Relationship between integral of a function and integral of square of function

Question

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 $$\int_{0}^\infty \omega^\alpha K_\nu\left(\omega\right)d\omega = 2^{\alpha-1} \Gamma\left(\frac{\alpha+1-\nu}{2}\right) \Gamma\left(\frac{\alpha+1+\nu}{2}\right)$$ 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 $$\int_{0}^\infty (\omega^\alpha K_\nu(\omega))^2d\omega$$

Answer 1

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 $\int_0^\infty g^2(x)~dx=\infty$.

Answer 2

If some wonder, i found the solution to my integral, with no obvious relation to the one of the function not squared :)

$$\int_{0}^\infty (\omega^\alpha K_\nu(\omega))^2d\omega = \frac{\sqrt{\pi}}{4} \frac{\Gamma(\alpha+1/2-\nu) \Gamma(\alpha+1/2)\Gamma(\alpha+1/2+\nu)}{\Gamma(\alpha+1)}$$