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48

Part of 2005 Today's Calculation Of Integral

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Today's Calculation of Integral 48

Source: 1969 Yokohama City University

6/19/2005
Evaluate
lim⁡n→∞(∫0πsin⁡2nxsin⁡xdx−∑k=1n1k)\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin ^ 2 nx}{\sin x}dx-\sum_{k=1}^n \frac{1}{k}\right)
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