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

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.

• First, split into two integrals, $\int_0^1$ and $\int_1^\infty$. Check what is the criteria for each to be convergent. – Dennis Gulko Jan 28 '20 at 21:25

## 1 Answer

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 $$p. Then $$\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.