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So I have the following integration:

$$ \frac{1}{2π}\int_{-1}^1\lvert\frac{e^{-(1+xi)}}{1+xi}\rvert^2 dx $$

and I was able to go until:

$$ \frac{1}{2πe^2}\int_{-1}^1\frac{e^{-2xi}}{1+x^2} dx $$

but I am not sure on how to solve the integration. I know if I had only the $1/(1+x^2)$ is would be $arctan(x)$. However there is the $e$ in the denominator that puzzles me.

Any idea on how to solve this?

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    $|e^{-1-xi}|=e^{-1}$...! if the absolute value encloses the whole expression inside the first integral. otherwise you are doomed to resort to sepcial functions like the so called sine integral – 2017-01-30
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    This looks like something you would solve easily using the residue theorem. – 2017-01-30
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    @tired is that because $e^{xi}$ is the unity circle, so the magnitude every ware is one? – 2017-02-01

1 Answers 1

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I found the solution thanks to @tired.

The reason is because: $e^{-1-xi}=e^{-1}*e^{xi}$.

Furthermore, we know that $e^{xi}=1$. This is true because we know that when x, in this instance, represent a point on the unity circle. The absolute value of this is the distance from the center to any point on the unity circle, which is always one regardless at which point you are looking at (that is why it is called the "unity" circle).

P.S. I know this is obvious for most people, but for me it was not and maybe it will not be for others as well. That is why I went in to detail with my answer.