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Q-0-x-c-c-x-dx-




Question Number 156470 by puissant last updated on 11/Oct/21
Q=∫_0 ^∞ (x^c /c^x )dx
Q=0xccxdx
Answered by puissant last updated on 11/Oct/21
posons c^x =e^u ⇒ e^(xlnc) =e^u ⇒ x=(u/(lnc))  ⇒ dx=(du/(lnc))  ⇒ Q=∫_0 ^∞ ((((u/(lnc)))^c )/e^u )×(du/(lnc)) = (1/((lnc)^(c+1) ))∫_0 ^∞ (u^c /e^u )du  =(1/((lnc)^(c+1) ))∫_0 ^∞ u^(c−1+1) e^(−u)  du           ∴∵ Q = ∫_0 ^∞ (x^c /c^x )dx=((Γ(c+1))/((lnc)^(c+1) ))..                   ..........Le puissant.........
posonscx=euexlnc=eux=ulncdx=dulncQ=0(ulnc)ceu×dulnc=1(lnc)c+10uceudu=1(lnc)c+10uc1+1eudu∴∵Q=0xccxdx=Γ(c+1)(lnc)c+1...Lepuissant

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