Infinity: For INT [infinity 0] 1/x^2 dx, am told I must do lim t-->infinity 1 - 1/t ?

[Vol]

Infinity: For INT [infinity 0] 1/x^2 dx, am told I must do lim t-->infinity 1 - 1/t ?

When you have a problem like INT [infinity 0] 1/x^2 dx, I am told that you cannot just integrate as normal, plug in infinity can get the answer. You have to do lim t-->infinity 1 - 1/t. Can anybody explain the reason for this? Is it because of the concept of approach somehow? Is it just a matter of good notation? Or is there some precise logical reason?

 

[pka]

When you have a problem like INT [infinity 0] 1/x^2 dx, I am told that you cannot just integrate as normal, plug in infinity can get the answer. You have to do lim t-->infinity 1 - 1/t. Can anybody explain the reason for this? Is it because of the concept of approach somehow? Is it just a matter of good notation? Or is there some precise logical reason?

You need to realize that 'infinity' is not a number, it is a concept. As such, it cannot be used as numbers are used.
You surely understand that \(\displaystyle \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{{x^2} - 9}} = \frac{1}{6}\).
We cannot just plug 3 in. Can we? The logic for improper integrals is similar.

 

[Steven G]

When you have a problem like INT [infinity 0] 1/x^2 dx, I am told that you cannot just integrate as normal, plug in infinity can get the answer. You have to do lim t-->infinity 1 - 1/t. Can anybody explain the reason for this? Is it because of the concept of approach somehow? Is it just a matter of good notation? Or is there some precise logical reason?

You have a similar problem at 0. You need to say lim as v-->0⁺.