Problem 4. Let $a$ be a real number, $a>1$. Determine the real numbers $b \geq 1$ such that

$$
\lim _{x \rightarrow \infty} \int_{0}^{x}\left(1+t^{a}\right)^{-b} \mathrm{~d} t=1
$$
Solution. Let $b \geq 1$ and the functions $f_{b}, F_{b}:[0, \infty) \rightarrow \mathbb{R}, f_{b}(x)=\left(1+x^{a}\right)^{-b}$ and $F_{b}(x)=$ $\int_{0}^{x} f_{b}(t) \mathrm{d} t$. Since $f_{b}$ is positive, it follows that $F_{b}$ is increasing, so the limit $I(b)=$ $\lim _{x \rightarrow \infty} F_{b}(x)$ exists.

$1 p$

Since $0 \leq f_{b}(x) \leq f_{1}(x)$ for all $x \geq 0$, it follows that

$$
\begin{aligned}
0 \leq F_{b}(x) & \leq F_{1}(x)=\int_{0}^{1}\left(1+t^{a}\right)^{-1} \mathrm{~d} t+\int_{1}^{x}\left(1+t^{a}\right)^{-1} \mathrm{~d} t \leq 1+\int_{1}^{x} t^{-a} \mathrm{~d} t \\
& =1+1 /(a-1)-x^{1-a} /(a-1) \leq a /(a-1)
\end{aligned}
$$

for all $x \geq 1$. Therefore, $I(b)$ is real.

We show that the function $b \mapsto I(b)$ is strictly decreasing, hence injective. Let $1 \leq b < c$. Since $f_{b}(x) > f_{c}(x)$ for all $x > 0$, it follows that

$$
\begin{aligned}
I(b) & =\int_{0}^{1} f_{b}(t) \mathrm{d} t+\lim _{x \rightarrow \infty} \int_{1}^{x} f_{b}(t) \mathrm{d} t>\int_{0}^{1} f_{c}(t) \mathrm{d} t+\lim _{x \rightarrow \infty} \int_{1}^{x} f_{b}(t) \mathrm{d} t \\
& \geq \int_{0}^{1} f_{c}(t) \mathrm{d} t+\lim _{x \rightarrow \infty} \int_{1}^{x} f_{c}(t) \mathrm{d} t=I(c)
\end{aligned}
$$

We show that $I(1+1 / a)=1$. Indeed,

$$
\begin{aligned}
\int_{0}^{x}\left(1+t^{a}\right)^{-1 / a} \mathrm{~d} t & =\int_{0}^{x} t^{\prime}\left(1+t^{a}\right)^{-1 / a} \mathrm{~d} t=\left.t\left(1+t^{a}\right)^{-1 / a}\right|_{0} ^{x}+\int_{0}^{x} t^{a}\left(1+t^{a}\right)^{-1-1 / a} \mathrm{~d} t \\
& =x\left(1+x^{a}\right)^{-1 / a}+\int_{0}^{x}\left(1+t^{a}\right)^{-1 / a} \mathrm{~d} t-F_{1+1 / a}(x)
\end{aligned}
$$

It follows that $F_{1+1 / a}(x)=x\left(1+x^{a}\right)^{-1 / a}$, hence $I(1+1 / a)=1$ and, therefore, $b=1+1 / a$.