- #1
dsr39
- 14
- 0
I am a little bit confused about dealing with integrals around singularities because my professor seems to treat some situations more rigorously than others.
We talked this integral and said
[tex] \int_{-\infty}^{\infty}\frac{1}{x^3} dx = undefined [/tex]
This seems a little bit unintuitive to me because clearly we have an odd function, so no matter what happens as you approach 0, you should get equal cancellations from the left and right sides... but since you cannot say EXACTLY what happens at x = 0, I can see calling this indeterminate.
But then when we came across this integral
[tex] \int_{-\infty}^{\infty}\frac{sin(x)}{x} dx [/tex]
and integrated right through the removable singularity without any problems. It seems hypocritical to deal with this one by just appealing to intuition and saying that the part near x=0 doesn't "blow up" so it has a negligible contribution when we don't in fact know what it does AT x=0.
Is there a good reason why in this case we can just integrate straight through the singularity even though we have an undefined value at x = 0?
We talked this integral and said
[tex] \int_{-\infty}^{\infty}\frac{1}{x^3} dx = undefined [/tex]
This seems a little bit unintuitive to me because clearly we have an odd function, so no matter what happens as you approach 0, you should get equal cancellations from the left and right sides... but since you cannot say EXACTLY what happens at x = 0, I can see calling this indeterminate.
But then when we came across this integral
[tex] \int_{-\infty}^{\infty}\frac{sin(x)}{x} dx [/tex]
and integrated right through the removable singularity without any problems. It seems hypocritical to deal with this one by just appealing to intuition and saying that the part near x=0 doesn't "blow up" so it has a negligible contribution when we don't in fact know what it does AT x=0.
Is there a good reason why in this case we can just integrate straight through the singularity even though we have an undefined value at x = 0?