Could someone give me some examples of functions that can't be integrated and perhaps suggest why they can't please?

I was told you couldn't integrate 2^x but then told you could...confused.

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- Mar 19th 2009, 01:37 AMGAdamsfunctions that can't be integrated
Could someone give me some examples of functions that can't be integrated and perhaps suggest why they can't please?

I was told you couldn't integrate 2^x but then told you could...confused. - Mar 19th 2009, 02:51 AMskeeter
$\displaystyle \int 2^x \, dx = \frac{2^x}{\ln{2}} + C$

some non-elementary integrals (courtesy of Mr. F) ...

Integration of Nonelementary Functions - Mar 19th 2009, 06:54 AMHallsofIvy
An example of a function that cannot be (Riemann) integrated is f(x)= 0 if x is rational, 1 if x is irrational.

No matter how you partition an interval from, say, a to b, every subinterval will contain some rational numbers and some irrational numbers. That means that for every possible partition, the "lower sum" will be 0 and the "upper sum" will be 1. Since the lower and upper sums do not converge to the same thing as we make the partitions finer and finer, the integral does not exist. - Mar 19th 2009, 07:33 AMShowcase_22
I'm pretty sure this one can't be integrated:

$\displaystyle f(x)=\frac{sin(x)}{x}$

Integration by parts yields an endless catastrophic cycle!! (Headbang) - Mar 19th 2009, 08:00 AMchisigma
In is interesting to establish if the following integral…

$\displaystyle \int_{-\infty}^{+ \infty} \frac {\sin \frac {1}{t}}{t} \cdot dt$

… exists or not… in my opinion it does exist… in yours?…

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 19th 2009, 08:46 AMShowcase_22
what's your reasoning for it existing?

In the case of $\displaystyle \int_{-\infty}^{+\infty} \frac{sin(t)}{t} dt$, integration by parts gives a sequence of sin(t)'s, cos(t)'s and $\displaystyle \frac{1}{t^{\alpha}}$'s.

Successive terms are smaller than the previous ones which suggests it's a convergent sequence. Of course, this isn't a proof but i'm pretty sure i've seen one somewhere.(Thinking) - Mar 19th 2009, 12:24 PMHallsofIvy
What exactly do you mean by "integrated"? Since that has a removable discontinuity at x= 0 and is continuous everywhere else, It certainly

**is**integrable over any bounded interval. And, since sin(x)/x goes to 0 as x goes to either infinity or negative infinity about as fast as 1/x, the integral probably does not exist, but I haven't checked that.

It might well be that the integral cannot be expressed in terms of elementary functions but that doesn't mean it is not integrable! - Mar 20th 2009, 04:03 AMGAdams
I should have given a context to my question...I was thinking in terms of: we are told (in college) that we would use a numerical method to find the area under a curve when we are unable to integrate...so I was confused about what/ when we wouldn't be able to integrate...errr...I'm still not too sure why I would need to use, say, the tapezium rule instead of integrating and finding exact area.

- Mar 20th 2009, 04:23 AMShowcase_22
ah right, great!

I did think that $\displaystyle \int\frac{sin(t)}{t}dt$ did have a solution btw.

With functions like this one, the function cannot be integrated (well, "can't be integrated easily" since the integral could exist but not be expressed as elementary functions. Thanks HallsOfIvy!) so it's much better to use the trapezium rule.

With easy functions to integrate there's no point using the trapezium rule. As you say, it's much easier to integrate the function between two limits. It is a common question on exams to use the trapezium rule to estimate the area under a graph, then integrate the function and find the error involved when using the trapezium rule.

I've thought of another one:

$\displaystyle f(x)=\begin{cases} x \ \ \ \ x \ rational \\ 0 \ \ \ \ x \ irrational \end{cases}$

This one can't be integrated since it's discontinuous everywhere apart from 0 (you might have to come up with a proof for this).