How Can We Simplify Evaluating the Improper Integral of Logarithmic Functions?

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In summary, when evaluating the improper integral ##\displaystyle \int_0^{ \infty} \log \left(1+ \frac{a^2}{x^2} \right) dx##, it is helpful to split it into two separate integrals, each with its own limit. This allows for a more manageable calculation, especially when finding an antiderivative may be cumbersome. However, if the goal is simply to find the correct answer, it is possible to use other notations and shortcuts. The key is to optimize notations when explaining the solution to others.
  • #1
Mr Davis 97
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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
 
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  • #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.
 
  • #3
JoePhysics said:
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.
It's x
 
  • #4
It's late here, but it does seem an elementary antiderivative exists. You could try deducing that and taking appropriate limits.
 
  • #5
JoePhysics said:
It's late here, but it does seem an elementary antiderivative exists. You could try deducing that and taking appropriate limits.
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?
 
  • #6
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.
 
  • #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
$$\log(1+a^2/x^2)=\log(a^2+x^2)-2\log x$$
 
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  • #8
Mr Davis 97 said:
I know that you should split the integral so that you have two clear improper integrals, rather than one that is doubly improper

I don't believe that there would be anything wrong with

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

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].
 
  • #9
mfb said:
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.
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
 
  • #10
Mr Davis 97 said:
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?

Mr Davis 97 said:
if you were doing this integral with pencil on paper, how would you write it out?

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.

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

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

[tex]
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)
[/tex]

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

[tex]
\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
[/tex]
[tex]
= \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)
[/tex]

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.

mfb said:
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.

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.
 
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  • #11
Mr Davis 97 said:
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
That's what I would do. You still have nontrivial limits, but they are not important for finding the antiderivative.
 

Related to How Can We Simplify Evaluating the Improper Integral of Logarithmic Functions?

What is an improper integral?

An improper integral is an integral where either the upper or lower limit of integration is infinite or the integrand has a vertical asymptote within the interval of integration.

How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you must evaluate the integral using limits. If the resulting limit is a finite number, then the integral converges. If the limit is infinite or does not exist, then the integral diverges.

What are some common methods used to evaluate improper integrals?

Some common methods used to evaluate improper integrals include splitting the integral into smaller intervals, using substitution or integration by parts, and using comparison tests.

Can improper integrals have multiple points of discontinuity?

Yes, improper integrals can have multiple points of discontinuity. In this case, the integral must be split into separate intervals and evaluated separately.

When should the comparison test be used to evaluate an improper integral?

The comparison test should be used when the integrand is difficult to integrate directly, or when the integrand has a similar form to a known convergent or divergent integral.

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