Finding $(p,q)$ such that $\frac{x^p}{1+x^q}$ is integrable on $(0,+\infty)$

Question

I'm trying to show that $f(x) = \frac{x^p}{1+x^q}$ is integrable on $(0,\infty)$ if and only if $p > -1$ and $q-p > 1$.

So on $[1,\infty)$ we can compare with $g(x) = x^{p-q}$ which is integrable when $q-p>1$.

Now I am having difficulty with the region $(0,1)$. As well as that, I'm not sure how to prove $f$ is not integrable on $[1,\infty)$ if $q-p \le 1$.

Thanks for any help

Answer 1

Generally, for $\alpha,\beta\in \mathbb{R}$ and $x\in (0,\infty)$, we have

$$\max \{x^\alpha,x^\beta\} < x^\alpha + x^\beta \leqslant 2 \max \{x^\alpha,x^\beta\},$$

and therefore

$$\frac{1}{2} \min \left\{\frac{1}{x^\alpha},\frac{1}{x^\beta}\right\} \leqslant \frac{1}{x^\alpha+x^\beta} < \min \left\{\frac{1}{x^\alpha},\frac{1}{x^\beta}\right\}.$$

Thus $\frac{1}{x^\alpha+x^\beta}$ is integrable over $[1,\infty)$ if and only if $\max \{\alpha,\beta\} > 1$, and it is integrable over $(0,1)$ if and only if $\min \{\alpha,\beta\} < 1$.

Answer 2

I am not good at math theory, so I will try to help you by evaluating the integral. Consider $$ \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx. $$ It is easy to notice that $$ \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx>0. $$ Now, let us evaluate the integral. Let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\Large\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\Large\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then \begin{align} \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\Large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\Large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\Large1-\frac{a}{b}-1}(1-y)^{\Large\frac{a}{b}-1}\ dy, \end{align} where the last integral in RHS is Beta function. $$ \text{B}(x,y)=\int_0^1t^{\ \Large x-1}\ (1-t)^{\ \Large y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Hence \begin{align} \int_0^\infty\dfrac{x^{\Large a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{\Large1-\frac{a}{b}-1}(1-y)^{\Large\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&=\large{\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}. \end{align} The last part uses Euler's reflection formula for Gamma function provided $\color{red}{0<a<b}$. Why?

Since $$ \int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx=\frac{\pi}{b\sin\left(\dfrac{a\pi}{b}\right)}>0 $$ then $a<b$ and $b>0$.