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Question Number 52550 by Tawa1 last updated on 09/Jan/19
∫0∞xex−1dx
Commented by maxmathsup by imad last updated on 11/Jan/19
letI=∫0∞xex−1dx⇒I=∫0∞xe−x1−e−xdx=∫0∞xe−x(∑n=0∞e−nx)dx=∑n=0∞(∫0∞xe−(n+1)xdx)=∑n=0∞AnAn=∫0∞xe−(n+1xdx=(n+1)x=t∫0∞tn+1e−tdtn+1=1(n+1)2∫0∞te−tdtbutbyparts∫0∞te−tdt=[−te−t]0+∞+∫e−tdt=[−e−t]0+∞=1⇒An=1(n+1)2⇒I=∑n=0∞1(n+1)2=∑n=1∞1n2=π26.
Answered by Smail last updated on 09/Jan/19
I=∫0∞xex−1dx=∫0∞xe−x1−e−xdx=∫0∞∑∞n=0xe−xe−nxdx=∑∞n=0∫0∞xe−(n+1)xdxbypartsu=x⇒u′=1v′=e−(n+1)x⇒v=1−(n+1)e−(n+1)xI=∑∞n=01n+1∫0∞e−(n+1)xdx=∑∞n=01(n+1)2=∑∞n=11n2=ζ(2)=π26∫0∞xex−1dx=π26
Commented by Tawa1 last updated on 09/Jan/19
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