Convergence of the improper integral $\int_{0}^{\infty}\frac{x^{p-1}}{1+qx}dx$

Question

I found that the following converges where ${0<p<1}$,and ${0<q}$, but I'm having some trouble where q is negative.

Because it has some "blow up" point, it seems to diverge, but i'm not sure.

Any help will be greatly appreciate.

$$\int \limits_{0}^{\infty}\frac{x^{p-1}}{1+qx}dx$$

Answer 1

Clearly $q$ should be greater or equal $0$.

Moreover, if $q=0$ we have $$\frac{x^{p-1}}{1+qx}=\frac1{x^{1-p}}\not\in L^1(0,\infty),\;\forall p$$ and if $q>0$ then $$\frac{x^{p-1}}{1+qx}\sim_\infty \frac1q\frac{1}{x^{2-p}}\in L^1([1,\infty))\iff 2-p>1\iff p<1$$ and $$\frac{x^{p-1}}{1+qx}\sim_0\frac1{x^{1-p}}\in L^1(0,1)\iff 1-p<1\iff 0<p$$

Conclusion The given integral exists if and only if $(0<p<1)\land (q>0)$.