| 20 | 匿名アカウント | 2022/01/09/17:40 |
Set $f_{n}(x) = \displaystyle\frac{nx^{s}}{1+nx}.$
Suppose that $1\le m<n.$ Then, since $x\in[1,\infty),$ it holds
$$\frac{f_{n}}{f_{m}}(x)=\frac{\frac{nx^{s}}{1+nx}}{\frac{mx^{s}}{1+mx}}=\frac{n(1+mx)}{m(1+nx)}=\frac{n+mnx}{m+mnx}>1$$
and we have $\{f_{n}\}$ is a pointwisely strictly monotonely increasing sequence w.r.t. $n.$
Hence we can apply Theorem $4-3$ and have
$$
\displaystyle\lim_{n\to\infty}\int_{1}^{\infty}f_{n}(x)dx
=\int_{1}^{\infty}\lim_{n\to\infty}f_{n}(x)dx
=\int_{1}^{\infty}f(x)dx.
$$
We conclude that
$$
\displaystyle\lim_{n\to\infty}\int_{1}^{\infty}f_{n}(x)dx
=\int_{1}^{\infty}f(x)dx
=\left[\dfrac{x^{s}}{s}\right]_{1}^{\infty}
=\left\{
\begin{array}{lcl}
-s^{-1} &\quad& s<0,\\
\infty &\quad& s\ge0.\\
\end{array}
\right.
$$
Here we assume $1^{r}\in\mathbb{R}$ where $r\in\mathbb{R},$ especially $r\in\mathbb{R}\backslash\mathbb{Z}.$
