# Homework Help: Improper integral question

1. Aug 29, 2010

### TsAmE

1. The problem statement, all variables and given/known data

Determine how large the number a has to be so that:

$$\int_{a}^{\infty} \frac{1}{1 + x^{2}} dx < 0.001$$

2. Relevant equations

None.

3. The attempt at a solution

I tried to evaluate the left hand side and got a final answer of:

$$a > \frac{\pi}{2} - \frac{1}{1000}$$

but the correct answer was 1000

I think these 2 steps in my working out might be the problem:

from $$tan(arctan(a)) > \frac{\pi}{2} - \frac{1}{1000}$$ to:

$$arctan(a) > \frac{\pi}{2} - \frac{1}{1000}$$

$$a > \frac{\pi}{2} - \frac{1}{1000}$$

Last edited: Aug 29, 2010
2. Aug 29, 2010

### phyzguy

Why did you write tan(arctan(a))? Since the indefinite integral is just arctan(x), and arctan(x) ->pi/2 as x->infinity, you should just write pi/2-arctan(a)<.001, or arctan(a)>pi/2-.001. Do you know how to solve this?

3. Aug 29, 2010

### Mentallic

The last 3 lines do not continue from each other at all. Firstly, how did you get tan(arctan(a))? How does this continue on to get just arctan(a) and then simply a?

4. Aug 29, 2010

### TsAmE

Sorry I made a mistake in the post. I have corrected the question.

5. Aug 29, 2010

### TsAmE

My last post was ambiguous. I meant I had corrected the mistake when I typed the question. Could someone please check my working?

6. Aug 29, 2010

### hunt_mat

Perform the integral to obtain:
$$\int_{a}^{\infty}\frac{dx}{1+x^{2}}=\tan^{-1}(\infty )-\tan^{-1}a=\frac{\pi}{2}-\tan^{-1}a<0.001$$
Then re-arrange to obtain:
$$\tan^{-1}a>\frac{\pi}{2}-0.001$$
Take tan to obtain the answer...

7. Aug 29, 2010

### Mentallic

I don't see any change in your original post. Your last 3 lines of working are still wrong.

8. Aug 30, 2010

### TsAmE

I reached the same answer as hunt as in:

arctan of $$> \frac{\pi}{2} - \frac{1}{1000}$$

and tried solving for a only to get:

$$a > \frac{\pi}{2} - \frac{1}{1000}$$

I dont see how I can get an answer of 1000 when there is a pi/2 - 0.001

My latex code is giving me problems when I try to edit it so thats probably why you didnt see my correction.

Last edited: Aug 30, 2010
9. Aug 30, 2010

### hunt_mat

So take tan of what I wrote down:
$$a>\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\frac{1}{\tan (1/1000)}\approx 1000$$
That is how you finish off the calculation.

10. Aug 30, 2010

### TsAmE

With respect to the inequality why is a > and not < ? Do you swap the sign around when you use the inverse trig function?

Also $$a>\tan\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =$$ 0.0274... on my calculator as opposed to 1000?

11. Aug 30, 2010

### hunt_mat

You use:
$$\tan x=\frac{\sin x}{\cos x}$$
And then you note:
$$\sin\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\cos\Bigg(\frac{1}{1000}\Bigg)$$
and
$$\cos\Bigg(\frac{\pi}{2}-\frac{1}{1000}\Bigg) =\sin\Bigg(\frac{1}{1000}\Bigg)$$
So you get the same as before, for small x, we have the following:
$$\tan x\approx x$$
Regarding the inequality, tan x is an increasing function, so the inequality remains the same, regarding your calculator, I think you may have entered the sum incorrectly.

12. Aug 30, 2010

### Mentallic

I don't understand why you think it would be < ?
You already had $arctan(a)>\pi/2-0.001$ so why would it suddenly change sign?

13. Aug 31, 2010

### TsAmE

Thanks I now see where I went wrong. One last thing I want to know is how would you know that $$\frac{1}{\tan (1/1000)}\approx 1000$$ without using a calculator, cause I am not allowed to use calculators in my tests.

14. Aug 31, 2010

### hunt_mat

Use a Maclaurin series to show that:
$$\tan x =x+o(x),\quad \sin x=x+o(x),\quad\cos x=1-\frac{x^{2}}{2}+o(x^{2})$$

15. Aug 31, 2010

### Mentallic

For small x, $x\approx tan(x)$ so $$\frac{1}{tan(1/1000)}\approx \frac{1}{1/1000}=1000$$