3
$\begingroup$

My question is kind of misleading but here is the thing. I know that $\int_{-\infty} ^{\infty}\frac{\sin x}{x}dx$ has an integral, as a lot of proofs are available here (and it is easily observable from its plot).
But what I don't get is that it has a removable singularity and hence no residue whatsoever in the upper semicircle (I am assuming a regular upper semicircular contour) and by this, its integral should be 0 [Relating to: removable singularities give integral as 0].

So what am I getting wrong here? I am sure I messed up my fundamentals somewhere. Please help me out.

PS: My inspiration for this question is a physics-based H.W. So less rigorous math would be appreciated. Thanks in advance.

$\endgroup$
2
  • 2
    $\begingroup$ What happens on the semicircle? (People seem to think too often that everything automatically tends to zero there. Nope!) $\endgroup$
    – metamorphy
    Jun 23, 2021 at 5:07
  • $\begingroup$ @metamorphy thanks. That makes sense. $\endgroup$
    – hawexrutile
    Jun 23, 2021 at 5:55

2 Answers 2

Reset to default
3
$\begingroup$

The infinite integral of $\frac{\sin x}{x}$ using complex analysis
You should learn here. Integral over "simple closed curve without singularities" should be zero by Cauchy-Goursat's Theorem. But integral on the real line or "semi-circle without real line" should not be like that, because they are not simple closed curves

$\endgroup$
1
  • 2
    $\begingroup$ I would add though, that the OP mentioned the residue theorem, and when integrating along the kind of closed curve as appears in the post the OP linked, it is the case that the integral along the entire closed curve vanishes. However, this of course does not mean that the integral along the real line vanishes, as is explained in the answer in the linked post. $\endgroup$
    – WhiteLake
    Jun 23, 2021 at 7:12
2
$\begingroup$

You have a singularity 'on' the simple closed contour, which should not be the case to apply Cauchy' s Theorem or Cauchy's Integral Formula!

$\endgroup$
1
  • $\begingroup$ I had the idea that removable singularities didn't contribute to contour integrals and hence didn't matter if it lied on the contour itself. $\endgroup$
    – hawexrutile
    Jun 23, 2021 at 7:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .