Does the Integral Diverge as a Tends to Infinity?

[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!

 

[Data]

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

 

[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.

 

[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!

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

 

[xanthym]

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!

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.


~~

 

[scholzie]

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

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

 

[xanthym]

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

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


~~

 

[Data]

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

 

[quasar987]

Ok, thanks everyone!