Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 52550 by Tawa1 last updated on 09/Jan/19

  ∫_0 ^( ∞)    (x/(e^x  − 1))  dx

0xex1dx

Commented by maxmathsup by imad last updated on 11/Jan/19

let I =∫_0 ^∞   (x/(e^x −1))dx ⇒ I =∫_0 ^∞   ((xe^(−x) )/(1−e^(−x) ))dx =∫_0 ^∞  xe^(−x) (Σ_(n=0) ^∞  e^(−nx) )dx  =Σ_(n=0) ^∞   (∫_0 ^∞   x e^(−(n+1)x) dx)=Σ_(n=0) ^∞ A_n      A_n =∫_0 ^∞   x e^(−(n+1x) dx =_((n+1)x=t )    ∫_0 ^∞   (t/(n+1)) e^(−t)   (dt/(n+1)) =(1/((n+1)^2 ))∫_0 ^∞   t e^(−t) dt but  by parts ∫_0 ^∞  t e^(−t) dt =[−t e^(−t) ]_0 ^(+∞)  +∫ e^(−t) dt =[−e^(−t) ]_0 ^(+∞) =1 ⇒A_n =(1/((n+1)^2 )) ⇒  I =Σ_(n=0) ^∞  (1/((n+1)^2 )) =Σ_(n=1) ^∞  (1/n^2 ) =(π^2 /6) .

letI=0xex1dxI=0xex1exdx=0xex(n=0enx)dx=n=0(0xe(n+1)xdx)=n=0AnAn=0xe(n+1xdx=(n+1)x=t0tn+1etdtn+1=1(n+1)20tetdtbutbyparts0tetdt=[tet]0++etdt=[et]0+=1An=1(n+1)2I=n=01(n+1)2=n=11n2=π26.

Answered by Smail last updated on 09/Jan/19

I=∫_0 ^∞ (x/(e^x −1))dx=∫_0 ^∞ ((xe^(−x) )/(1−e^(−x) ))dx  =∫_0 ^∞ Σ_(n=0) ^∞ xe^(−x) e^(−nx) dx  =Σ_(n=0) ^∞ ∫_0 ^∞ xe^(−(n+1)x) dx  by parts  u=x⇒u′=1  v′=e^(−(n+1)x) ⇒v=(1/(−(n+1)))e^(−(n+1)x)   I=Σ_(n=0) ^∞ (1/(n+1))∫_0 ^∞ e^(−(n+1)x) dx  =Σ_(n=0) ^∞ (1/((n+1)^2 ))  =Σ_(n=1) ^∞ (1/n^2 )=ζ(2)=(π^2 /6)  ∫_0 ^∞ (x/(e^x −1))dx=(π^2 /6)

I=0xex1dx=0xex1exdx=0n=0xexenxdx=n=00xe(n+1)xdxbypartsu=xu=1v=e(n+1)xv=1(n+1)e(n+1)xI=n=01n+10e(n+1)xdx=n=01(n+1)2=n=11n2=ζ(2)=π260xex1dx=π26

Commented by Tawa1 last updated on 09/Jan/19

God bless you sir

Godblessyousir

Terms of Service

Privacy Policy

Contact: info@tinkutara.com