# Math Help - Are all functions integrable?

1. ## Are all functions integrable?

Hi all,

Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??

2. Sorry, you'll have to define what you mean FAR better than that.

3. Originally Posted by sreejithpdas
Hi all,

Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??
No, a function is not defined if it has an infinite amount of discontiunities on the interval it is being integrated on..

Also I could interpret this as

Neither

$\int_0^{1}\frac{dx}{x^2}$ or $\int_1^{\infty}\frac{dx}{\sqrt{x}}$

are "not integrable" because they diverge...

it all depends on how you define integrable

4. Originally Posted by sreejithpdas
Hi all,

Are all functions 'Integrable'? Is there any function existing which is not integrable? can anyone give me an example??
No. Dirichlet's function is an example.

Originally Posted by Mathstud28
No, a function is not defined if it has an infinite amount of discontiunities on the interval it is being integrated on..
That is wrong. A function can have infinitely many discontinouties and still be integrable.

it all depends on how you define integrable
Your examples with improper integrals are bad examples because "integrable" is defined on closed bounded intervals.

5. Originally Posted by ThePerfectHacker
No. Dirichlet's function is an example.

That is wrong. A function can have infinitely many discontinouties and still be integrable.
I thought that...hmm...well...I guess you cannot believe everything you hear! Thanks TPH!

6. Originally Posted by ThePerfectHacker
That is wrong. A function can have infinitely many discontinouties and still be integrable.
This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?

Also, I believe $\int e^xtan(x)\,dx$ is another integral that can't be solved...

That integral came up in my Differential Equations class when solving $y^{//}+y=tan(x)$ (this was the correct equation). He gave us $y^{//}-y=tan(x)$, and we wasted 45 minutes trying to figure that one out... :'(

7. Originally Posted by Chris L T521
This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?
I do not know the theory behind Laplace transforms so I cannot say. But I am assuming that the necessacity of "piecewise continous" can be relaxed and you can use "integrable" instead. The reason why books do not use that is because they are engineering books and engineering students do not know what "integrable" means. Also, Laplace Transforms are not a good example for the above question because the poster was asking about proper integrables, i.e. integrals over finite closed intervals.

8. Originally Posted by Chris L T521
This is the case with Laplace Transforms, right (the functions must be piecewise continuous in order to apply the transform)?

Also, I believe $\int e^xtan(x)\,dx$ is another integral that can't be solved...

That integral came up in my Differential Equations class when solving $y^{//}+y=tan(x)$ (this was the correct equation). He gave us $y^{//}-y=tan(x)$, and we wasted 45 minutes trying to figure that one out... :'(
Well

$\int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx$

Multiply out and integrate termwise

9. Originally Posted by Chris L T521
Also, I believe $\int e^xtan(x)\,dx$ is another integral that can't be solved...
Originally Posted by Mathstud28
Well

$\int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx$

Multiply out and integrate termwise
I think Chris meant that you cannot find the integral in closed form. But "integrable" does not mean having a closed form. In fact, "integrable" is used only for definite integrals. Integrable means that the Riemann Sums approximate the function well. So we can take a limiting value and call that to be the integral. Though not always it is possible.

10. Originally Posted by Mathstud28
Well

$\int{e^{x}\tan(x)dx}=\int\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}x^{2n+1}}{(2n+2)!}\bigg]dx$

Multiply out and integrate termwise
argh...series comes to the rescue again...

But that looks nasty to integrate termwise...

11. Originally Posted by ThePerfectHacker
I think Chris meant that you cannot find the integral in closed form. But "integrable" does not mean having a closed form. In fact, "integrable" is used only for definite integrals. Integrable means that the Riemann Sums approximate the function well. So we can take a limiting value and call that to be the integral. Though not always it is possible.
I have always wondered this

Is this integrable?

$\int_0^{0}\bigg(sign(x)\bigg)'dx$?

EDIT: I call the signum function sign(x) just in case that was unclear

12. Originally Posted by Mathstud28
I have always wondered this

Is this integrable?

$\int_0^{1}\bigg(sign(x)\bigg)'dx$?

EDIT: I call the signum function sign(x) just in case that was unclear
You cannot write that. The sign function is not differenciable at 0. So the derivative there makes no sense.

13. Originally Posted by Mathstud28
I have always wondered this

Is this integrable?

$\int_0^{0}\bigg(sign(x)\bigg)'dx$?

EDIT: I call the signum function sign(x) just in case that was unclear
Well, wouldn't that give you zero? Since the upper and lower bounds are the same??

14. Originally Posted by ThePerfectHacker
You cannot write that. The sign function is not differenciable at 0. So the derivative there makes no sense.
But I thought $sign(0)$ was a point on the unit circle in the complex plane...why is it not differentiable

15. Originally Posted by Chris L T521
Well, wouldn't that give you zero? Since the upper and lower bounds are the same??
Assuming you could integrate it (which apparently you cant)

it would give

$sign(0)-sign(0)$

which apparently I have come to know is synonomous with

$\pm-\pm$

but that cant be right

Page 1 of 2 12 Last