# integrability question

• Sep 2nd 2011, 04:45 AM
transgalactic
integrability question
3b) show an example for a function wich is bounded and exists in [a,b]
but its not integrabile.
prove that its not integrabile by definition
3c)calculate$\displaystyle \int\frac{ln(x^{2}+1}{x^{4}}dx$
• Sep 2nd 2011, 06:16 AM
HallsofIvy
Re: integrability question
What have you done on this?
• Sep 2nd 2011, 06:38 AM
transgalactic
Re: integrability question
i dont know how to solve th first part
but i know some laws for proving integrability
(*) for some dividing P
E is the set of uppers sums, T is the set of lower sums
if infE=supT then its defferentiable.

(*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.
s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.
in each subsection if inf{S(p)-s(p)}=0 then its defferentiable
or if $\displaystyle S(p)-s(p)<\epsilon$

ok i tried this one:
f(x)=1 on x=0
f(x)=0 on other wise

for every diving of the section s(p)=0 so is supT=0 too
but S(p) is obviosly not sero so here is my example.
so $\displaystyle inf{S(p)-s(p)}\leq 0$ but i dont know how to express S(p)
?

regarding the second one i thought of using by parts
but its not solving it
• Sep 2nd 2011, 11:37 AM
e^(i*pi)
Re: integrability question
For the second part integration by parts is a good start (you'll then want to try partial fractions)

$\displaystyle u = \ln(x^2+1) \rightarrow du = \dfrac{2x}{x^2+1} dx$

$\displaystyle \frac{dv}{dx} = x^{-4} \rightarrow v = -\frac{1}{3x^3}$

$\displaystyle \int \dfrac{\ln(x^2+1)}{x^4} dx = -\dfrac{\ln(x^2+1)}{3x^3} - \int \dfrac{2x}{3x^3(x^2+1)}dx$

$\displaystyle = -\dfrac{\ln(x^2+1)}{3x^3} - \dfrac{2}{3} \int \dfrac{1}{x^2(x^2+1)} dx$

Now use partial fractions.

Hint: $\displaystyle \dfrac{1}{x^2(x^2+1)} = \dfrac{(x^2 +1) - x^2}{x^2(x^2+1)}$
• Sep 2nd 2011, 12:04 PM
transgalactic
Re: integrability question
Quote:

Originally Posted by e^(i*pi)
For the second part integration by parts is a good start (you'll then want to try partial fractions)

$\displaystyle u = \ln(x^2+1) \rightarrow du = \dfrac{2x}{x^2+1} dx$

$\displaystyle \frac{dv}{dx} = x^{-4} \rightarrow v = -\frac{1}{3x^3}$

$\displaystyle \int \dfrac{\ln(x^2+1)}{x^4} dx = -\dfrac{\ln(x^2+1)}{3x^3} - \int \dfrac{2x}{3x^3(x^2+1)}dx$

$\displaystyle = -\dfrac{\ln(x^2+1)}{3x^3} - \dfrac{2}{3} \int \dfrac{1}{x^2(x^2+1)} dx$

Now use partial fractions.

Hint: $\displaystyle \dfrac{1}{x^2(x^2+1)} = \dfrac{(x^2 +1) - x^2}{x^2(x^2+1)}$

$\displaystyle \frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}$+C/(1+x^2)
did i splitted it correctly by theory
?
• Sep 2nd 2011, 12:11 PM
Siron
Re: integrability question
Use the hint of e^(ipi) which is very useful:
$\displaystyle \frac{1}{x^2(x^2+1)}=\frac{(x^2+1)-x^2}{x^2(x^2+1)}=\frac{x^2+1}{x^2(x^2+1)}-\frac{x^2}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}$

EDIT:
If you would use the traditional method then it has to be:
$\displaystyle \frac{1}{x^2(x^2+1)}=\frac{Ax+B}{x^2}+\frac{Cx+D}{ x^2+1}=\frac{(Ax+B)(x^2+1)+(Cx+D)(x^2)}{x^2(x^2+1) }=\frac{Ax^3+Bx^2+Ax+B+Cx^3+Dx^2}{x^2(x^2+1)}=$
$\displaystyle \frac{(A+C)x^3+(B+D)x^2+Ax+B}{x^2(x^2+1)}$

