Question Number 102985 by DGmichael last updated on 12/Jul/20

Answered by mathmax by abdo last updated on 12/Jul/20

at form of serie    let f(x) =∫_0 ^x  (e^t /((1+t^2 )^2 ))dt  if o≤x<1  we  have  (1/(1+u)) =Σ_(n=0) ^∞  (−1)^n  u^(n )  ⇒by derivation we get  −(1/((1+u)^2 )) =Σ_(n=1) ^∞  n(−1)^n  u^(n−1)  ⇒(1/((1+u)^2 )) =Σ_(n=1) ^∞  n(−1)^(n−1)  u^(n−1)  ⇒  (1/((1+t^2 )^2 )) =Σ_(n=1) ^∞  n(−1)^(n−1)  t^(2n−2)   ⇒f(x) =∫_0 ^(x ) e^t (Σ_(n=1) ^(∞ ) n(−1)^(n−1 ) t^(2n−2) )dt  =Σ_(n=1) ^∞  n(−1)^(n−1)  ∫_0 ^(x ) t^(2n−2)  e^t  dt  =Σ_(n=1) ^∞  n(−1)^(n−1)  U_n   U_n =∫_0 ^x  t^(2n−2 )  e^t  dt   by parts  u^′  =t^(2n−2)  and v =e^t  ⇒  U_n =[(1/(2n−1)) t^(2n−1)  e^t ]_0 ^x  −∫_0 ^x  (1/(2n−1)) t^(2n−1)  e^(t ) dt  =((x^(2n−1)  e^x )/(2n−1)) −(1/(2n−1)) ∫_0 ^x  t^(2n−1)  e^t  dt and  ∫_0 ^x  t^(2n−1)  e^t  dt =[(t^(2n) /(2n)) e^t ]_0 ^x  −∫_0 ^x  (t^(2n) /(2n))e^t  dt =((x^(2n)  e^x )/(2n)) −(1/(2n)) ∫_0 ^x  t^(2n)  e^t  dt  =((x^(2n)  e^x )/(2n))−(1/(2n))U_(n+1)  ⇒U_n =((x^(2n−1)  e^x )/(2n−1))−(1/(2n−1)){((x^(2n)  e^x )/(2n))−(1/(2n)) U_(n+1) }  =((x^(2n−1)  e^x )/(2n−1))−((x^(2n)  e^x )/((2n−1)(2n))) +(1/(2n(2n−1))) U_(n+1)   ...be continued

Answered by maths mind last updated on 12/Jul/20

∫(e^x /(t+x^2 ))dx=f(t)  =(1/(2i(√t)))∫{−(e^x /(x+i(√t)))+(e^x /(x−i(√t)))}dx  u=x+i(√t)⇒du=dx  =−(1/(2i(√t)))∫(e^(u−i(√t)) /u)du+(1/(2i(√t)))∫(e^(u+i(√t)) /u)du  =2Re∫(e^(u+i(√t)) /u)du  =2Re∫(e^u /u){cos((√t))+isin((√t)))du  =2∫(e^u /u)cos((√t))du=2cos((√t))E_i (u)  =2cos((√t))E_i (x+i(√t))+c=f(t)  we want−f′(1)  f′(t)=−2sin((√t))E_i (x+i(√t))+(i/(√t))cos((√t)).(e^(x+i(√t)) /(x+i(√t)))  −f′(1)=2sin(1)E_i (x+i)−((cos(1)e^x (cos(1)+isin(1)))/(x+i))+c  =∫(e^x /((1+x^2 )^2 ))dx