So I was going through some questions, and these two made me curious:

1) An integrable function is always bounded

2) If a function is defined on an unbounded interval, then it cannot be integrable.

I looked through the fundamental theorem of calculus, but I just don't seem to understand the connection with the boundaries, open or closed. I did see, however, that the theorem starts with : f is continuous on an interval a,b, which I assume is bounded.

Can somebody please explain if integrable functions can indeed be UNBOUNDED?

  • $\begingroup$ Welcome to math stack exchange! The function $$f(x)=\frac{1}{x^2+1}$$ is defined on $\mathbb R$, but the integral $$\int_{-\infty}^\infty f(x)dx$$ exists $\endgroup$ – Peter Jun 18 '18 at 13:32
  • $\begingroup$ Clearly that function you wrote down is an example of an unbounded function ? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:34
  • $\begingroup$ No, it is defined on an unbound interval, but not unbounded. It is a counterexample for $2)$ $\endgroup$ – Peter Jun 18 '18 at 13:38
  • $\begingroup$ So it is an example of an integrable function which is not bounded? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:39
  • $\begingroup$ No, the function is bounded. To refute $1)$ , you need another function. $\endgroup$ – Peter Jun 18 '18 at 13:40

1) Even with the (improper) Riemann integral, an integrable function need not be bounded.


The function $f$ defined as $\;f(x)=\begin{cases}\dfrac1{\sqrt x}&\text{if }\; x>0\\f(0)=&\text{whatever you want} \end{cases}$ is integrable on $[0,1]$ and its integral is $$\int_0^1f(x)\,\mathrm d x=\lim_{\varepsilon\to 0} \int_{\varepsilon}^1\dfrac1{\sqrt x}\,\mathrm d x=\lim_{\varepsilon\to 0}2\sqrt x\,\biggr|_\varepsilon^1=2.$$

2) The function $\dfrac 1{1+x^2}$ is defined on $\mathbf R$, and integrable on this interval:$$\int_{-\infty}^{+\infty}\dfrac {\mathrm d x}{1+x^2}=\arctan x\,\biggr|_{-\infty}^{+\infty}=\frac\pi 2-\Bigl(-\frac\pi2\Bigr)=\pi.$$

  • $\begingroup$ these are two examples of unbounded integrable functions ? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:47
  • $\begingroup$ The second one is an example of an integrable function defined on an unbounded interval, but it is bounded . $\endgroup$ – Bernard Jun 18 '18 at 13:50
  • $\begingroup$ So how should i answer this question : An integrable function is always bounded $\endgroup$ – Sara Saletti Jun 18 '18 at 13:53
  • $\begingroup$ Are you saying that integrable functions always have to be bounded? I am confused $\endgroup$ – Sara Saletti Jun 18 '18 at 13:54
  • $\begingroup$ Okay , maybe i got this. Both questions are false. for the first question , i use 1/square root of x as an example, for the second question i use the second equation you gave me , correct? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:57

Yes, an integrable function can be unbounded. For example, the function $1/\sqrt{x}$ on the domain (0,1] is unbounded but the integral has a finite value.

  • $\begingroup$ the function 1/x√ is an example of an integrable function which is NOT bounded? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:42
  • $\begingroup$ @SaraSaletti Integrable over the given interval. I am not sure but doesn't "integrable" mean "integrable over $\mathbb R$ ? $\endgroup$ – Peter Jun 18 '18 at 13:45
  • $\begingroup$ Yes!!!!!!!!!!!!!!! $\endgroup$ – Sara Saletti Jun 18 '18 at 13:45
  • $\begingroup$ For me, $f:x\mapsto \begin{cases}\frac{1}{\sqrt x}&(0,1]\\ whatever&x=0\end{cases}$ is not Riemann integrable on $[0,1]$, but the improper integral converges. Since $\frac{1}{\sqrt x}$ is not under-bouded, $\sum_{i=0}^{n}(x_{i+1}-x_i)\min_{[x_i,x_{i+1}]}f=-\infty $ for all subdivision $\sigma :0=x_0<...<x_n=1$. And for me, indeed, Riemann integrable function are bounded. $\endgroup$ – Todd Jan 25 '20 at 15:35

I suppose that you are talking about the Riemann integral here. If so, yes, the concept is defined only for bounded functions defined on intervals which are closed and bounded, that is, intervals of the type $[a,b]$, with $a<b$.

If $f$ is unbounded or if the interval is unbounded, we get the so-called improper integrals. For instance, we sey that $\frac1{x^2}$ is integrable on $[1,+\infty)$, because the limit$$\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}$$exists and we define$$\int_1^{+\infty}\frac{\mathrm dx}{x^2}=\lim_{M\to\infty}\int_1^M\frac{\mathrm dx}{x^2}=1.$$But this is an extension of the concept of Riemann integral.

  • $\begingroup$ Well , the question just asks if an integrable function is always bounded. Doesn't actually state which type of integrable function . Becuase an integrable function can be both improper or riemman , correct? $\endgroup$ – Sara Saletti Jun 18 '18 at 13:37
  • $\begingroup$ @SaraSaletti I strongly suspect that whoever asked the question is expecting an affirmative answer. $\endgroup$ – José Carlos Santos Jun 18 '18 at 13:40
  • $\begingroup$ Yes indeed. This is a past exam question, where the answers are : true or false. $\endgroup$ – Sara Saletti Jun 18 '18 at 13:41
  • $\begingroup$ I am confused at this point . $\endgroup$ – Sara Saletti Jun 18 '18 at 13:54
  • $\begingroup$ @SaraSaletti Since this is an exam question, the answer depends upon the way your teacher defined the concept, right?! And it seems that you are the only person around who can determine that. $\endgroup$ – José Carlos Santos Jun 18 '18 at 13:56

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