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let-f-n-x-e-nx-2e-2nx-with-x-from-0-1-calculate-0-f-n-x-dx-and-n-0-0-f-n-x-dx-2-find-S-x-n-0-f-n-x-and-0-S-x-dx-




Question Number 46610 by maxmathsup by imad last updated on 29/Oct/18
let f_n (x)=e^(−nx) −2e^(−2nx)   with x from[0,+∞[  1)calculate ∫_0 ^∞  f_n (x)dx  and Σ_(n=0) ^∞  (∫_0 ^∞  f_n (x)dx)  2) find S(x)=Σ_(n=0) ^∞  f_n (x)  and ∫_0 ^∞  S(x)dx
letfn(x)=enx2e2nxwithxfrom[0,+[1)calculate0fn(x)dxandn=0(0fn(x)dx)2)findS(x)=n=0fn(x)and0S(x)dx
Commented by maxmathsup by imad last updated on 05/Nov/18
1)∫_0 ^∞   f_n (x)dx =∫_0 ^∞  (e^(−nx) −2e^(−2nx) )dx =∫_0 ^∞  e^(−nx) dx−2 ∫_0 ^∞  e^(−2nx) dx  =[−(1/n) e^(−nx) ]_0 ^∞  −2 [−(1/(2n)) e^(−2nx) ]_0 ^∞  =(1/n) −2((1/(2n))) =0  ⇒Σ_(n=0) ^∞ (∫_0 ^∞ f_n (x)dx)=0
1)0fn(x)dx=0(enx2e2nx)dx=0enxdx20e2nxdx=[1nenx]02[12ne2nx]0=1n2(12n)=0n=0(0fn(x)dx)=0
Commented by maxmathsup by imad last updated on 05/Nov/18
2) we have S(x)=Σ_(n=0) ^∞  e^(−nx) −2Σ_(n=0) ^∞  e^(−2nx)   =Σ_(n=0) ^∞  (e^(−x) )^n  −2 Σ_(n=0) ^(∞ )  (e^(−2x) )^n    and for x>0 we get  S(x)=(1/(1−e^(−x) )) −2 (1/(1−e^(−2x) ))  ∫ S(x)dx =∫  (dx/(1−e^(−x) )) −2 ∫  (dx/(1−e^(−2x) )) +c  but  ∫  (dx/(1−e^(−x) )) =∫  (e^x /(e^x −1)) dx =ln∣e^x −1∣  ∫   ((2dx)/(1−e^(−2x) )) =∫ ((2e^(2x) )/(e^(2x) −1)) dx =ln∣e^(2x) −1∣ ⇒∫ S(x)dx=ln∣((e^x −1)/(e^(2x) −1))∣ +c  =ln∣(1/(e^x  +1))∣ +c ⇒∫_0 ^∞  S(x)dx =[−ln(e^x  +1)]_0 ^(+∞)  =−∞ .
2)wehaveS(x)=n=0enx2n=0e2nx=n=0(ex)n2n=0(e2x)nandforx>0wegetS(x)=11ex211e2xS(x)dx=dx1ex2dx1e2x+cbutdx1ex=exex1dx=lnex12dx1e2x=2e2xe2x1dx=lne2x1S(x)dx=lnex1e2x1+c=ln1ex+1+c0S(x)dx=[ln(ex+1)]0+=.

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