say, $a>-1$ then $b>a+1$ show that the following function is convergent:
$$\int_0^\infty\frac{x^a}{1+x^b}dx$$ ,
i already rack my my mind for this , but i just can't get proper answer , surely some values of $a$ and $b$ make possible to obtain $arctan$ that convergence even in infinity . i tried plug-in some number within the demanded condition into calculator and it show the convergence with combination of $ln$ that beautifully leaving only constant and $arctan$
but i can't bring myself to mathematically prove this , then i found an answer that use comparison theorem with $$ \frac{x^a}{x^b}>\frac{x^a}{1+x^b}$$ , working the integral like usual, and from the condition is clear that we would has negative power because both $a$ and $b$ is at least differs by 1 . but what i can't accept is the solution just plug zero in the later part of the integral , so it basically turn like this
$$\int_0^\infty x^{a-b}dx=\lim_{d\rightarrow\infty}\vert\frac{x^{a-b+1}}{a-b+1}-0\vert_0^d$$ from here since the power negative of course the result tent to zero (convergent) so is there any explanation for that zero in later part ? instead of calculating it properly. cause we all know zero that had negative power tend to infinity , thanks in advanced
ps, if by any chance my post is unclear , i can provide link for the answer
Let $g(x)=x^{a}$ if $x<1$ and $\frac 1{x^{b-a}}$ if $x \geq 1$. Show that $g$ is integrable by splitting the integral into $(0,1)$ and $(1,\infty)$. Now verify that $\frac{x^{a}} {1+x^{b}} \leq g(x)$ for all $x$,