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Question Number 27003 by Joel578 last updated on 01/Jan/18

$$\underset{1/8}{\stackrel{1/2}{\int }}\lfloor \ln \lceil \frac{1}{x}\rceil \rfloor dx$$

Commented by abdo imad last updated on 01/Jan/18

$$ifyoumeantheintegralI={\int }_{\frac{1}{8}}^{\frac{1}{2}}ln\left[\frac{1}{x}\right]dxwedothechangement$$$$\frac{1}{x}=t\Rightarrow I=-{\int }_8^{2}ln\left[t\right]\frac{dt}{t^2}={\int }_2^8\frac{ln\left(\left[t\right]\right)}{t^2}dt=\sum _{k=2}^7{\int }_k^{k+1}\frac{lnk}{t^2}dt$$$$=\sum _{k=2}^{k=7}ln\left(k\right){[-\frac{1}{t}]}_k^{k+1}=\sum _{k=2}^{k=7}(\frac{1}{k}-\frac{1}{k+1})ln\left(k\right)$$$$=\sum _{k=2}^7\frac{ln\left(k\right)}{k(k+1)}=\frac{ln\left(2\right)}{6}+\frac{ln\left(3\right)}{12}+\frac{ln\left(4\right)}{20}+\frac{ln\left(5\right)}{30}+\frac{ln\left(6\right)}{42}+\frac{ln\left(7\right)}{56}.$$

Commented by Joel578 last updated on 02/Jan/18

$$thankyouverymuch$$

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