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Question Number 150627 by puissant last updated on 14/Aug/21
∫01(−1)E(1x)dxx
Answered by puissant last updated on 14/Aug/21
∫01(−1)E(1x)xdx∀x∈]0;1[,E(1x)existeetE(1x)=k⇔k⩽1x<k+1⇔x∈]1k+1;1k]soitX∈]0;1],∃!n/1n+1<X⩽1n∧limX→0+n=+∞ona:∫X1(−1)E(1x)xdx=∫X1n(−1)nxdx+∑n−1k=1∫1k+11k(−1)kxdx=(−1)nln(1nX)+∑n−1k=1(−1)kln(k+1k)ornn+1<nX⩽1⇒limX→0+ln(1nX)=0(−1)kln(1+1k)converge.D′oulimX→0+∫X1(−1)E(1x)dx=ln(2π)...
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