Let $u(x)=(x^{\alpha}+x^{\beta})^{-1}$ $x,\alpha,\beta>0$.
For which $p\geq 1$ is true that $u\in\mathcal L^p(\lambda^1,(0,\infty))$?
My attempt $$\|u\|_p^p=\int_{(0,\infty)}(x^{\alpha}+x^{\beta})^{-p}\lambda(dx)$$
Then I divided the domain in two, leading to
$$=\int_{(0,1)} \bigg(\frac{1}{x^{\alpha}+x^{\beta}}\bigg)^{p}\lambda(dx)+\int_{[1,\infty)} \bigg(\frac{1}{x^{\alpha}+x^{\beta}}\bigg)^{p}\lambda(dx) \color{red}{*}$$
It's easy to see that $$\frac{1}{x^{\alpha}+x^{\beta}}\leq \frac{1}{2\cdot\min\{x^{\alpha},x^{\beta}\}}$$
Now for $x\in(0,1)$
$$\frac{1}{x^{\alpha}+x^{\beta}}\leq \frac{1}{2\cdot\min\{x^{\alpha},x^{\beta}\}}=\color{red}{\frac{1}{2\cdot x^{\max\{\alpha,\beta\}}}},$$ while for $x\in[1,\infty)$
$$\frac{1}{x^{\alpha}+x^{\beta}}\leq \frac{1}{2\cdot\min\{x^{\alpha},x^{\beta}\}}=\color{green}{\frac{1}{2\cdot x^{\min\{\alpha,\beta\}}}},$$
Now I can write that
$$\color{red}{*}\leq \int_{(0,1)} \bigg(\color{red}{\frac{1}{2\cdot x^{\max\{\alpha,\beta\}}}}\bigg)^p\lambda(dx)+\int_{[1,\infty)}\bigg(\color{green}{\frac{1}{2\cdot x^{\min\{\alpha,\beta\}}}}\bigg)^p\lambda(dx)$$
So basically we have to select $p\geq 1$ in such a way that both integrals are finite.
The first term is finite iff $$-\max\{\alpha,\beta\}p>-1$$ i.e. $$p<\frac{1}{\max\{\alpha,\beta\}}$$
And the second term will be finite iff
$$-\min\{\alpha,\beta\}p<-1$$ which implies $$p>\frac{1}{\min\{\alpha,\beta\}}$$
But then
$$\frac{1}{\min\{\alpha,\beta\}}<\frac{1}{\max\{\alpha,\beta\}}$$
which is a contradiction.
I am sure I've made some mistake in my calculation but I am not able to spot it.
Thanks in advance.
HINT: I didnt spot your possible mistake (maybe some of the chosen bounds is too weak?) but you can assume that $\alpha \leqslant \beta $, then you can took common factor and the integrand becomes $x^{-p\alpha }(1+x^{\beta -\alpha })^{-p}$. This simplifies slightly the analysis and you can see that
$$ \int_{(0,\infty )}\frac1{x^{\alpha p}(1+x^{\beta -\alpha } )^p}\,\mathrm d x<\infty \\ \iff \int_{(0,1)}\frac1{x^{\alpha p}}\,\mathrm d x<\infty \quad \text{ and }\quad \int_{[1,\infty )}\frac1{x^{\beta p }}\,\mathrm d x<\infty $$ what gives you $p\in(\tfrac{1}{\beta },\tfrac{1}{\alpha })\cap [1,\infty )$.