Relationship between integral of a function and integral of square of function Is there a relationship between the results of $$\int_{0}^\infty f(x)dx$$ and $$\int_{0}^\infty f(x)^2dx$$
if f is positive definite?
EDIT: I'm adding more details concerning the specific functions i'm dealing with.
So far I have found $$\int_{0}^\infty \omega^\alpha K_\nu(\omega)d\omega = 2^{\alpha-1} \Gamma(\frac{\alpha+1-\nu}{2}) \Gamma(\frac{\alpha+1+\nu}{2})$$
with $\alpha=0, \frac12, 1, \frac32 ,...$ and $\nu=0, \frac12, 1$, and $K_\nu$ is the modified Bessel function of 2nd kind.
I would like to find $$\int_{0}^\infty (\omega^\alpha K_\nu(\omega))^2d\omega$$
 A: If we talk about Lebesgue-integrals, we can consider
$$
\int_0^\infty f(x)~dx=\int_{\{x\geq 0~:~f(x)\leq 1\}}f(x)~dx+
\int_{\{x\geq 0~:~f(x)> 1\}}f(x)~dx.
$$
So you see
$$
\int_{\{x\geq 0~:~f(x)\leq 1\}}f(x)~dx\geq \int_{\{x\geq 0~:~f(x)\leq 1\}}f^2(x)~dx
$$
and
$$
\int_{\{x\geq 0~:~f(x)> 1\}}f(x)~dx\leq \int_{\{x\geq 0~:~f(x)> 1\}}f^2(x)~dx.
$$
So you can't really compare the integrals because the inequality are different. Moreover consider the sequence of functions $f_n:[0,\infty)\to[0,\infty)$ defined by 
$$
f(x)=\begin{cases}
\frac1{n^{3/4}} & x\leq n\\0 & x>n
\end{cases}
$$
Then you get $$
\int_0^\infty f_n(x)~dx=n^{1/4}\to\infty\text{ for }n\to\infty
$$
while
$$
\int_0^\infty f_n^2(x)~dx=n^{-1/2}\to0\text{ for }n\to\infty
$$
But you can also do it the other way. Define
$$
g_n(x)=\begin{cases}
n^{3/4} & x\leq \frac1n\\ 0 & x>\frac1n
\end{cases}.
$$
Then you get $$
\int_0^\infty g_n(x)~dx=n^{-1/4}\to 0\text{ for }n\to\infty
$$
while
$$
\int_0^\infty g_n^2(x)~dx=n^{1/2}\to\infty\text{ for }n\to\infty
$$
So you see that there is no simple relation between $\int_0^\infty f(x)~dx$ and $\int_0^\infty f^2(x)~dx$.
You get go even further. For $$
f(x)=\begin{cases}\frac1x & x>0\\0 & x=0\end{cases}
$$
you get $\int_0^\infty f(x)~dx=\infty$ while $\int_0^\infty f^2(x)~dx=1$ and for
$$
g(x)=\begin{cases}\frac1{\sqrt{x}} & x\leq 1\\0 & x>1\end{cases}
$$
you get $\int_0^\infty g(x)~dx=\frac23$ while $\int_0^\infty g^2(x)~dx=\infty$.
A: If some wonder, i found the solution to my integral, with no obvious relation to the one of the function not squared :)
$$\int_{0}^\infty (\omega^\alpha K_\nu(\omega))^2d\omega = \frac{\sqrt{\pi}}{4} \frac{\Gamma(\alpha+1/2-\nu) \Gamma(\alpha+1/2)\Gamma(\alpha+1/2+\nu)}{\Gamma(\alpha+1)}$$
