functions which can&#x27;t be integrated

yeah, i would like to know too

:smile:

Reply 3

Dekota

My thread;

http://www.thestudentroom.co.uk/t166224.html

:smile:

Reply 4

C4>O7

5

e^{x^2} in non-integrable.

Reply 5

It partly depends what you mean by non-integrable. For example

exp(-x^2/2)

can't, in the sense that its integral can be written in a nice elementary way, but it can be integrated as the values appear in normal distribution tables.

There are though "non-measurable" functions that can be integrated FULL STOP, not just that their integrals are too nasty to write in an elementary way.

Reply 6

Kernel

1

I have a question: if a function isn't anything's derivative (ie nothing you differentiate will give this function), is it impossible to integrate it?

Reply 7

dvs

As Neapolitan mentioned, there are functions that can't be expressed in terms of nice functions (also called "elementary functions", like polynomials, sin, cos, log, exp, etc.). That doesn't mean they aren't integrable in the literal sense, though. In general functions that are continuous (on a set I) are integrabale (on I). So using that result/theorem, we see that e^(x^2) is in fact integrable.

On the other hand, there are functions that can't be integrated. Like for instance the function:

Unparseable latex formula:

f(x)=\left\{\begin{array}{ll}1 & \mbox{ if } x \mbox{ is rational}\\0 & \mbox{ if } x \mbox{ is irrational}\end{array}\right.

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Kernel

I have a question: if a function isn't anything's derivative (ie nothing you differentiate will give this function), is it impossible to integrate it?

Not necessarily.

Reply 8

Trevor 12345

3

rpotter

I would like to know too, some1 told me that the majority of functions cant be integrated

An infinte number can be, and an infinite number cannot be !! (but which of the infinites is larger....)

Reply 9

Dekota

Mrm.

An infinte number can be, and an infinite number cannot be !! (but which of the infinites is larger....)

The infinite of the functions that cannot be! :p:

Reply 10

AlphaNumeric

I would suspect that both the number of functions which can and number which can't are uncountably infinite (since you can get polynomials with real coefficents). Wether the uncountable infinity of the unintegrable ones is a larger infinity than the uncountable of the integrables ones is a question better left to the pure whizzes like Neapolitan, I have no idea where to even begin.

Reply 11

dvs

I remember reading somewhere that there is a bijection from the set of non-integrable functions to the set of integrable functions.

Reply 12

dvs

On the other hand, there are functions that can't be integrated. Like for instance the function:

Unparseable latex formula:

f(x)=\left\{\begin{array}{ll}1 & \mbox{ if } x \mbox{ is rational}\\0 & \mbox{ if } x \mbox{ is irrational}\end{array}\right.

I think most mathematicians would say that was integrable as it is zero except on a null set. That is Lebesgue integrable.

It is though the case that the function isn't Riemann integrable as its upper and lower integrals aren't equal.

Reply 13

AlphaNumeric

dvs

I remember reading somewhere that there is a bijection from the set of non-integrable functions to the set of integrable functions.

Really? Though my knowledge of pure maths is .... well it's crap, I have to admit I struggle to see how that can be even vaguely possible. It's like saying "For every nice function, there is a unique non-nice function", while I would have thought the fact something is non integrable is a bit more subtle than being the output of a specific mapping.

Reply 14

AlphaNumeric

Really? Though my knowledge of pure maths is .... well it's crap, I have to admit I struggle to see how that can be even vaguely possible. It's like saying "For every nice function, there is a unique non-nice function", while I would have thought the fact something is non integrable is a bit more subtle than being the output of a specific mapping.

No, this is true. Well, partly it depends on what you mean by integrable, but suppose we were consider functions f:[0,1] -> R.

There are c^c such functions where c is the cardinality of the continuum.

There are c^c functions from the Cantor set to R, and if we extend each of these to be 0 elsewhere on [0,1] then we have c^c (Lebesgue) integrable functions as they are all almost everywhere zero.

On the other hand, given any non-integrable function f, then adding it to each of these c^c integrable functions produces c^c non-integrable functions.

Hence there are c^c of each type (integrable/non-integrable) and a bijection between the two.

Reply 15

dvs

Neapolitan

I think most mathematicians would say that was integrable as it is zero except on a null set. That is Lebesgue integrable.

It is though the case that the function isn't Riemann integrable as its upper and lower integrals aren't equal.

Yeah, I thought this discussion was about Riemann integrability.

There are c^c such functions where c is the cardinality of the continuum.

I'm probably wrong, but aren't there 2^c such functions?

Reply 16