That means:
$\displaystyle B=1$ therefore $\displaystyle D=-B \Leftrightarrow D=-1$ and $\displaystyle A=C=0$
And so:
$\displaystyle \frac{1}{x^2(x^2+1)}=\frac{1}{x^2}-\frac{1}{x^2+1}$
• Sep 2nd 2011, 12:16 PM
transgalactic
Re: integrability question
ok i could take care of this integral from here
regarding the first part:
i know some laws for proving integrability

(*) for some dividing P

E is the set of uppers sums, T is the set of lower sums

if infE=supT then its defferentiable.

(*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

in each subsection if inf{S(p)-s(p)}=0 then its defferentiable

or if $\displaystyle S(p)-s(p)<\epsilon$

ok i tried this one:

f(x)=1 on x=0

f(x)=0 on other wise

for every diving of the section s(p)=0 so is supT=0 too

but S(p) is obviosly not sero so here is my example.

so $\displaystyle inf{S(p)-s(p)}\leq 0$ but i dont know how to express S(p)

?
• Sep 2nd 2011, 02:29 PM
mr fantastic
Re: integrability question
Quote:

Originally Posted by transgalactic
$\displaystyle \frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}$+C/(1+x^2)
did i splitted it correctly by theory
?

At the level you appear to be studying at, you are most surely expected to know how to perform a partial fraction decomposition. You can at least work backwards from your answer and see if it works.

And please show all of your working in your first posting of future questions. We should not have to be asking for it.
• Sep 2nd 2011, 10:05 PM
transgalactic
Re: integrability question
my last post states that i have solved the integral and moved to the next one.

regarding the first part:

i know some laws for proving integrability

(*) for some dividing P

E is the set of uppers sums, T is the set of lower sums

if infE=supT then its integrible.

(*) S(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

s(p) is the sum of each supremum of the function in subsection mutiplied by the length of the subsection.

in each subsection if inf{S(p)-s(p)}=0 then its defferentiable

or if $\displaystyle S(p)-s(p)<\epsilon$

ok i tried this one:

f(x)=1 on x=0

f(x)=0 on other wise

for every diving of the section s(p)=0 so is supT=0 too

but S(p) is obviosly not sero so here is my example.

so $\displaystyle inf{S(p)-s(p)}\leq 0$ but i dont know how to express S(p)

?
• Sep 3rd 2011, 01:14 PM
emakarov
Re: integrability question
Quote:

Originally Posted by transgalactic
3b) show an example for a function wich is bounded and exists in [a,b]
but its not integrabile.
prove that its not integrabile by definition

It is known that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere, i.e., the set of its points of discontinuity has measure zero. A set has measure zero if it can be covered by intervals whose combined length can be made arbitrarily small; however, it is not necessary to understand this precisely for this problem. Every finite set and even every countable infinite set has measure zero, so the function you tried: f(x)=1 on x=0 f(x)=0 otherwise, is integrable (it has a single point of discontinuity). You need a lot more points of discontinuity. Consider Dirichlet function, which is nowhere continuous. It is easy to show that the upper sum, say, on [0, 1] is always 1 and the lower sum is 0.

Quote:

(*) for some dividing P
E is the set of uppers sums, T is the set of lower sums
if infE=supT then its defferentiable.
Why are you saying "differentiable" instead of "integrable"?
• Sep 3rd 2011, 01:50 PM
transgalactic
Re: integrability question
dirichlet functio is defined
D(x) =1 for rational x
D(x) =0 for irrational x

$\displaystyle S(p)=\sum_{i=1}^{n}m_{i}(x_{i}-x_{i-1})=0$ so supT=0
$\displaystyle S(p)=\sum_{i=1}^{n}M_{i}(x_{i}-x_{i-1})=1$ so infE=1
supT is not equal infE so its not integrible
correct?
• Sep 3rd 2011, 02:04 PM
emakarov
Re: integrability question
To say whether this is correct, I need to know what are $\displaystyle m_i$, $\displaystyle M_i$, T and E, as well as why all $\displaystyle m_i=0$ and all $\displaystyle M_i=1$.