For which $p$ is $\frac{1}{x^a+x^b}$ in $\cal{L}^p$? Let $f(x)=\frac{1}{x^a+x^b}$ with $x,a,b>0$. For which $p\ge1$ is $f$ in $\cal{L}^p(\lambda)$ over the interval $(0,\infty)$? Here $\lambda$ is the one dimensional Lebesgue measure.
Attempt: 
We are given, for $x>0$, that $x^t \in {\cal{L}}^1((0,1),\lambda))$ if and only if $t>-1$ and $x^t \in {\cal{L}}^1([1,\infty),\lambda))$ if and only if $t<-1$. $(*)$
Consider $p=1$ first.
Suppose $a\le b$. For $0<x<1$ we have 
$\frac{1}{2x^a}\le \frac{1}{x^a+x^b}\le \frac{1}{x^a}$, 
and for $x\ge1 $ we have
$\frac{1}{2x^b}\le \frac{1}{x^a+x^b}\le \frac{1}{x^b}$.
Hence  $\frac{1}{x^a+x^b}$ is in ${\cal{L}}^1((0,\infty),\lambda)$ if and only if both $a<1$ and $b>1$ from $(*)$.
Now let $p\ge1$.
Raising the two inequalities above to the $p$th power we see that we need both $ap<1$ and $bp>1$ from $(*)$.
We conclude that:
Given $a<1$ and $b>1$, $\frac{1}{x^a+x^b}$ is in ${\cal{L}}^p((0,\infty),\lambda)$ for $\frac{1}{b}<p<\frac{1}{a}$.
Is this correct?
Edit: After reading Jack's answer, I believe that the conclusion is as follows.
Given the values of $0<a\le b, f(x)$ will be in ${\cal{L}}^p(\lambda,(0,\infty))$ for $p\ge1$ provided we can find such a value of $p\ge1$ satisfying both inequalities $ap<1$ and $bp>1$. 

Your work is indeed correct although you do not need to consider the cases $p=1$ and $p>1$ separately. In both cases the scenarios are identical.
