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Question Number 207906 by nachosam last updated on 30/May/24
help∫1∞x−ln(x)dx
Answered by Berbere last updated on 30/May/24
∫0∞(et)(−t)etdt=∫0∞e−t2+tdt=∫0∞e−(t−12)2+14dt=e14∫−12∞e−x2dx=e14∫0∞e−x2dx+e14∫−120e−x2dxA=e14.π2−e14π2erf(−12)erf(x)=2π∫0xe−t2dt⇒erf(−x)=−erf(x)A=π2e14(1+erf(12))
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