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for what $p$ and $q$ Is $\int_0^\infty \frac{dx}{x^p + x^q}$ convergent?

Answer: $(p-1)(q-1) \lt 0$

I need help. I don’t know how to get this answer.

I thought maybe I could solve this by trying different cases. Making $q=p$ Made the integral divergent so they have to be different.

Thus, if $p \lt q$ (or the opposite), this would result in and integral like this one: $$\int_0^\infty \frac{dx}{x^p(1+x^{q-p})}$$ which could be solved by partial fractions if I’d knew the value of $q-p$.

I don’t know how to proceed from here. I’d appreciate any help.

Note: The book first asked for an integral in such way to be solved:$$\int_0^\infty\frac{du}{u^{1/2}+u^{3/2}}= \pi$$So the answer holds for this example.

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    $\begingroup$ First, split into two integrals, $\int_0^1$ and $\int_1^\infty$. Check what is the criteria for each to be convergent. $\endgroup$ – Dennis Gulko Jan 28 '20 at 21:25
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You can assume w.l.o.g. that $p\leq q$. $$\int_0^{\infty}\frac{1}{x^p+x^q}\,dx=\int_0^{1}\frac{1}{x^p+x^q}\,dx+\int_1^{\infty}\frac{1}{x^p+x^q}\,dx $$ Both integrals on the RHS need to be convergent. You're right about the case $p=q$ since $\frac 12\int_0^{1}\frac{1}{x^p}\,dx$ is convergent for $p<1$ while $\frac 12\int_1^{\infty}\frac{1}{x^p}\,dx$ is convergent for $p>1$ and both conditions can't be true. Let $p<q$. Then $$\frac{1}{x^p+x^q}=\frac{1}{x^p(1+x^{q-p})}\sim\frac{1}{x^p}\,\,(x\to 0^+) \\ \frac{1}{x^p+x^q}=\frac{1}{x^q(1+x^{p-q})}\sim\frac{1}{x^q}\,\,(x\to\infty)$$ So the two integrals on the RHS are convergent if and only if $\int_0^{1}\frac{1}{x^p}\,dx$ and $\int_1^{\infty}\frac{1}{x^q}\,dx$ are convergent, respectively. That gives us $p<1$ and $q>1$. Hence, the answer is $\min(p,q)<1$ and $\max(p,q)>1$. $(p-1)(q-1)<0$ is another way to state the answer.

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