# Thread: square of a function is not integrable

1. ## square of a function is not integrable

Is there a function f: R->R is integrable but f^2 is not? why?

2. ## Re: square of a function is not integrable

Originally Posted by parklover
Is there a function f: R->R is integrable but f^2 is not? why?
Theorem: If $f$ is an integrable function in $[a,b]$ then $f^2$ is an integrable function in $[a,b]$.

P.S. In response to the to reply #3.
It depends on how one defines integral.
Most often Riemann integrals require bounded functions.
That example is not a bounced function.
The term is improper integral.

3. ## Re: square of a function is not integrable

Consider $f(x) =\begin{cases} \frac 1{\sqrt{|x|}}&\mbox{if } -1\leq x<0 \mbox{ or }01 \end{cases}$ and choose any value you want for $x=0$.

4. ## Re: square of a function is not integrable

Originally Posted by girdav
Consider $f(x) =\begin{cases} \frac 1{\sqrt |x|}&\mbox{if } -1\leq x<0 \mbox{ or }01 \end{cases}$ and choose any value you want for $x=0$.
sorry, I am slow. why is it not integrable？i think ln (x)

5. ## Re: square of a function is not integrable

The function $f$ is integrable but not its square. As you noticed, a primitive of $\frac 1x$ is $\ln x$, but if you compute $\int_{\varepsilon}^1\frac 1xdx$, you will find $-\ln \varepsilon$, and it has an infinite limit as $\varepsilon\to 0$.

6. ## Re: square of a function is not integrable

Originally Posted by parklover
sorry, I am slow. why is it not integrable？i think ln (x)
The function $\frac{1}{\sqrt{|x|}}$ is not Riemann integrable on $[-1,0]$.
However as an improper integral $\int_{ - 1}^0 {\frac{{dx}}{{\sqrt {\left| x \right|} }}}=2$.
But $\int_{ - 1}^0 {\frac{{dx}}{{ {\left| x \right|} }}}$ does not exist even as an improper integral.