# Why does $\int_{-\infty} ^{\infty}\frac{\sin x}{x}dx$ have a definite integral?

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.

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

The infinite integral of $\frac{\sin x}{x}$ using complex analysis