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Question Number 140310 by liberty last updated on 06/May/21
provethat∫0∞lnxx2+1dx=0
Answered by qaz last updated on 06/May/21
∫0∞lnx1+x2dx=∫∞0−lnx1+x2(−1)dx=−∫0∞lnx1+x2dx=0
Commented by TheSupreme last updated on 06/May/21
t=1xdx=−1t2dt∫0∞ln(x)1+x2dx=∫∞0ln(1t)1+1t2(−1t2)dt=∫∞0ln(t)t2+1dtI=−I=0
Answered by mathmax by abdo last updated on 06/May/21
∫0∞lnxx2+1dx=−12Re(ΣRes(fai))withf(z)=log2zz2+1f(z)=log2z(z−i)(z+i)⇒Res(f,i)=log2i2i=(log(eiπ2))22i=(iπ2)22i=−π28iRes(f,−i)=log2(−i)2i=log2(e−iπ2)2i=−π28i⇒ΣRes=iπ28+iπ28=iπ24⇒Re(ΣRes)=0⇒∫0∞logxx2+1dx=0
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