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Question Number 115302 by mnjuly1970 last updated on 24/Sep/20

$$...nicemath...$$$$find$$$${lim}_{n\to \mathrm{\infty }}\left\{\underset{k=1}{\sum ^n}{\int }_{k-1}^k{tan}^{-1}\left(\frac{nx-nk}{kx+n^2}\right)dx\right\}$$$$$$

Answered by Olaf last updated on 24/Sep/20

$$\arctan a-\arctan b=\arctan \left(\frac{a-b}{1+ab}\right)$$$$\mathrm{Let}a=\frac{x}{n}\mathrm{and}b=\frac{k}{n}$$$$\arctan \frac{x}{n}-\arctan \frac{k}{n}=\arctan \left(\frac{\frac{x}{n}-\frac{k}{n}}{1+\frac{xk}{n^2}}\right)$$$$\arctan \frac{x}{n}-\arctan \frac{k}{n}=\arctan \left(\frac{nx-nk}{kx+n^2}\right)$$$${\mathrm{I}}_k\left(x\right)={\int }_{k-1}^k\arctan \left(\frac{nx-nk}{kx+n^2}\right)dx$$$${\mathrm{I}}_k\left(x\right)={\int }_{k-1}^k[\mathrm{arct}an\left(\frac{x}{n}\right)-\arctan \left(\frac{k}{n}\right)]dx$$$${\mathrm{I}}_k\left(x\right)={\left[x\mathrm{arct}an\left(\frac{x}{n}\right)\right]}_{k-1}^k...$$$$-{\int }_{k-1}^{k}x\frac{\frac{1}{n}}{1+{\left(\frac{x}{n}\right)}^2}dx-\arctan \frac{k}{n}$$$${\mathrm{I}}_k\left(x\right)=k\mathrm{arct}an\frac{k}{n}-(k-1)\arctan \left(\frac{k-1}{n}\right)...$$$$-{\left[\frac{1}{2}\ln (1+{\left(\frac{x}{n}\right)}^2)\right]}_{k-1}^k-\arctan \frac{k}{n}$$$${\mathrm{I}}_k\left(x\right)=k\arctan \frac{k}{n}-(k-1)\arctan \left(\frac{k-1}{n}\right)...$$$$-\frac{1}{2}\ln \frac{1+{\left(\frac{k}{n}\right)}^2}{1+{\left(\frac{k-1}{n}\right)}^2}-\arctan \frac{k}{n}$$$$\underset{k=1}{\sum ^n}{\mathrm{I}}_k\left(x\right)=n\arctan \left(1\right)-\frac{1}{2}\ln 2-\underset{k=1}{\sum ^n}\arctan \left(\frac{k}{n}\right)$$$$\underset{k=1}{\sum ^n}{\mathrm{I}}_k\left(x\right)=\frac{\pi }{4}n-\frac{1}{2}\ln 2-\underset{k=1}{\sum ^n}\arctan \left(\frac{k}{n}\right)$$$$tobecontinued...$$$$$$

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