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

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.

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

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' ?

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TD
Homework Helper
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: $$\mathop {\lim }\limits_{n \to \infty } \int_1^\infty {\frac{1} {{n^2 + x^2 }}dx}$$ ?

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

(btw: how do you get the graphics?)

TD
Homework Helper
Theraven1982 said:
Yes, that's it!
And the limit and integral should be interchanged to solve this problem.

(btw: how do you get the graphics?)
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?

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?

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

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

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

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

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

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