integrate ( x + 1/X) and the range of the limit is 1 and 0?

Question

x2/2 is for [x], how about 1/x?

Answer 1

For one thing, this is an improper integral, because our integral boundaries are 0 and 1 and 1/x is not defined at 0. Therefore, we must convert accordingly.

Integral (0 to 1, x + 1/x) dx

To convert this accordingly, we need to find the limit of t to 1, as t approaches 0 from the right.

lim Integral (t to 1, x + 1/x) dx
t -> 0+

Now, solve the integral. The integral of x is (1/2)x^2, and the integral of 1/x is ln|x|, so we have

lim [ (1/2)x^2 - ln|x| ] {evaluated from t to 1}
t -> 0+

lim [ { (1/2)(1)^2 - ln|1| } - { (1/2)t^2 - ln|t| }
t -> 0+

ln(1) = 0, so

lim [ (1/2) - (1/2)t^2 + ln|t| ]
t -> 0+

As t approaches 0 from the right, ln(t) approaches -infinity, while the other values are a fixed value (1/2 is 1/2, and (1/2)t^2 as t approaches 0 is just 0, so our form is [1/2 + (-infinity)]).

Therefore, the limit doesn't exist.

Answer 2

The integral is x^2/2 +ln IxI which can not be taken between 0 and 1 because at x=>0 lnIxI diverges to - infinity