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Evaluate the integral 1 I=∫ |d cos(ln (1/x))/dx| dx, ,n∈Z (positive integer). e^(−2nπ) Since ln(1/x)=−lnx and cosine is even, differentiating gives d cos(lnx)/dx=-sin(lnx)/x. Using u=−lnx, the integral becomes 2nπ ∫ ∣sinu∣du. 0 Because ∣sinu∣ has period π, with area 2 over each period, the interval contains 2n complete ...
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