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-n-x-n-

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} (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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