let ∀n∈N: I_n (f(x))= the n^(th) antiderivate of f(x) with I_0 =f(x) find the formula for the constants a_n , b_n of I_n (ln x)=a_n x^n ln x +b


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Question Number 173150 by Frix last updated on 07/Jul/22

$let \forall n \in N : I_{n} (f (x)) = the n^{th} antiderivate$ $of f (x) with I_{0} = f (x)$ $find the formula for the constants a_{n}, b_{n} of$ $I_{n} (\ln x) = a_{n} x^{n} \ln x + b_{n} x^{n}$
	Answered by aleks041103 last updated on 07/Jul/22
	$I_(n+1) (ln x)=∫I_n (ln x)dx= =∫(a_n x^n ln x + b_n x^n )dx= =a_n ∫x^n ln x dx + b_n ∫x^n dx ∫x^n ln x dx=∫ln x d((x^(n+1) /(n+1)))= =(x^(n+1) /(n+1))ln x−(1/(n+1))∫x^(n+1) (dx/x)= =(1/(n+1))x^(n+1) ln x −(1/((n+1)^2 ))x^(n+1) ⇒I_(n+1) (ln x)=a_n [(1/(n+1))x^(n+1) ln x −(1/((n+1)^2 ))x^(n+1) ]+(b_n /(n+1))x^(n+1) = =((a_n /(n+1)))x^(n+1) ln x + ((b_n /(n+1))−(a_n /((n+1)^2 )))x^(n+1) = =a_(n+1) x^(n+1) ln x + b_(n+1) x^(n+1) ⇒ { ((a_(n+1) =(a_n /(n+1)))),((b_(n+1) =(b_n /(n+1))−(a_n /((n+1)^2 )))) :} (1/a_(n+1) )=(n+1)(1/a_n )⇒(1/a_n )=const.n!⇒a_n =((const)/(n!)) I_0 (ln x)=ln x⇒a_0 =1⇒a_n =(1/(n!)) ⇒b_(n+1) =(b_n /(n+1))−(1/((n+1)!(n+1))) (n+1)b_(n+1) −b_n =−(1/((n+1)!)) b_n =(c_n /(n!)) ⇒(n+1)(c_(n+1) /((n+1)!))−(c_n /(n!))=−(1/((n+1)!))=−(1/((n+1) n!)) ⇒c_(n+1) −c_n =−(1/(n+1)) ⇒c_n =const.−Σ_(i≥1) ^n (1/i) c_0 =const.=0⇒c_n =−Σ_(i≥1) ^n (1/i)=−H_n ⇒b_n =−(H_n /(n!)) ⇒ { ((a_n =(1/(n!)))),((b_n =−(H_n /(n!)) , H_(n≥1) =Σ_(k=1) ^n (1/k), H_0 =0)) :}$
	$I_{n + 1} (l n x) = \int I_{n} (l n x) d x =$ $= \int (a_{n} x^{n} l n x + b_{n} x^{n}) d x =$ $= a_{n} \int x^{n} l n x d x + b_{n} \int x^{n} d x$ $\int x^{n} l n x d x = \int l n x d (\frac{x^{n + 1}}{n + 1}) =$ $= \frac{x^{n + 1}}{n + 1} l n x - \frac{1}{n + 1} \int x^{n + 1} \frac{d x}{x} =$ $= \frac{1}{n + 1} x^{n + 1} l n x - \frac{1}{{(n + 1)}^{2}} x^{n + 1}$ $\Rightarrow I_{n + 1} (l n x) = a_{n} [\frac{1}{n + 1} x^{n + 1} l n x - \frac{1}{{(n + 1)}^{2}} x^{n + 1}] + \frac{b_{n}}{n + 1} x^{n + 1} =$ $= (\frac{a_{n}}{n + 1}) x^{n + 1} l n x + (\frac{b_{n}}{n + 1} - \frac{a_{n}}{{(n + 1)}^{2}}) x^{n + 1} =$ $= a_{n + 1} x^{n + 1} l n x + b_{n + 1} x^{n + 1}$ $\Rightarrow {\begin{cases} a_{n + 1} = \frac{a_{n}}{n + 1} \\ b_{n + 1} = \frac{b_{n}}{n + 1} - \frac{a_{n}}{{(n + 1)}^{2}} \end{cases}$ $\frac{1}{a_{n + 1}} = (n + 1) \frac{1}{a_{n}} \Rightarrow \frac{1}{a_{n}} = c o n s t . n! \Rightarrow a_{n} = \frac{c o n s t}{n!}$ $I_{0} (l n x) = l n x \Rightarrow a_{0} = 1 \Rightarrow a_{n} = \frac{1}{n!}$ $\Rightarrow b_{n + 1} = \frac{b_{n}}{n + 1} - \frac{1}{(n + 1)! (n + 1)}$ $(n + 1) b_{n + 1} - b_{n} = - \frac{1}{(n + 1)!}$ $b_{n} = \frac{c_{n}}{n!}$ $\Rightarrow (n + 1) \frac{c_{n + 1}}{(n + 1)!} - \frac{c_{n}}{n!} = - \frac{1}{(n + 1)!} = - \frac{1}{(n + 1) n!}$ $\Rightarrow c_{n + 1} - c_{n} = - \frac{1}{n + 1}$ $\Rightarrow c_{n} = c o n s t . - \underset{i ⩾ 1}{\sum^{n}} \frac{1}{i}$ $c_{0} = c o n s t . = 0 \Rightarrow c_{n} = - \underset{i ⩾ 1}{\sum^{n}} \frac{1}{i} = - H_{n}$ $\Rightarrow b_{n} = - \frac{H_{n}}{n!}$ $\Rightarrow {\begin{cases} a_{n} = \frac{1}{n!} \\ b_{n} = - \frac{H_{n}}{n!}, H_{n ⩾ 1} = \underset{k = 1}{\sum^{n}} \frac{1}{k}, H_{0} = 0 \end{cases}$
	Commented by Frix last updated on 08/Jul/22

	$thank you!$
	Commented by Tawa11 last updated on 11/Jul/22

	$Great sir$
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