Integral \int_{1}^{\infty} \frac{1}{n^2+x^2}dx

1. Aug 8, 2005

Theraven1982

Hello,

I'm trying to calculate the following integral,
in the limit that n goes to infinity:

$$\int_{1}^{\infty} \frac{1}{n^2+x^2}dx$$

I would be interested in the answer, but more in the way the answer can be obtained.
I'm rehearsing an old exam, but the first question is already a problem :(.

Any help would be greatly appreciated,

edit: i already looked on the internet, and in textbooks on how to accomplish this. But i only find integrals in which n doesn't go to infinity, but is a constant.

Last edited: Aug 8, 2005
2. Aug 8, 2005

Zurtex

Integrate from 1 to t and then limit t to infinity. It's actually a pretty easy example

3. Aug 8, 2005

TD

Perhaps this is more clear:

$$\int_{1}^{\infty} \frac{1}{n^2+x^2}dx$$

If you can find the indefinite, just take the limit afterwards. You should think arctan-wise here...

4. Aug 8, 2005

Theraven1982

Forgot to mention the following; in the problem it said the limit and integral should be interchanged.
I know the integral of

$$\frac{1}{(n^2+x^2)}=\frac{1}{a}ArcTan(x/a).$$

But is this just 'it' ?

Last edited: Aug 8, 2005
5. Aug 8, 2005

TD

I was confused here for a second, I thought you had to use the limit for your boundary to go to infinity. So it's n that has to go infinity as well?

So you mean: $$\mathop {\lim }\limits_{n \to \infty } \int_1^\infty {\frac{1} {{n^2 + x^2 }}dx}$$ ?

Last edited: Aug 8, 2005
6. Aug 8, 2005

Theraven1982

Yes, that's it!
And the limit and integral should be interchanged to solve this problem.

(btw: how do you get the graphics?)

7. Aug 8, 2005

TD

Well I see you already know how to write $$\LaTeX$$ code, just put it between [ tex ] [ /tex ] -tags, without the spaces then

Btw, I find 0 as answer, would that be correct?

8. Aug 8, 2005

Theraven1982

I don't know if it is correct, i think when you would put it in mathematica, it'll give 0.
But why? It certainly makes sense, and the answer isn't unexpected... But how to prove it? Can you just fill in the following formula:

$$\int_{1}^{\infty} \frac{1}{(n^2+x^2)}=\frac{1}{a}ArcTan(x/a)$$

This gives zero, but is this the way to go?

Last edited: Aug 8, 2005
9. Aug 8, 2005

TD

Well, I didn't use interchanging limit and integral, but I get 0 as well this way:

$$\begin{gathered} \mathop {\lim }\limits_{n \to \infty } \left( {\int_1^\infty {\frac{1} {{n^2 + x^2 }}dx} } \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{t \to \infty } \int_1^t {\frac{1} {{n^2 + x^2 }}dx} } \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1} {n}\arctan \left( {\frac{x} {n}} \right)} \right]_1^t } \right) \hfill \\ \\ = \mathop {\lim }\limits_{n \to \infty } \left( {\mathop {\lim }\limits_{t \to \infty } \left( {\frac{1} {n}\arctan \left( {\frac{t} {n}} \right) - \frac{1} {n}\arctan \left( {\frac{1} {n}} \right)} \right)} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{\pi } {{2\left| n \right|}} - \frac{1} {n}\arctan \left( {\frac{1} {n}} \right)} \right) = 0 \hfill \\ \end{gathered}$$

10. Aug 8, 2005

MalleusScientiarum

Both the integral and the limit should be uniformly convergent, so you can interchange the two. Definite integral over zero is always zero.

11. Aug 8, 2005

Theraven1982

Yes, this is just the way of filling in the integral; but i don't know what changes when you first take the limit. You could equally well take the n^2 outside of the integral:

$$\int_{1}^{\infty} \frac{1}{n^2}\frac{1}{1+(x/n)^2}$$

from which it can also be seen.
But maybe i should be satisfied with this

12. Aug 8, 2005

Theraven1982

So you just say: if n goes to infinity, the formula goes to zero, and therefore the integral goes to zero?

edit
Thanks to all who replied; really appreciated.

W.

Last edited: Aug 8, 2005