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


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

$calculate \int_{20}^{+ \infty} \frac{dx}{{(x - 18)}^{19} {(x - 19)}^{18}}$
	Answered by Dwaipayan Shikari last updated on 19/Jul/20
	$∫_(20) ^(+∞) (((x−19)dx)/((x^2 −37x+342)^(19) )) (1/2)∫_(20) ^(+∞) ((2x−37)/((x^2 −37x+342)^(19) ))dx−(1/((x^2 −37+342)^(19) )) −(1/(36))[(1/((x^2 −37x+342)^(18) ))]_(20) ^(+∞) −(1/2)∫(1/((x−18)^(19) (x−19)^(19) )) (1/(36)).(1/2^(18) )−{(1/2)∫_(20) ^(+∞) (1/((x−18)^(19) (x−19)^(19) ))}→I_a (1/(36.2^(18) ))−I_a continue......$
	$\int_{20}^{+ \infty} \frac{(x - 19) d x}{{(x^{2} - 37 x + 342)}^{19}}$ $\frac{1}{2} \int_{20}^{+ \infty} \frac{2 x - 37}{{(x^{2} - 37 x + 342)}^{19}} d x - \frac{1}{{(x^{2} - 37 + 342)}^{19}}$ $- \frac{1}{36} {[\frac{1}{{(x^{2} - 37 x + 342)}^{18}}]}_{20}^{+ \infty} - \frac{1}{2} \int \frac{1}{{(x - 18)}^{19} {(x - 19)}^{19}}$ $\frac{1}{36} . \frac{1}{2^{18}} - {\frac{1}{2} \int_{20}^{+ \infty} \frac{1}{{(x - 18)}^{19} {(x - 19)}^{19}}} \to I_{a}$ $\frac{1}{36 . 2^{18}} - I_{a} c o n t i n u e . . . . . .$
	Answered by mathmax by abdo last updated on 19/Jul/20
	$A =∫_(20) ^(+∞) (dx/((x−18)^(19) (x−19)^(18) )) changement x−19 =t give A =∫_1 ^(+∞) (dt/((t+1)^(19) t^(18) )) =∫_1 ^∞ (dt/(((t/(t+1)))^(18) (t+1)^(37) )) we do the changement (t/(t+1)) =u ⇒ t =ut +u ⇒(1−u)t =u ⇒t =(u/(1−u)) ⇒(dt/du) =((1−u+u)/((1−u)^2 )) =(1/((1−u)^2 )) t+1 =(u/(1−u)) +1 =((u+1−u)/(1−u)) =(1/(1−u)) ⇒ A =∫_(1/2) ^1 (du/((1−u)^2 u^(18) ((1/(1−u)))^(37) )) =∫_(1/2) ^1 (((1−u)^(37) )/((1−u)^2 u^(18) ))du =∫_(1/2) ^1 (((1−u)^(35) )/u^(18) ) du =−∫_(1/2) ^1 (((u−1)^(35) )/u^(18) )du =−∫_(1/2) ^1 ((Σ_(k=0) ^(35 ) C_(35) ^k u^k (−1)^(35−k) )/u^(18) )du =Σ_(k=0) ^(35 ) (−1)^k C_(35) ^k ∫_(1/2) ^1 u^(k−18) du =Σ_(k=0 and k≠17) ^(35) (−1)^k C_(35) ^k [(1/(k−17)) u^(k−17) ]_(1/2) ^1 −C_(35) ^(17) [ln∣u∣]_(1/2) ^1 A=Σ_(k=0and k≠17) ^(35) (((−1)^k C_(35) ^k )/(k−17)){1−(1/2^(k−17) )} −ln(2)C_(35) ^(17)$
	$A = \int_{20}^{+ \infty} \frac{dx}{{(x - 18)}^{19} {(x - 19)}^{18}} changement x - 19 = t give$ $A = \int_{1}^{+ \infty} \frac{dt}{{(t + 1)}^{19} t^{18}} = \int_{1}^{\infty} \frac{dt}{{(\frac{t}{t + 1})}^{18} {(t + 1)}^{37}} we do the changement \frac{t}{t + 1} = u \Rightarrow$ $t = ut + u \Rightarrow (1 - u) t = u \Rightarrow t = \frac{u}{1 - u} \Rightarrow \frac{dt}{du} = \frac{1 - u + u}{{(1 - u)}^{2}} = \frac{1}{{(1 - u)}^{2}}$ $t + 1 = \frac{u}{1 - u} + 1 = \frac{u + 1 - u}{1 - u} = \frac{1}{1 - u} \Rightarrow$ $A = \int_{\frac{1}{2}}^{1} \frac{du}{{(1 - u)}^{2} u^{18} {(\frac{1}{1 - u})}^{37}} = \int_{\frac{1}{2}}^{1} \frac{{(1 - u)}^{37}}{{(1 - u)}^{2} u^{18}} du$ $= \int_{\frac{1}{2}}^{1} \frac{{(1 - u)}^{35}}{u^{18}} du = - \int_{\frac{1}{2}}^{1} \frac{{(u - 1)}^{35}}{u^{18}} du = - \int_{\frac{1}{2}}^{1} \frac{\sum_{k = 0}^{35} C_{35}^{k} u^{k} {(- 1)}^{35 - k}}{u^{18}} du$ $= \sum_{k = 0}^{35} {(- 1)}^{k} C_{35}^{k} \int_{\frac{1}{2}}^{1} u^{k - 18} du$ $= \sum_{k = 0 and k \neq 17}^{35} {(- 1)}^{k} C_{35}^{k} {[\frac{1}{k - 17} u^{k - 17}]}_{\frac{1}{2}}^{1} - C_{35}^{17} {[\ln ∣ u ∣]}_{\frac{1}{2}}^{1}$ $A = \sum_{k = 0 and k \neq 17}^{35} \frac{{(- 1)}^{k} C_{35}^{k}}{k - 17} {1 - \frac{1}{2^{k - 17}}} - \ln (2) C_{35}^{17}$
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