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

  1. Aug 8, 2005 #1

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

    [tex] \int_{1}^{\infty} \frac{1}{n^2+x^2}dx [/tex]

    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. jcsd
  3. Aug 8, 2005 #2


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    Integrate from 1 to t and then limit t to infinity. It's actually a pretty easy example :smile:
  4. Aug 8, 2005 #3


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    Perhaps this is more clear:

    [tex]\int_{1}^{\infty} \frac{1}{n^2+x^2}dx[/tex]

    If you can find the indefinite, just take the limit afterwards. You should think arctan-wise here...
  5. Aug 8, 2005 #4
    Forgot to mention the following; in the problem it said the limit and integral should be interchanged.
    I know the integral of

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

    But is this just 'it' ?
    Last edited: Aug 8, 2005
  6. Aug 8, 2005 #5


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    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: [tex]\mathop {\lim }\limits_{n \to \infty } \int_1^\infty {\frac{1}
    {{n^2 + x^2 }}dx} [/tex] ?
    Last edited: Aug 8, 2005
  7. Aug 8, 2005 #6
    Yes, that's it!
    And the limit and integral should be interchanged to solve this problem.

    (btw: how do you get the graphics?)
  8. Aug 8, 2005 #7


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

    Btw, I find 0 as answer, would that be correct?
  9. Aug 8, 2005 #8
    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:

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

    This gives zero, but is this the way to go?
    Last edited: Aug 8, 2005
  10. Aug 8, 2005 #9


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

    \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} [/tex]
  11. Aug 8, 2005 #10
    Both the integral and the limit should be uniformly convergent, so you can interchange the two. Definite integral over zero is always zero.
  12. Aug 8, 2005 #11
    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:

    [tex] \int_{1}^{\infty} \frac{1}{n^2}\frac{1}{1+(x/n)^2} [/tex]

    from which it can also be seen.
    But maybe i should be satisfied with this
  13. Aug 8, 2005 #12
    So you just say: if n goes to infinity, the formula goes to zero, and therefore the integral goes to zero?

    Thanks to all who replied; really appreciated.

    Last edited: Aug 8, 2005
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