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Question Number 87013 by mathmax by abdo last updated on 01/Apr/20
calculate ∫_0 ^∞ (e^(−[x]) /(x+1))dx
calculate∫0∞e−[x]x+1dx
Commented by mathmax by abdo last updated on 02/Apr/20
I =∫_1 ^∞ (e^(−[x]) /(x+1))dx ⇒ I =Σ_(n=1) ^∞ ∫_n ^(n+1) (e^(−n) /(x+1))dx =Σ_(n=1) ^∞ e^(−n) [ln(x+1)]_n ^(n+1) =Σ_(n=1) ^∞ e^(−n) {ln(n+1)−ln(n)} =Σ_(n=1) ^∞ e^(−n) ln(((n+1)/n)) =Σ_(n=1) ^∞ e^(−n) ln(1+(1/n)) and this serie is convergent.
I=∫1∞e−[x]x+1dx⇒I=∑n=1∞∫nn+1e−nx+1dx=∑n=1∞e−n[ln(x+1)]nn+1=∑n=1∞e−n{ln(n+1)−ln(n)}=∑n=1∞e−nln(n+1n)=∑n=1∞e−nln(1+1n)andthisserieisconvergent.
Commented by mathmax by abdo last updated on 02/Apr/20
forgive the Q is calculate ∫_1 ^(+∞) (e^(−[x]) /(x+1))dx
forgivetheQiscalculate∫1+∞e−[x]x+1dx

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