# Does there exist a function that is differentiable but not integrable? or integrable but not differentiable?

It has become very complicated to me to find out a function which is differentiable but not integrable or integrable but not differentiable.

• Your question asks for the same thing twice. Should the second be "integrable but not differentiable"? – Chris Eagle Apr 2 '12 at 13:42
• Also, the word is "differentiable". – Chris Eagle Apr 2 '12 at 13:42
• Please see if any of the answers you have received to previous questions are satisfactory and (if so) click the check mark on them to accept. – Ross Millikan Apr 2 '12 at 13:42
• Yes, second one is "integrable but not differentiable" – GAURAV SHARMA Apr 2 '12 at 13:49

Differentiable Function that is not Integrable: $f(x)=\frac{1}{x}$ is differentiable in $(0,1)$, however it's not integrable in this interval. $$\int_{0}^{1} f(x) \mathrm dx=\int_{0}^{1} \frac{1}{x} \mathrm dx$$ $$=\ln 1- \ln 0$$ That's Undefined.

Well, If you are thinking Riemann integrable, Then every differentiable function is continuous and then integrable!

However any bounded function with discontinuity in a single point is integrable but of course it is not differentiable!

• Will u please elaborate it with example? – GAURAV SHARMA Apr 2 '12 at 13:52
• Yes I can but you should work more on that to create an example by yourself ( for the second)! For the first one we are saying there is no to way find a function differentiable which is not integrable a Riemann! – checkmath Apr 2 '12 at 14:00
• it is worth mentioning Volterra's function – Zubin Mukerjee Apr 21 at 15:46

This answer interprets the notion of "integrable" on an interval as the function having a definite integral between any pair of points of that interval. Sometimes a notion of integrable is defined more restrictively and also requires this indefinite integral to have a finite limit as one or both points tend to an end of the interval; as OP does not clearly want to address such questions, I will not do so here either.

To remove some possible confusion in the given answers: for your first question the answer is just "it is not possible to find any differentiable function that is not integrable". The reason is that differentiable (everywhere) implies continuous (everywhere) implies Riemann integrable (on any connected part of the domain) implies Lebesgue integrable or any other sense of "integrable" you may be interested in. For the other direction there exist examples, as described in the other answers (basically if a function has discontinuities, it will not be differentiable in the points of discontinuity, but it might still be integrable (depending somewhat on what sense of "integrable" one uses), and it will certainly be so if the discontinuities occur in a discrete set of points only).

• Continuous implies integrable on compact domains. $f(x)=x$ isn't an integrable function from $\mathbb{R}$ to $\mathbb{R}$. – Chris Eagle Apr 2 '12 at 15:42

well the Weierstrass function is integrable but not differentiable. and if a function is differentiable then the set of discontinuity has measure zero so than it is integrable

• Jason A question, for you Find a function of differentiable fuction that converges pointwisely to a not Riemman integrable function! – checkmath Apr 2 '12 at 14:05

for a function that is differentiable but not integrable(here integrable refers to not having elementary antiderivatives......

$x^x, \frac{\sin{x}}{x},\frac{cosx}{x},\frac{1}{\ln(x)}$,etc. However, the derivative too may cease to exist at certain point(s) and/or region(s). Still working on the 2nd part...

• Welcome to MSE! It helps readability to format answers using MathJax (see FAQ). Regards – Amzoti Aug 21 '13 at 4:46
• Note that this is not the usual definition of 'integrable', which does have a technical meaning. – Steven Stadnicki Aug 21 '16 at 6:00

I don't think there is such function that is differentiable but not Riemann integrable. Riemman integral should be on a closed inteval [a,b]. For the function $y=\frac{1}{x}$, it is integrable on [0+$\delta$, 1-$\delta$].

However, we can say it is differentiable on (0,1), but not improperly integrable on (0, 1)