niceintegral
Integral involving the Residue theorem
Today we present here the proof of this nice integral. The solution involves Complex Analysis, geometric series, even functions and trigonometry
∫∞0cos√xe2π√x−1dx=∫∞0[∞∑n=1e−2πn√xcos√x]dx=∞∑n=1∫∞0e−2πn√xcos√xdx=∞∑n=1∫∞0e−2πn√x(ei√x+e−i√x2)dx=∞∑n=1∫∞0e√x(i−2πn)+e√x(−i−2πn)2dx=12∞∑n=1∫∞0e√x(i−2πn)dx+12∞∑n=1∫∞0e√x(−i−2πn)dx=12∞∑n=1e√x(i−2πn)(−1+(i−2πn)√x)(i−2πn)2|∞0+12∞∑n=1e√x(−i−2πn)(−1+(−i−2πn)√x)(−i−2πn)2|∞0=∞∑n=11(i−2πn)2+∞∑n=11(−i−2πn)2=∞∑n=1[1(i−2πn)2+1(−i−2πn)2]=2+∞∑n=0[1(i−2πn)2+1(−i−2πn)2]=1+12∞∑n=−∞1(i−2πn)2+12∞∑n=−∞1(−i−2πn)2=1−12Res(πcotπz(i−2πz)2,i2π)−12Res(πcotπ(−i−2πz)2,−i2π)=1−12limz→i2π[ddz(z−i2π)2(πcotπz)(i−2πz)2]−12limz→−i2π[ddz(z+i2π)2(πcotπz)(−i−2πz)2]=1−18πlimz→i2πddzcotπz−18πlimz→−i2πddzcotπz=1+18csc2(i2)+18csc2(−i2)=1−18(2ie−12−e12)2−18(2ie12−e−12)2=1−e(e−1)2
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