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Question Number 143860 by mathmax by abdo last updated on 19/Jun/21
calculate∫0∞xe−x2log(1+ex)dx
Answered by mathmax by abdo last updated on 20/Jun/21
Φ=∫0∞xe−x2log(1+ex)dx⇒Φ=[−12e−x2log(1+ex)]0∞−∫0∞−12e−x2×ex1+exdx=log22+12∫0∞e−x21+e−xdxbut∫0∞e−x21+e−xdx=∫0∞e−x2∑n=0∞e−nxdx=∑n=0∞∫0∞e−x2−nxdx=∑n=0∞∫0∞e−(x2+2n2x+n24−n24)dx=∑n=0∞en24∫0∞e−(x+n2)2dx=∑n=0∞en24∫n2∞e−z2dzifweputerf(λ)=∫λ∞e−z2dzweget∫0∞e−x21+e−xdx=∑n=0∞en24erf(n2)⇒Φ=log22+12∑n=0∞erf(n2)en24
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