis book!
    Last edited by TheCoffeeMachine; November 16th 2010 at 11:41 AM.
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  2. #2
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    Quote Originally Posted by TheCoffeeMachine View Post
    My Question (1): why are these identities not used and instead the arctangent ones are?
    The obvious answer is that this approach takes a nice simple trigonometric integral and replaces it by something involving complex logarithms, which are difficult beasts to tame.

    Quote Originally Posted by TheCoffeeMachine View Post
    But using the relation found above gives:

    \displaystyle \int_{0}^{\infty}\frac{1}{x^2+1}\;{dx} = \frac{1}{2i}\int_{-1}^{1}\frac{1}{t}\;{dt}, which doesn't converge, since the logarithmic function is not defined on [-1, 1].

    Question (2): why is this? I probably need to pick up a complex analysis book!
    A complex analysis book would be a good investment.

    The reason things look wrong here is that when you make the substitution t = \frac{x-i}{x+i}, as x goes from 0 to ∞, t goes from 1 to 1. But it does not go from 1 to 1 along the real axis, it goes (anticlockwise) round the unit circle in the complex plane. The integral of 1/t is log(t), and the values of log(t) at 1 and 1 are \pi i and 2\pi i. Taking the integral along that contour, you get the result \displaystyle\frac1{2i}\int_{-1}^1\frac1t\,dt = \frac\pi2, which is what you would expect from \displaystyle\int_0^\infty\!\!\!\frac1{x^2+1}\,dx.
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  3. #3
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    Thank you, Opalg.
    Quote Originally Posted by Opalg View Post
    The obvious answer is that this approach takes a nice simple trigonometric integral and replaces it by something involving complex logarithms, which are difficult beasts to tame.
    Well understood!
    A complex analysis book would be a good investment.
    Feel free to recommend one, please.
    The reason things look wrong here is that when you make the substitution t = \frac{x-i}{x+i}, as x goes from 0 to ∞, t goes from –1 to 1. But it does not go from –1 to 1 along the real axis, it goes (anticlockwise) round the unit circle in the complex plane. The integral of 1/t is log(t), and the values of log(t) at –1 and 1 are \pi i and 2\pi i. Taking the integral along that contour, you get the result \displaystyle\frac1{2i}\int_{-1}^1\frac1t\,dt = \frac\pi2, which is what you would expect from \displaystyle\int_0^\infty\!\!\!\frac1{x^2+1}\,dx.
    That's really amazing!
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