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I have to evaluate the principal value of the integral $\displaystyle\int_{-\infty}^\infty\frac{\mathrm{e}^{-2\mathrm{i}x}}{x^2+1}\,\mathrm{d}x$ and I have to use residues.

First, I thought that because of the exponent, there is an infinite number of poles, so I cannot use a semicircle as a parametrization from $-R$ to $R$, where $R\to\infty$ in the limit. This would imply that the residues would become an infinite series.

Next, I considered a rectangular contour shifted by $\frac12\pi$. I thought this meant that the $\gamma_3$ parameter was just the $\gamma_1$ parameter multiplied by a factor $\mathrm{e}^\pi$. The $\gamma_2$ and $\gamma_4$ go to zero when $R\to\infty$. However, $\gamma_1$ and $\gamma_3$ do not. So then, I have to calculate the residue of the function, which is the residue of $\frac14\pi$ and of $\mathrm{i}$. The residue of $\frac14\pi$ gives me trouble. If I use that $\textrm{Res}=\frac{P(x)}{Q'(x)}$ defined at $x=\frac14\pi$, I do not get a value.

Ice Tea
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Danielvanheuven
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  • Not very sure what you mean but I think there are only two poles of $\frac{e^{-2iz}}{z^2+1}$, and obviously they are at $z=i$ and $z=-i$. Hence a semicircle is exactly the perfect way to use. – MonkeyKing Jan 13 '20 at 18:53
  • You are very confused. The exponential function is holomorphic in the entire plane (hence, an entire function). The function $e^{-2iz}{z^2+1}$ has simple poles at $z=\pm i$. – Ted Shifrin Jan 13 '20 at 19:43
  • why does it not have poles at $\frac{1}{4}\pi + \frac{1}{2}n\pi$ the function $\frac{e^{ax}}{1+e^x}$ has them at $(2n+1)\pi i$ what is different? – Danielvanheuven Jan 13 '20 at 20:11

2 Answers2

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The poles should only be at x = i, x = - i. Recalling that $e^0 = 1$

Form the usual semi-circle and then isolate those points in the usual way. Then use the residue theorem.

LiquidMikerrs
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  • Well, not the "usual" semicircle, perhaps. – Ted Shifrin Jan 13 '20 at 19:31
  • oke because i thought if you used a semicircle that you cald have any $x= \frac{1}{4}\pi+\frac{1}{2}n\pi$ since by use of complex exponentials those are all 0. So then you get that those are all poles. I saw that for another example in the book so i was not sure if that was the case for this one aswell. Since it is a principal value integral p.v. – Danielvanheuven Jan 13 '20 at 19:37
  • I'm not sure what you are asking there. Poles happen at singularities. Exponential functions are entire. Therefore the only issue is the 1/(x^2+1). Please note your argument for x on another function is comparing apples and oranges. – LiquidMikerrs Jan 13 '20 at 21:17
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The poles are at $x=i$ and $x=-i$. By using a contour with a semi-circle:

$$\Gamma=\mathbb{R}\cup \{Re^{i\phi}\}_{\phi=\pi}^{2\pi}$$

and letting $R\to\infty$, we have

$$\oint_\Gamma f(z)\;\mathrm{d}z=2\pi i \mathrm{Res}\left(f, z=-i\right)=-\int_{-\infty}^\infty f(x)\;\mathrm{d}x+\lim_{R\to\infty}\int_0^\pi f(Re^{j\phi})iRe^{i\phi}\;\mathrm{d}\phi$$

where $f(z)=\frac{e^{-2iz}}{z^2+1}$. The integral along the semi-circle satisfies

$$\left|\int_{\pi}^{2\pi} \frac{e^{-2iRe^{i\phi}}}{R^2 e^{2i\phi}+1}iRe^{i\phi}\;\mathrm{d}\phi\right|\le \int_{\pi}^{2\pi} \frac{Re^{2R\sin\phi}}{\sqrt{1+R^4+2R^2\cos(2\phi)}}\;\mathrm{d}\phi\le \int_{\pi}^{2\pi}\frac{R}{|1-R^2|}\;\mathrm{d}\phi=\frac{\pi R}{|1-R^2|}\to0$$

i.e., it goes to $0$ as $R\to\infty$. Therefore we have

$$\int_{-\infty}^\infty f(x)\;\mathrm{d}x=-2\pi i\mathrm{Res}\left(f, z=-i\right)=-2\pi i\frac{e^{-2}}{-2i}=\pi e^{-2}.$$

wimi
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  • In the book i saw the example $\frac{e^{ax}}{1+e^x}$ which has poles at all the points $z=(2n+1)\pi i$ why does that function have infinity amount of poles but this one does not but also has the exponential in the p.v. integral what is different? – Danielvanheuven Jan 14 '20 at 06:13
  • also you got $e^{-2}$ it should be $e^2$ right since you have $i^2$ so that is minus times the minus already there so +? – Danielvanheuven Jan 14 '20 at 06:16
  • @Danielvanheuven I was solving a different problem, sorry. I have fixed it now. Note that now I need the "lower" semicircle from $\pi$ to $2\pi$, because I need to ensure that the exponent in $e^{2R\sin\phi}$ is negative. The pole I need to use now is $z=-i$, which now gives $e^{-2}$. – wimi Jan 14 '20 at 06:50
  • @Danielvanheuven the function with denominator $1+e^x$ has poles when $e^x=-1$, which occurs infinitely many times. This is because of the $e^x$ in the denominator, which does not occur in the function of this question. The function in this question has poles when $x^2=-1$. – wimi Jan 14 '20 at 06:53