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If we consider

$$\int_{0}^{\infty} \frac{dx}{1+x^2}$$

Using complex contour integration only.

We choose a contour in the TOP HALF plane.

From the poles $z = \pm i$ only, the pole: $z=i$ is inside the contour.

Are we **still considering ** the pole $z= -i$

But, is the residue of $z=-i$ = 0? Because it is outside the contour??

Bottomline: Why do we only have to consider the pole inside the contour rather than outside? How does that give the actual integral?

Amad27
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1 Answers1

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You don't have to pay attention to any poles outside the contour. The residue at $-i$ is not zero, but it doesn't contribute because only poles inside the contour contribute. (This all assumes no poles are on the contour itself.)

There are plenty of proofs of the residue theorem, so you can look them up yourself.

Ian
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    But why dont we have to pay attention to the poles outside the contour? – Amad27 Dec 23 '14 at 19:42
  • What about $\sin(x)/x$ ? with two semi circle, $z=0$ outside contour – Amad27 Dec 23 '14 at 19:45
  • $\sin(z)/z$ is holomorphic on the whole plane $\mathbb{C}$ when it is continuously extended to $z=0$. – Ian Dec 23 '14 at 19:45
  • It depends on what you want to take for granted. If you take for granted the theorem about equality of contour integrals over homotopic paths, then you can prove the residue theorem by deforming your contour to some small circles around each residue which are joined by line segments. – Ian Dec 23 '14 at 19:48
  • Here they are using the pole OUTSIDE the contour? I thought this was illegal according to the residue theorem or we are not supposed to do contour integration with poles outside the contour itself. – Amad27 Dec 23 '14 at 19:48
  • http://math.stackexchange.com/questions/1079114/contour-integration-when-pole-is-outside-the-contour – Amad27 Dec 23 '14 at 19:49
  • @Amad27 It's hard to know what you mean in that link you posted as figure 7.5 doesn't appear there, but no: poles outside a contour do not play any role in evaluating the integral. – Timbuc Dec 23 '14 at 20:25
  • @Timbuc, they are using Fig7-8. THey say that soon, 7-5 was a semicircle ful – Amad27 Dec 23 '14 at 20:42
  • @Amad27 An immediately they say the integral equals zero, and then they let that contour around zero to tend to zero and the value is what is written there. Perhaps you can use the lemma and, in particular, its corollary in the first answer here: http://math.stackexchange.com/questions/83828/definite-integral-calculation-with-0-pole-and/184874#184874 – Timbuc Dec 23 '14 at 20:47
  • @Timbuc, ah - I see. so they are saying if the small radius, $\epsilon \to 0$ then the residue is $0$? – Amad27 Dec 23 '14 at 21:06
  • @Amad27 For any positive value of the small radius, there is no residue, so the contour integral is zero. As the small radius tends to zero, the part of the contour on the real axis approaches the region of integration that you want for the real problem. – Ian Dec 23 '14 at 21:17
  • @Ian, are you talking about this: Cauchy's Integral Formula from DaveNine's answer here: http://math.stackexchange.com/questions/1079114/contour-integration-when-pole-is-outside-the-contour?noredirect=1#comment2196104_1079114 ? – Amad27 Dec 24 '14 at 08:50