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The questions is:

Let $f$ be a function defined as $f(x) = (-1)^n/n $ for $x \in [n, n+1), n \in \mathbb{N}.$

Show that $lim_{n\to\infty}$ $\int_{[1,n]}\ f $ exists

Also, is $f$ integrable on $[1,\infty) $

Now, this function seems to closely mirror a sequence of simple (or step) function, however I do not know if it can be written as such and if that is even worth exploring. Any assistance on where to begin would be helpful. Please feel free to ask for any further clarification as well.

Brian Tung
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Tadpole
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  • Please clarify your notation. Is $$f(x) = \begin{cases}\frac{(-1)^{\lfloor x \rfloor}}{\lfloor x \rfloor} & x\ge 1\0&\text{else}\end{cases}$$ or does $f$ depend on $n$? Do you mean $$\lim_{n\to\infty} \int_1^n f(x)\ \mathrm dx = 0$$ or what? Also, what do you know about conditions on integrability? – AlexR Apr 12 '15 at 23:44
  • I believe $f$ depends on n; I wrote up the question as it was written. Also, up to now we know that the Lebesgue integral is defined as $\int f$ = $\int f^+$ - $\int f^-$. We can approximate a function with a sequence of simple functions; etc. – Tadpole Apr 12 '15 at 23:50
  • If $f$ depends on $n$, the question is of little sense. As for the last thing: This will be sufficient to answer if $f$ is integrable over $[1,\infty]$ if the question actually asks about the function I guessed in my comment. You might want to take a look at the MathJax help to see how to properly typeset your question. – AlexR Apr 12 '15 at 23:52
  • This is what has been troubling me: the function doesn't make much sense to me, and I'm unsure how to work with it. In class, the graph of this function was a sequence of step functions that progressively converged towards the x-axis, i.e.: It appears to tend towards 0. Not sure if this helps – Tadpole Apr 12 '15 at 23:57
  • I edited the original question; this is how it looks exactly – Tadpole Apr 13 '15 at 00:05
  • Note that my function can also be written as $$f(x) = \begin{cases}(-1)^n / n & x\in [n,n+1), n\in\mathbb N\ 0 & \text{else}\end{cases}$$ – AlexR Apr 13 '15 at 00:07
  • To clarify, the functions in your last comment and in your first are equivalent? – Tadpole Apr 13 '15 at 00:50
  • Yes, the first is just a bit more concise. – AlexR Apr 13 '15 at 00:51
  • Thanks for the assistance. I'll see if this helps in solving the problem! – Tadpole Apr 13 '15 at 00:52

1 Answers1

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It seems to me that since this is a step function (piecewise constant, with each piece extending between integers) this is better written as a sum:

$$ \int_{x=1}^\infty f(x) \, dx = \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln 2 $$

ETA: I suppose I had better explain where that last step comes from. One can write, for $z \not= 1$,

$$ \frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots $$

This is somewhat dicey, but if we take the (indefinite) integral of both sides, we get

$$ -\ln (1-z) = C + z + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \cdots = C + \sum_{n=1}^\infty \frac{z^n}{n} $$

and by inspection at $z = 0$ we can quickly determine $C = 0$. The expression on the right converges absolutely only on the open interval $z \in (-1, 1)$, but it does converge conditionally at $z = -1$, so we write (with some trepidation!)

$$ -\ln 2 = -\ln (1-(-1)) = \sum_{n=1}^\infty \frac{(-1)^n}{n} $$

as desired. The general expression for $z$ is the Mercator series (q.v.).

Brian Tung
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  • So the integral of the function is -ln(2)?. Is there any possibility that we can write this function as a simple function? It seems to me that this is possible, and it comes down to how we define the characteristic function. Thus far, I have that if we note that $\int f$ = $\int f^+$ - $\int f^-$, then the simple function approximation for $f^+$ is $\frac 1n$ * characteristic equation (where x is 1 if x is even, 0 if odd). The problem I have is I'm not sure if this characteristic equation is valid or if this is worth doing to solve the problem. – Tadpole Apr 14 '15 at 18:07
  • As I say, I think the function (being stepwise constant as it is) is far simpler written as a series summation. – Brian Tung Apr 14 '15 at 18:12
  • One more question: is the limit in this problem the same as the integral? These seem, at least to me, to be the same. However, perhaps not because these are stated as two different questions. – Tadpole Apr 15 '15 at 01:15
  • When an integral has infinity as one of its limits, it is implicitly a limit. We write $\int_{x=1}^\infty f(x) , dx$ as a shorthand for $\lim_{t \to \infty} \int_{x=1}^t f(x) , dx$. – Brian Tung Apr 15 '15 at 01:42