# Definite Integral of 1/x from 0 to 1

1. Oct 8, 2014

### hddd123456789

Hello,

Just wondering about something, given that ln(1)=0, then the below should hold true?

$\int_0^1 1/x=0$

But the entire graph from 0 to 1 of 1/x is positive, unbounded in fact to positive infinity. So is there an intuitive explanation for why the area under the graph of 1/x from 0 to 1, while being seemingly positive infinity, is actually zero? I assume it has something to do with ln(0) and thus 1/0 being undefined?

2. Oct 8, 2014

### gopher_p

$\int_0^1 1/x\ dx$ is a divergent improper integral, so it is not true that $\int_0^1 1/x\ dx=0$. The region bounded by the graph of $y=1/x$, the $x$-axis, the $y$-axis, and the horizontal line $x=1$ does not have finite area.

On the other hand, $\int_0^1 1/\sqrt{x}\ dx$ is a convergent improper integral with$\int_0^1 1/\sqrt{x}\ dx=2$. One way of interpreting that fact is to say that the region bounded by the graph of $y=1/\sqrt{x}$, the $x$-axis, the $y$-axis, and the horizontal line $x=1$ does have finite area. So this does give us an example of an unbounded region (can't draw a circle around it) that has finite area.

3. Oct 8, 2014

### Staff: Mentor

One definition of ln(x) is the following:
$$\ln(x) = \int_1^x \frac{dt}{t}$$

From this definition, it's easy to see that ln(1) = 0.

The integral you show is an improper integral, since the integrand is not defined at the left endpoint of the interval.

4. Oct 8, 2014

### hddd123456789

Got it, so is this one of those things that are just defined to be the way they are? I mean even though the definition I used was of an improper integral, just looking at the graph of 1/x really gives a counter-intuitive interpretation of what is going on: starting at 1, I'm seeing the area increasing unbounded all the way to 0, and yet the overall area must equal zero by definition. Even taking definite integrals from something like 10^-10, 10^-50, 10^-100 to 1 gives larger and larger values. Weird.

Last edited: Oct 8, 2014
5. Oct 8, 2014

### Staff: Mentor

You're missing the point. The integral you show has an infinite area and does not represent ln(1).

6. Oct 8, 2014

### hddd123456789

Nvm then I don't get it lol, I still haven't gotten to improper integrals yet; so you're saying that $\int 1/x=ln(x)$ is simply false over the interval [0,1]?

7. Oct 8, 2014

### Staff: Mentor

Yes, because the integrand is not defined over that interval.

In general we have this:
$$\int \frac{dx}{x} = \ln|x| + C$$

The above says that ln|x| is an antiderivative of 1/x.
But to evaluate a definite integral, the integrand has to be defined and continuous at every point of a closed interval. If the integrand is undefined at an endpoint, you can use limits to determine whether the integral converges to some value (or not). Your integral diverges (the limit doesn't exist) so couldn't possibly be equal to ln(1).

8. Oct 8, 2014

### PeroK

You're also missing that $ln(0)$ is undefined. Even if you take, loosely, that $ln(0) = -\infty$ then:

$\int_{0}^{1} 1/x=[ln(x)]_{0}^{1} = ln(1) - ln(0) = \infty$

Which, loosely, implies that the area is infinite. So, I'm not sure how you got the definite integral to be 0 in the first place.

9. Oct 13, 2014

### Char. Limit

I think I see what's going on here. hddd, your problem is that you're assuming all definitions that involve definite integrals have the lower endpoint of the integral starting at zero. From what I can tell, you're working off the following definition:

$$ln(x) \colon = \int_{0}^{x} \frac{1}{t} dt$$

The problem is, this definition is wrong. The integral is actually defined such that

$$ln(x) \colon = \int_{1}^{x} \frac{1}{t} dt$$

Note the different endpoints in each integral. Using this knowledge, you can easily see that ln(1) = 0, as the integral from a point to the same point of any function is zero!

10. Oct 14, 2014

### hddd123456789

Yeah I see that now, thanks.

Yeah, it seems I was letting the geometric description over-rule the definitions. Can I still find it a bit weird that the geometric description holds for $\int_b^1 1/x=ln(1)-ln(b)$ for b as close as you want to, but not equal to, zero lol, or am I still missing something?

11. Oct 14, 2014

### Staff: Mentor

No, you're not missing anything (other than the dx in the integral.

Another way to look at this is:
$$\int_1^b \frac{dx}{x} = \ln(b) - \ln(1) = \ln(b)$$
As long as b > 0, the above holds, and b can be smaller than 1, equal to 1, or larger than 1.
If you switch the limits of integration, it changes the sign of the result, so
$$\int_b^1 \frac{dx}{x} = -\int_1^b \frac{dx}{x} = - \ln(b)$$

If b is close to zero (but positive), its ln will be negative, so -ln(b) will be a large, positive number.