2019 ROMANIAN MATHEMATICAL OLYMPIAD

We show that $b_0=1+1/a$ is the only real number satisfying the required condition. Let $b\ge 1$, and let $f_b,F_b:[0,\infty )\to \mathbb{R}$, $f_b(x)=(1+x^a{)}^{-b}$ and $F_b(x)={\int }_0^x{f}_b(t)dt$. Since $f_b$ is positive, $F_b$ is increasing, so the limit $I(b)={\lim }_{x\to \infty }{F}_b(x)$ exists. Since $0\le f_b(x)\le f_1(x)$ on the ray $x\ge 0$, $$\begin{array}{rl}0\le F_b(x)& \le F_1(x)={\int }_0^1(1+t^a{)}^{-1}dt+{\int }_1^x(1+t^a{)}^{-1}dt\le 1+{\int }_1^x{t}^{-a}dt\\ & =1+1/(a-1)-x^{1-a}/(a-1)\le a/(a-1),\end{array}$$ whatever $x\ge 1$, so $I(b)$ is a real number.

$$\begin{array}{rl}I(b)& ={\int }_0^1{f}_b(t)dt+\underset{x\to \infty }{\lim }{\int }_1^x{f}_b(t)dt>{\int }_0^1{f}_c(t)dt+\underset{x\to \infty }{\lim }{\int }_1^x{f}_b(t)dt\\ & \ge {\int }_0^1{f}_c(t)dt+\underset{x\to \infty }{\lim }{\int }_1^x{f}_c(t)dt=I(c).\end{array}$$

Finally, we show that $I(1+1/a)=1$. To this end, integrate by parts to write $$\begin{array}{rl}{\int }_0^x(1+t^a{)}^{-1/a}dt& =t(1+t^a{)}^{-1/a}{|}_0^x+{\int }_0^x{t}^a(1+t^a{)}^{-1-1/a}dt\\ & =x(1+x^a{)}^{-1/a}+{\int }_0^x(1+t^a{)}^{-1/a}dt-F_{1+1/a}(x).\end{array}$$

Consequently, $F_{1+1/a}(x)=x(1+x^a{)}^{-1/a}$, so $I(1+1/a)=1$, and $b_0=1+1/a$, by injectivity.