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Homework Help: Funny Definite Integral

  1. Jul 5, 2010 #1

    Char. Limit

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    Gold Member

    So, I was playing on Wolfram Alpha, and I managed to come up with this:


    In Tex, I believe this is...

    [tex]\int_{i\infty}^{\pi}e^{i x} dx = i[/tex]

    However, I have more than one problem with it, and I want to know if my problems are actually problems. First, the bounds. Can you multiply a transfinite number by i? Would the answer make any sense whatsoever? And can you integrate from an imaginary point to a real point?

    Actually, those bounds are the only problems I have... But they do look problematic. Can someone tell me if this integral is a real integral?
    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 5, 2010 #2
    The function [itex]f(z) = e^{i z}[/itex] has an absolute value:

    |f(z)| = |e^{i z}| = e^{\Re(i z)} = e^{-\Im{z}}

    which tends to zero as [itex]\Im{z} \rightarrow +\infty[/itex]. The lower bound on your integral is exactly like that. Also, the function is entire. Therefore, the integral

    F(z) = \int_{\gamma}{f(z') \, dz'}

    has the same value for all contours [itex]\gamma[/itex] starting from an infinitely high point in the upper half--plane and ending anywhere in the complex plane [itex]z[/itex]. If [itex]z = x + i y[/itex], it is convenient to choose the contour as:

    \gamma_{1}: \ z = i t, \infty > t \ge y, \ dz = i \, dt \\

    \gamma_{2}: \ z = t + i y, 0 \le t \le x, \ dz = dt


    F(z) = \int_{\infty}^{y}{e^{i i t} \, i \, dt} + \int_{0}^{x}{e^{i (t + i y)} \, dt}

    F(z) = -i \, \int_{y}^{\infty}{e^{-t} \, dt} + e^{-y} \, \int_{0}^{x}{e^{i t} \, dt}

    F(z) = -i \, \left.(-e^{-t})\right|^{\infty}_{0} + e^{-y} \, \left.\frac{e^{i t}}{i}\right|^{x}_{0}

    F(z) = -i + e^{-y} \frac{e^{i x} - 1}{i} = e^{-y} \, \sin x - i \, [ 1 + e^{-y} \, (\cos x - 1) ]
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