improper integration: int [0, 3] [ 1 / (1 - x) ] dx

[Kaitlin]

Looking at the int(1/(1-x))dx (from 0 to 3), there is a discontinuity at x = 1, so the integral is improper.

We split the integral up, going from 0 to 1 and from 1 to 3.

If the first integral is -oo, why does it not matter what the second integral equals? Couldn't the second integral be +oo and "perhaps" they cancel to get 0?

I looked at a graph and it appears the area from 0 to 1 is -oo and the area from 1 to 2 is +oo, and also it looks symmetrical, so graphically it seems that the integral would converge? Thank you!!!

 

[stapel]

You are assuming that "infinity" is always the same size, just like | -3 | = | +3 | = 3. But this is very much not the case! "Infinity" is not a number! :shock:

Eliz.

 

[Kaitlin]

improper integration

But here the graph is symmetrical from 0 to 1 and 1 to 2 so it seems the infinity here is the same, right?

 

[galactus]

The integral is improper because the integrand has no bound( "approaches negative infinity" ) as x approaches 1 from the right.

\(\displaystyle \L\\\int_{0}^{1}\frac{1}{1-x}dx+\int_{1}^{3}\frac{1}{1-x}dx\)

If either one diverges, then we say \(\displaystyle \L\\\int_{a}^{b}f(x)\) diverges.

\(\displaystyle \L\\\int_{1}^{3}\frac{1}{1-x}dx\\=\lim_{L\to\1^{+}}\int_{L}^{2}\frac{1}{1-x}dx\\=\lim_{L\to\1^{+}}\left[-ln|1-x|\right]_{L}^{3}\\=\lim_{L\to\1^{+}}\left[-ln|-2|+ln|1-L|\right]\\=\lim_{L\to\1^{+}}ln|1-L|=-\infty\)

Therefore, the limit does not exist and it diverges.

You are looking at the 'symmetry' of the region. Infinity is not defined. You can not have them 'cancel' one another out like that.
Looking at the graph, we can see there is an asymptote at x=1, but we can not perceive a finite limit and use symmetry of the region.

Don't let the vertical line a calculator draws for the vertical asymptote throw you off.

One that does converge is the Gaussian curve. It has area 1, but has a horizontal asymptote at the x-axis. It draws infintely close to x-axis, but never touches it.