"Let ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$, where n$\in$N. If (20)I10 = $\alpha$I9 + $\beta$I8, for natural numbers $\alpha$ and $\beta$, then $\alpha$ $-$ $\beta$ equals to ___________."

Question

Accepted Answer

"${I_n} = 2\int\limits_1^e {{x^{19}}{{(\ln x)}^n}\,.\,dx}$

$$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $$

${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$

$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$

Putting n = 10, we get

$20{I_{10}} = ({e^{20}} - 10{I_9})$ ...... (1)

Putting n = 9, we get

$20{I_9} = {e^{20}} - 9{I_8}$ ....... (2)

Subtracting (2) from (1), we get

$20{I_{10}} = 10{I_9} + 9{I_8}$

By comparing with (20)I₁₀ = $\alpha$I₉ + $\beta$I₈, we get

$\alpha$ = 10, $\beta$ = 9 $\Rightarrow$ $\alpha$ $-$ $\beta$ = 1"

Let ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$, where n$\in$N. If (20)I₁₀ = $\alpha$I₉ + $\beta$I₈, for natural numbers $\alpha$ and $\beta$, then $\alpha$ $-$ $\beta$ equals to ___________.

Solution

About this question

Drill 25 more like these. Every day. Free.

Let ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$, where n$\in$N. If (20)I10 = $\alpha$I9 + $\beta$I8, for natural numbers $\alpha$ and $\beta$, then $\alpha$ $-$ $\beta$ equals to ___________.

Solution

About this question

More Definite Integration questions

Drill 25 more like these. Every day. Free.

Let ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$, where n$\in$N. If (20)I₁₀ = $\alpha$I₉ + $\beta$I₈, for natural numbers $\alpha$ and $\beta$, then $\alpha$ $-$ $\beta$ equals to ___________.