# Does the improper integral exist?

I need to find a continuous and bounded function $\mathrm{f}(x)$ such that the limit $$\lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x$$ doesn't exist.

I thought about $\mathrm{f}(x) = \sin x$ but I am not sure if the fact that we divide by $T$ may some how make it converge to zero.

What do you think ?

• "such that the limit" what? exists? does not exist? – user2566092 Apr 10 '14 at 15:42
• $sin(x)$ does not work since the antiderivative $-cos(x)$ is also bounded for all $T$. Hence the limit is zero... – Thorben Apr 10 '14 at 16:00
• Such a function exists, but is a pain (for me) to write down. Basically the function can be +1 so that the integral hits ${1 \over 2}$ then transition to -1 so that the integral hits $-{1 \over 2}$, etc, etc. – copper.hat Apr 10 '14 at 16:02

$\sin(x)$ won't do, but it's only a tad trickier. $$f(x) = \sin\left(\ln(x)\right)$$ should do the job (you can integrate it exactly by elementary techniques and then show it is $\mathcal{O}(T)$).

Note that the function is bounded and continuous in $(0,+\infty)$.

• $\int_0^T f(x)dx=-1/2T(cos(ln(T))-sin(ln(T)))$... divide that by $T$ and you get $-1/2(cos(ln(T))-sin(ln(T)))\rightarrow -1$ hence the limit exists for this $f(x)$. Maybe I missed something... – Thorben Apr 10 '14 at 16:53
• @Thorben: As $T \to \infty$, that oscillates between $\pm {1 \over 2 \sqrt{2}}$. I think this is a nice solution. – copper.hat Apr 10 '14 at 19:12
• This only needs a minor tweak to make $f$ be continuous at $x=0$. – copper.hat Apr 10 '14 at 19:24
• @copper.hat you are absolutly right! As I look again at the function I have no idea how I obtained this limit... sorry for that! – Thorben Apr 10 '14 at 20:05
• @Thorben: No problem, I wouldn't have caught it if not for symbolic integration... – copper.hat Apr 10 '14 at 20:07

The idea here is simple, but the details laborious. Keep the function at $+1$ long enough so the average hits some positive value, then keep the function at $-1$ long enough so the average hits some negative value and repeat.

Let $t_n = 2^{n+2}-4$. Let $f_n(t) = \begin{cases} t-t_n, & t \in [t_n,t_n+1] \\ 1 , & t \in (t_n+1,t_{n+1}-1) \\ 1-(t-(t_{n+1}-1)) , & t \in [t_{n+1}+1, t_n] \\ 0, \text{otherwise}\end{cases}$.

Note that $f_n$ has support $(t_n,t_{n+1})$, and $\int_0^{t_{n+1}} f_n = t_n+3$.

Let $\phi(t) = \sum_{k=0}^\infty (-1)^kf_k(t)$. Then \begin{eqnarray} \int_0^{t_{n+1}} \phi_n &=& \sum_{k=0}^n (-1)^k (t_k+3) \\ &=& \sum_{k=0}^n (-1)^k (2^{k+2}-1) \\ &=& \sum_{k=0}^n (-2)^{k+2}- \sum_{k=0}^n (-1)^k \\ &=& {4 \over 3} (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \end{eqnarray} Then \begin{eqnarray} \sigma_n &=& {1 \over t_{n+1} }\int_0^{t_{n+1}} \phi_n \\ &=& {4 \over 3} { (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+3}-4} \\ &=& {1 \over 3}{ (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+1}-1} \\ &=& {1 \over 3} { {1 \over 2^{n+1} } - (-1)^{n+1} - {1 \over 2^{n+2} (1 - (-1)^{n+1} }\over 1 - {1 \over 2^{n+1} } } \end{eqnarray} We see that $\liminf_n \sigma_n = - {1 \over 3}$, $\limsup_n \sigma_n = {1 \over 3}$, hence the limit does not exist.

The integral has to diverge to infinity if this limit is to not exist, (of course any other limit will give zero overall). Given the type $\dfrac{\infty}{\infty}"$ limit, if we apply L'Hopitals; $$\displaystyle\lim_{T\to \infty}\dfrac{1}{T}\displaystyle\int_{0}^{T}f(x)\ dx = \lim_{T\to \infty} f(T)$$

This makes it clear that the limit doesn't exist iff $f$ is not bounded, so there is no suitable function.

• If the function has a limit as $t \to \infty$ this is true, but it doesn't have to have such a limit. It is straightforward, if tedious, to create a bounded $f$ that doesn't have an average limit. – copper.hat Apr 10 '14 at 18:34
• "The integral has to diverge to infinity if this limit is to not exist". Nope. "does not converge " is not synonymous to "diverges to infinity". – leonbloy Apr 10 '14 at 21:12

What about $\mathrm{f}(x)=T$?

$$\lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, \int_0^T T~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, T^2=\infty$$

• What does this mean? $T$ is not a constant. – copper.hat Apr 10 '14 at 18:43
• $T$ does not depend on $x$. Or am I misunderstanding something? – zighalo Apr 10 '14 at 18:53
• This is genius!! So simple and brilliant. – rubik Apr 10 '14 at 18:54
• @copper.hat: See http://www.wolframalpha.com/input/?i=limit+of+1%2FT+*+%28integral+from+0+to+T+of+T+with+respect+to+x%29+as+T+-%3E+inf&a=*C.T-_*Variable-&a=UnitClash_*T.*Teslas.dflt-- – rubik Apr 10 '14 at 18:54
• @zighalo: $f(x)$ is taken to be defined prior to the creation of the expression containing the limit which binds the formal parameter $T$. At the earlier time, $T$ is a free variable. At the later time, the formal parameter of the limit must be unbound in its context, so cannot be any symbol already in the context. If you set $f(x)=T$, then $T$ cannot be the formal parameter in the limit; it must be some other free-at-that-time variable. – Eric Towers Apr 10 '14 at 19:42