# Show the integral diverges

1. Mar 19, 2005

### quasar987

Here's the problem:

Consider a > 0. Show that the integral

$$\int_{a+1}^{\infty} \frac{1}{1/a-1/x}dx$$

diverges. Do so without using the fact that the primitive of 1/x is ln x.

I have no clue what to do!

2. Mar 19, 2005

### Data

What's $$\lim_{x\rightarrow \infty} \frac{1}{\frac{1}{a} - \frac{1}{x}}$$?

Last edited: Mar 19, 2005
3. Mar 19, 2005

### scholzie

Evaluate the integral as
$$\lim_{b\rightarrow \infty}\int_{a+1}^{b}\frac{1}{\frac{1}{a}-\frac{1}{x}}\ {dx}$$
$$=\lim_{b\rightarrow \infty}\int_{a+1}^{b}a-\frac{a^2}{a-x}\ dx$$
...
... you can do the intermediary steps here... you'll work out that you get something like the following:
$$\lim_{b\rightarrow \infty}ab-a^2\log{a-b}-a^2-a+\log{1}$$

as you know, $ab$ goes to infinity as b goes to infinity. ln(a-b) also becomes undefined as you subtract infinity from any number, since log is only defined for positive numbers.

4. Mar 19, 2005

### Data

Answering my query is a whole lot easier, and, unlike scholzie's method, does not require you to break the rules of the question!

5. Mar 20, 2005

### xanthym

Note that since {a > 0}:

$$:(1): \ \ \ \ \ 0 \ \ < \ \ a \ \ = \ \ \frac {1} {1/a} \ \ < \ \ \frac{1} {1/a-1/x} \ \ \ \ \color{blue} \mathbb \forall \ \ x \ \ge \ (a + 1)$$

$$:(2): \ \ \ \ \ \ \Longrightarrow \ \ \ \ 0 \ \ < \ \ \int_{a+1}^{\infty} a \ dx \ \ < \ \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx$$

$$:(3): \ \ \ \ \ \ \Longrightarrow \ \ \ \ 0 \ \ < \ \ \left [ ax \right ]_{a + 1}^{\infty} \ \ < \ \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx$$

$$:(4): \ \ \ \ \color{red} \Longrightarrow \ \ (\infty) \ < \ \int_{a+1}^{\infty} \frac {1} {1/a-1/x} \ dx$$

Thus, subject integral diverges.

~~

Last edited: Mar 20, 2005
6. Mar 20, 2005

### scholzie

Didnt see the bit about not using the antiderivative ln(x)... I was only trying to help...

7. Mar 20, 2005

### xanthym

No problem ... several techniques are now available for comparison.

~~

Last edited: Mar 20, 2005
8. Mar 20, 2005

### Data

Yep, nothing wrong with posting alternate methods of solution, I was just trying to avoid getting him confused.

9. Mar 20, 2005

### quasar987

Ok, thanks everyone!

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