Let $f(u)$ be a function such that $\lim_{u \rightarrow 0} f(u)=0$ e.g. $f(u)={u}$. How would I evaluate $$ \lim_{t \rightarrow 0} \int_{0}^t \frac{1}{f(u)}du $$ Is this always equal to zero?

My initial instinct is to use L'Hospital but not sure how or even if it makes sense.

Edit: Assume $f$ is continuous around $0$. Use the Lebesgue integral. I think also assume that $\frac{1}{f(u)}$ is integrable (if this is not required then ignore). Hence, the example $f(u)=u$ is not a good one because it is not integrable.

  • $\begingroup$ there seems to be two limits here. how are you defining the improper integral $\int_0^t \frac{1}{f(u)} du$ in the first place? $\endgroup$ – abel Jan 24 '15 at 14:34
  • $\begingroup$ Lebesgue integration (so that dominated convergence theorem can be used if needed)... I actually didn't realize at first that I needed to specify how I am defining it. $\endgroup$ – user103828 Jan 24 '15 at 14:48

If $\frac{1}{|f|}$ is integrable near $0$ then $\int_{0}^{t}\frac{1}{f}=\int_{0}^{1}\frac{\chi_{[0,t]}}{f}$, where $\chi_{[0,t]}$ is the indicator function of $[0,1]$. We have that $\frac{\chi_{[0,t]}}{f}\to0$ and $\left|\frac{\chi_{[0,t]}}{f}\right|\leq\left|\frac{1}{f}\right|$. We can apply dominated convergence theorem to get $$\lim_{t\to0}\int_{0}^{t}\frac{1}{f}=\lim_{t\to0}\int_{0}^{1}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}\lim_{t\to0}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}0=0$$

It remains to study the case in which $\frac{1}{f}$ is integrable but $\frac{1}{|f|}$ is not.

If we assume this is Henstock-Kurzweil integral, then we don't need $\frac{1}{|f|}$ to be integrable to apply dominated convergence. It is enough to know that $0\leq \frac{\chi_{[0,t]}}{f}\leq \frac{1}{f}$, and that $0$ and $\frac{1}{f}$ are integrable, which we have.

So, it remains to study the case when a less general integral definition is used, e.g. Riemann integral.

If it is Riemann integral ($\int_{0}^{t}\frac{1}{f}$ exists as an improper Riemann integral) then it is Henstock integrable and their values coincide.

  • $\begingroup$ Thanks for the response. I'm not sure I completely understand how to use this... so assuming DCT works then $\lim_{t \rightarrow 0} \int_{0}^t \frac{1}{f(u)}du= \int_{0}^1 (\lim_{t \rightarrow 0}\frac{\chi_[0,t]}{f(u)})du=0$. Ok now I see... so if DCT works then the limit is $0$. $\endgroup$ – user103828 Jan 24 '15 at 14:42
  • $\begingroup$ So if for example $f(u)=u$ so that $\frac{1}{u}$ is not integrable near $0$ i.e. $\int_{0}^{t}\frac{1}{u}du =\ln(t)|_{0}^t = -\infty$ then is there anything we can do? does it then not make sense to ask what the limit would be? $\endgroup$ – user103828 Jan 24 '15 at 14:51
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    $\begingroup$ @user103828 Well, in that case the integral inside the limit wasn't even defined to begin with. I guess a minimal assumption should be that $\int_{0}^{t}\frac{1}{f}$ exists in some definition of integral. $\endgroup$ – Pp.. Jan 24 '15 at 14:54

If $f(u)\to 0$ as $u\to0$ then supposing $f$ continous in $]0,R[$ (is it? what hypotesis on $f$?) then we have $1/f(u)\to\infty$ as $u\to0$.

Now by integral mean value thm $\forall t>0\;\exists u_t\in]0,t[$, s.t. $$ \frac1t\int_0^t\frac1{f(u)}\,du=\frac1{f(u_t)} $$

Then $$ \int_0^t\frac1{f(u)}\,du=\frac t{f(u_t)} $$

Then it's clear that $f(u_t)\to0$ as $t\to 0$, so the value of $$ \lim_{t\to0}\frac t{f(u_t)} $$ depends on $f$.

  • $\begingroup$ does not $u_t$ depend on $t$ too? $\endgroup$ – abel Jan 24 '15 at 14:51
  • $\begingroup$ $u_t$ depends obviously on $f$ and even by the domain, so in this case, even by $t$ $\endgroup$ – Joe Jan 24 '15 at 14:52
  • $\begingroup$ yes, you can assume f is continuous around 0. $\endgroup$ – user103828 Jan 24 '15 at 14:54
  • $\begingroup$ Notice that the dependence on $f$ can be a constant dependence (like being zero always, for example). Notice also that if $\frac{1}{f}$ is integrable near zero, then it should tend to zero slower than $t$. $\endgroup$ – Pp.. Jan 24 '15 at 15:00
  • $\begingroup$ Thanks. So what would happen if $f(u)=u$... then this is equal to $\lim_{t \rightarrow 0}\frac{t}{u_t}$? Did I need to assume at the very least integrability of $\frac{1}{f(u)}$? $\endgroup$ – user103828 Jan 24 '15 at 15:03

Since $f(0)=0$ this is an improper integral so it should be treated as such. If $f(u)=u$ then you won't get $0$. You take the limit after you find an antiderivative, which, in this case, is a natural logarithm.

  • $\begingroup$ So when $f(u)=\frac{1}{u}$ then I get $\lim_{t\rightarrow 0}\ln(u)|_{u=0}^{t}=?$ $\endgroup$ – user103828 Jan 24 '15 at 14:33
  • $\begingroup$ Yes, if this is a Riemann integral (I assume that this is what it is) then this limit does not exist in the reals. $\endgroup$ – Angelo Jan 24 '15 at 15:45

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