Show $\lim_{x\to +\infty} \int_1^x \frac 1{x^\alpha+t^\beta}\;\mathrm{d} t = 0$.

Question

Show that $$ \lim_{x\to +\infty} \int_1^x \frac 1{x^\alpha+t^\beta}\;\mathrm{d} t = 0 $$ if $\alpha, \beta >0$ and $\max{\alpha, \beta} > 1$.

Answer 1

Let $\beta >1$. Let $\epsilon >0$ and choose $M$ such that $\int_M^{\infty} \frac 1 {x^{\alpha}+t^{\beta}} dt \leq \int_M^{\infty} \frac 1 {t^{\beta}} dt <\epsilon$. Note that $\int_1^{M} \frac 1 {x^{\alpha}+t^{\beta}} dt \leq \frac {M-1} {x^{\alpha}} \to 0$ so $\int_1^{M} \frac 1 {x^{\alpha}+t^{\beta}} dt <\epsilon$ for $x$ sufficiently large.

Next consider the case when $\alpha >1$. Here we have the obvious bound $\frac {x-1} {x^{\alpha}}$ which tends to $0$.