Something about the property of the integral bother mes

[flyingpig]

Homework Statement

[tex]\int_{0}^{\infty } e^-^x dx = -\frac{1}{e^x} \Biggr|_0^\infty = 0 + 1 = 1[/tex]

Notice that I abused [tex]\frac{1}{\infty} = 0[/tex].

My question is, when we compute integrals, why do we ignore the fact that [tex]\frac{1}{\infty} = 0[/tex] is not a limit?

 

[SammyS]

A proper way to do this is:

[tex]\int_{0}^{\infty } e^{-x}\,dx[/tex]

[tex]=\lim_{t\to\infty}\ \int_{0}^{\,t} e^{-x}\,dx[/tex]
​

 

[flyingpig]

But in most cases, we just throw it out. We don't even care about the limit anymore.

 

[SammyS]

But we have matured, & we (A·S·S-U-ME) that we know what we mean by 1/∞ =0. (even when we don;t!) LOL!

Keep up the questioning! -even if you frustrate me & others. You're obviously trying to figure this stuff out !

 

[dexdt]

well an integral from x to infinity is not really a "riemann integral", it is a limiting process of riemann integrals

 

[osnarf]

Technically speaking, you can't really throw it out. What SammyS said is the definition of the improper integral.

e^infinity isn't a specific number, so you can't plug in infinity to the primitive. If you don't write it down that's one thing (still a bad habit, though), as long as you understand that that is still what's going on.

For instance,if the problem is finding the integral of (1/x)dx from 0 to 1. You can't just plug 0 into the primitive because 1/(0^2) isn't defined as a number (1/0 doesn't equal infinity. however, we can make 1/x as large as we possibly want by making x sufficiently close to 0, so we say the limit of this approaches infinity); you have to take the limit.