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I Evaluating improper integrals

  1. Mar 20, 2017 #1
    I am asked to evaluate the following integral: ##\displaystyle \int_0^{ \infty} \log \left(1+ \frac{a^2}{x^2} \right) dx##. So of course this is an improper integral, but I am confused about how to go writing out the integral. From previous courses, I know that you should split the integral so that you have two clear improper integrals, rather than one that is doubly improper, so something like ##\displaystyle \int_0^1 \log \left(1+ \frac{a^2}{x^2} \right) dx + \displaystyle \int_1^{ \infty} \log \left(1+ \frac{a^2}{x^2} \right) dx##, and then we're supposed to write this out with limits: ##\displaystyle \lim_{t \rightarrow 0} \int_t^1 \log \left(1+ \frac{a^2}{x^2} \right) dx + \lim_{s \rightarrow \infty} \int_1^s \log \left(1+ \frac{a^2}{x^2} \right) dx##, and we evaluate each improper integral separately. This is how I learned to do it, but it all seems very cumbersome.

    Is there a better way of doing this? I feel like doing it this way just takes an unnecessarily long time
    Last edited: Mar 20, 2017
  2. jcsd
  3. Mar 20, 2017 #2
    Just to be clear, is the variable of integration ##x## or is it ##a##? That's why the ##\mathrm{d}x## or ##\mathrm{d}a## notation is there.
  4. Mar 20, 2017 #3
    It's x
  5. Mar 20, 2017 #4
    It's late here, but it does seem an elementary antiderivative exists. You could try deducing that and taking appropriate limits.
  6. Mar 20, 2017 #5
    I know that an antiderivative exists, and I know what it is. My question is, it seems very cumbersome to write out everything in terms of limits, so is there a better way?
  7. Mar 21, 2017 #6


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    With an antiderivative you don't need all those limits. Technically they are there, but you don't have to write it like that every time - unless your prof explicitly asks for that.
  8. Mar 21, 2017 #7
    integrate it by parts
    $$\int\log(1+a^2/x^2)dx=x\log(1+a^2/x^2)-\int xd\log(1+a^2/x^2)$$
    or if you need only to prove the convergence use
    $$\log(1+a^2/x^2)\sim a^2/x^2\quad x\to\infty,\quad a\ne 0$$ and for ##x## close to zero
    $$\log(1+a^2/x^2)=\log(a^2+x^2)-2\log x$$
    Last edited: Mar 21, 2017
  9. Mar 21, 2017 #8
    I don't believe that there would be anything wrong with

    \underset{t\to 0^+}{\lim_{s\to\infty}} \int\limits_t^s \log\Big(1 + \frac{a^2}{x^2}\Big) dx

    The relevant thing is that you don't parametrize the both limits with the same parameter. For example don't integrate over set [itex][\frac{1}{s},s][/itex].
  10. Mar 21, 2017 #9
    For you personally, if you were doing this integral with pencil on paper, how would you write it out? Would you first note that the integral is improper, then calculate the antiderivative separately, and then use that to calculate the value of he improper integral? This would seem like a better way that doing it all in one line with the limits and everything
  11. Mar 21, 2017 #10
    Can you specify what your objective really is? Do you simply want to find out a correct answer to the integral problem with as little work as possible, or are you interested to know how to explain and show the solution to other readers in as neat way as possible, while simultaneously avoiding writing anything that would be incorrect or against common rules?

    If your objective is to simply find a correct answer, of course you can use what ever special notations you want in your own notes. There isn't going to be need for others to understand your tricks.

    If your objective is to explain your solution to others, then it will be an art of optimizing notations.

    Chopping the calculation into pieces is usually a good idea. If I was given the task of explaining a solution without writing anything that would be wrong, I would first state that we can check that the derivative formula

    D_x\Big(x\ln\Big(1 + \frac{a^2}{x^2}\Big) - 2a\arctan\Big(\frac{a}{x}\Big)\Big) = \ln\Big(1 + \frac{a^2}{x^2}\Big)

    is right for [itex]x>0[/itex], and then I would state that based on this we know that

    \int\limits_0^{\infty} \ln\Big(1 + \frac{a^2}{x^2}\Big)dx = \underset{t\to 0^+}{\lim_{s\to\infty}} \int\limits_t^s \ln\Big(1+ \frac{a^2}{x^2}\Big)dx
    = \lim_{s\to\infty}\Big(s\ln\Big(1+ \frac{a^2}{s^2}\Big) - 2a\arctan\Big(\frac{a}{s}\Big)\Big) - \lim_{t\to 0^+}\Big(t\ln\Big(1 + \frac{a^2}{t^2}\Big) - 2a\arctan\Big(\frac{a}{t}\Big)\Big)

    is right too. Of course you can add more steps between these, but the point is, that these are reasonably nice formulas, and there is nothing wrong or against the rules in these.

    Perhaps it looked like that from a distance, but when I took a closer look, some of these limits turned out to be nontrivial. Out of the four terms, two were almost like substitutions, one could be handled nicely with Taylor series, and one came the nicest with l'Hopital's rule, so in the end it seems that you should keep the limits visible.
    Last edited: Mar 22, 2017
  12. Mar 22, 2017 #11


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    That's what I would do. You still have nontrivial limits, but they are not important for finding the antiderivative.
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