ผมเข้าใจว่าอาจารย์Punkน่าจะลืมลบพจน์ $\displaystyle{3\left(1-\frac{\pi}{4}\right)}$ ออกครับ
ถ้าลบพจน์นั้นออกด้วยคำตอบก็เท่ากันครับ
My Solution
$$I=\int_{0}^{1}\left( r\left\lfloor\frac{1}{x}\right\rfloor -\bigg\lfloor\frac{r}{x}\bigg\rfloor\right)dx,\forall r\in\mathbb{R}^{+}$$
Lemma : $\displaystyle{L=\lim_{n\rightarrow\infty}\left[\frac{1}{\lfloor an\rfloor+1}+\frac{1}{\lfloor an\rfloor+2}+\cdots+\frac{1}{\lfloor bn\rfloor}\right]=\ln\frac{b}{a},\forall a,b\in\mathbb{R},b>a>0}$
Proof :
$\displaystyle{L=\lim_{n\rightarrow\infty}\frac{1}{n}\left[\frac{1}{\frac{\lfloor an\rfloor}{n}+\left(\frac{1}{n}\right)}+\frac{1}{\frac{\lfloor an\rfloor}{n}+\left(\frac{2}{n}\right)}+\cdots+\frac{1}{\left(\frac{\lfloor bn\rfloor}{n}\right)}\right]=\lim_{n\rightarrow\infty}\int_{\frac{\lfloor an\rfloor}{n}}^{\frac{\lfloor bn\rfloor}{n}}\frac{dx}{x}}$
$\displaystyle{x-1<\lfloor x\rfloor\leq x\rightarrow\lim_{n\rightarrow\infty}\ln\left(\frac{bn-1}{an}\right)<L<\lim_{n\rightarrow\infty}\ln\left(\frac{bn}{an-1}\right)\rightarrow L=\ln\frac{b}{a}}$
Case : $r\in (0,1)\rightarrow I=-r\ln r$
$$I=\int_{0}^{1}\left( r\left\lfloor\frac{1}{x}\right\rfloor -\bigg\lfloor\frac{r}{x}\bigg\rfloor\right)dx=\lim_{n\rightarrow\infty}\int_{\frac{1}{n}}^{1}\left( r\left\lfloor\frac{1}{x}\right\rfloor -\bigg\lfloor\frac{r}{x}\bigg\rfloor\right)dx$$
$$=\lim_{n\rightarrow\infty}r\int_{\frac{1}{n}}^{1}\bigg\lfloor\frac{1}{x}\bigg\rfloor dx-\lim_{n\rightarrow\infty}\int_{\frac{1}{n}}^{1}\bigg\lfloor\frac{r}{x}\bigg\rfloor dx=r\lim_{n\rightarrow\infty}\left(\int_{rn}^{n}\frac{\lfloor x\rfloor}{x^{2}}dx-\int_{r}^{1}\frac{\lfloor x\rfloor}{x^{2}}dx\right)$$
$$=r\lim_{n\rightarrow\infty}\sum_{k=\lfloor rn\rfloor}^{n-1}\left[-\frac{k}{x}\right]_{k}^{k+1}=r\lim_{n\rightarrow\infty}\left[\frac{1}{\lfloor rn\rfloor+1}+\frac{1}{\lfloor rn\rfloor+2}+\cdots+\frac{1}{n}\right]=-r\ln r$$
Case : $r=1\rightarrow I=0$
Case : $r>1\rightarrow I=rH(\lfloor r\rfloor)-\lfloor r\rfloor-r\ln r$
$$I=\int_{0}^{1}\left( r\left\lfloor\frac{1}{x}\right\rfloor -\bigg\lfloor\frac{r}{x}\bigg\rfloor\right)dx=r\lim_{n\rightarrow\infty}\left(\int_{1}^{r}\frac{\lfloor x\rfloor}{x^{2}}dx-\int_{n}^{rn}\frac{\lfloor x\rfloor}{x^{2}}dx\right)$$
$$=r\left(\sum_{k=1}^{\lfloor r\rfloor-1}\left[-\frac{k}{x}\right]_{k}^{k+1}+\int_{\lfloor r\rfloor}^{r}\frac{\lfloor r\rfloor}{x^{2}}dx-\ln r\right)=rH(\lfloor r\rfloor)-\lfloor r\rfloor-r\ln r$$