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I am studying the famous integral of $\dfrac{\sin x}{x}$ in complex analysis. My lecturer integrated $e^{iz}/z$ over a semi-circle with $0$ (the origin), taken out by a small semi-circle.

He asked us to think about why he chose $e^{iz}/z$ and not $e^{-iz}/z$.

My guess is it is something to do with wanting the contour to be traversed anti-clockwise and not clockwise. Is this along the right lines?

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    $\begingroup$ Probably. Among these two, it's just taste that decides. Most people's taste prefers the upper half-plane with a positively oriented contour, however. $\endgroup$
    – Daniel Fischer
    Mar 27, 2014 at 22:29
  • $\begingroup$ So choosing to integrate $e^{-iz}/z$ would have the same outcome, just travelling the opposite way on the contour? $\endgroup$
    – teddyt55
    Mar 27, 2014 at 22:32
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    $\begingroup$ No, if you take $e^{-iz}/z$, you must take a different contour. The large semicircle in the lower half-plane. $\endgroup$
    – Daniel Fischer
    Mar 27, 2014 at 22:33

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You choose the sign so that on the half-plane when you trace the contour, the exponential function decays instead of blowing up. For the upper half-plane:

$$e^{i(x+iy)}=e^{ix}e^{-y}$$

The minus ensures that the integral along the contour at $y\to +\infty$ tends to $0$.

If you choose the lower half-plane, you need to switch the sign.

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  • $\begingroup$ I think I understand, but when you say $y-> infinty$ do you mean R goes to infinity? $\endgroup$
    – teddyt55
    Mar 27, 2014 at 22:36
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    $\begingroup$ Well if $R$ goes to infinity, then $y=R\sin\phi$ also goes to infinity everywhere except on the real line, which you get the integral that you want to calculate. $\endgroup$
    – orion
    Mar 27, 2014 at 22:43
  • $\begingroup$ Thanks orion, it is an interesting question I think. So does $e^{-y}$ go to infinity faster than $e^{ix}$ ? $\endgroup$
    – teddyt55
    Mar 27, 2014 at 23:17
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    $\begingroup$ Well $e^{ix}$ oscillates, it doesn't converge to anything. $e^{-y}$ of course tends to $0$ when $y$ goes to infinity. And a product of bounded oscillatory function and a function that goes to zero, also goes to zero, so it always works. $\endgroup$
    – orion
    Mar 27, 2014 at 23:21
  • $\begingroup$ I knew that, (I am a fool!) Thanks again Orion. I get it now! :-D $\endgroup$
    – teddyt55
    Mar 27, 2014 at 23:22

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