Question Number 104028 by mathmax by abdo last updated on 19/Jul/20

calculate  ∫_(20) ^(+∞)   (dx/((x−18)^(19) (x−19)^(18) ))

Answered by Dwaipayan Shikari last updated on 19/Jul/20

∫_(20) ^(+∞) (((x−19)dx)/((x^2 −37x+342)^(19) ))  (1/2)∫_(20) ^(+∞) ((2x−37)/((x^2 −37x+342)^(19) ))dx−(1/((x^2 −37+342)^(19) ))  −(1/(36))[(1/((x^2 −37x+342)^(18) ))]_(20) ^(+∞) −(1/2)∫(1/((x−18)^(19) (x−19)^(19) ))  (1/(36)).(1/2^(18) )−{(1/2)∫_(20) ^(+∞) (1/((x−18)^(19) (x−19)^(19) ))}→I_a   (1/(36.2^(18) ))−I_a     continue......

Answered by mathmax by abdo last updated on 19/Jul/20

A =∫_(20) ^(+∞)  (dx/((x−18)^(19) (x−19)^(18) ))  changement x−19 =t give  A =∫_1 ^(+∞)  (dt/((t+1)^(19) t^(18) )) =∫_1 ^∞  (dt/(((t/(t+1)))^(18)  (t+1)^(37) ))  we do the changement (t/(t+1)) =u ⇒  t =ut +u ⇒(1−u)t =u ⇒t =(u/(1−u)) ⇒(dt/du) =((1−u+u)/((1−u)^2 )) =(1/((1−u)^2 ))  t+1 =(u/(1−u)) +1 =((u+1−u)/(1−u)) =(1/(1−u)) ⇒  A =∫_(1/2) ^1    (du/((1−u)^2  u^(18) ((1/(1−u)))^(37) ))  =∫_(1/2) ^1  (((1−u)^(37) )/((1−u)^2  u^(18) ))du   =∫_(1/2) ^1  (((1−u)^(35) )/u^(18) ) du =−∫_(1/2) ^1  (((u−1)^(35) )/u^(18) )du =−∫_(1/2) ^1  ((Σ_(k=0) ^(35 )  C_(35) ^k  u^k (−1)^(35−k) )/u^(18) )du  =Σ_(k=0) ^(35 )  (−1)^k  C_(35) ^k   ∫_(1/2) ^1  u^(k−18)  du  =Σ_(k=0 and k≠17) ^(35) (−1)^k  C_(35) ^k  [(1/(k−17)) u^(k−17) ]_(1/2) ^1   −C_(35) ^(17)  [ln∣u∣]_(1/2) ^1   A=Σ_(k=0and k≠17) ^(35)   (((−1)^k  C_(35) ^k )/(k−17)){1−(1/2^(k−17) )} −ln(2)C_(35) ^(17)