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What does $\int_{0}^{\infty} e^{y(iu-\alpha)}dy = ?$ Please note $i$ is a complex variable, $\alpha$ and $u $ are constants.

I know this integral evaluates to:

$$\left.\frac{e^{y(iu-\alpha)}}{iu-\alpha}\right|_0^\infty$$

I'm unsure how this evaluates due to the terms inside the parenthesis of the exponetial for which $\infty$ will have to be distributed to. So you end up with a $-\infty$ and $+\infty$ term inside that exponential, which is throwing me off. How do you evaluate and make sense out of this?

Did
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user11460
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  • What about $u$ and $\alpha$? Are they complex? Real? Do you know anything else about them? – David Nov 10 '14 at 04:53
  • @David - I just edited my question to add more info about them. Yes they can be treated like real constants. – user11460 Nov 10 '14 at 17:21

1 Answers1

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I am assuming that $\alpha$ is real.

Write $e^{y(iu-\alpha)} =e^{iuy}e^{-y\alpha} $.

If $\alpha>0$, $e^{-y\alpha} \to 0 $, so the limit at $\infty$ is $0$.

If $\alpha=0$, $e^{iuy} =\cos(uy)+i\sin(uy) $ oscillates as $y \to \infty$, so the limit does not exist.

If $\alpha < 0$, it blows up as $y \to \infty$.

marty cohen
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