Question

Let \[{I_n} = \int\limits_1^e {{x^{19}}{{(\log \left| x \right|)}^n}dx} \] , where \[n \in N\] . If \[20{I_{10}} = \alpha {I_9} + \beta {I_8}\] , for natural numbers \[\alpha \] and \[\beta \], then \[\alpha - \beta \] equal to ______________.

Answer 1

Hint : We are given an integral \[{I_n} = \int\limits_1^e {{x^{19}}{{(\log \left| x \right|)}^n}dx} \] where \[n \in N\]. Also, \[20{I_{10}} = \alpha {I_9} + \beta {I_8}\]. We have to find the value of \[\alpha - \beta \]. We will first integrate the given integral using integrating by parts, then we will find the values of \[{I_{10}},{I_9}\], then find the relation between these integrals. Then we will compare the found relation with \[20{I_{10}} = \alpha {I_9} + \beta {I_8}\] to find the values of \[\alpha ,\beta \], then we will subtract them to find the value of \[\alpha - \beta \].

Formula used : Following formulas will be useful for solving this question
\[\int {uv = u\int {v - u'\int v } } \]

Complete step-by-step Solution :
Let us assume \[\log \left| x \right| = \ln x\]
In order to use the formula for integration by parts i.e.,\[\int {uv = u\int {v - \int {u'v} } } \], we will consider here,
\[
  u = \ln x, \\
  v = {x^{19}} \\
 \]
 On integrating the given integral with the help of integration by parts, we get
\[
  {I_n} = \int\limits_1^e {{x^{19}}{{(\log \left| x \right|)}^n}dx} \\
   = \left. {{{(\ln x)}^n} \times \dfrac{{{x^{20}}}}{{20}}} \right|_1^e - \int\limits_1^e {n\dfrac{{{{(\ln x)}^{n - 1}}}}{x} \times \dfrac{{{x^{20}}}}{{20}}dx} \\
   = \left. {{{(\ln x)}^n} \times \dfrac{{{x^{20}}}}{{20}}} \right|_1^e - \int\limits_1^e {n{{(\ln x)}^{n - 1}} \times \dfrac{{{x^{19}}}}{{20}}dx} \\
 \]
As, we know from the given that \[{I_n} = \int\limits_1^e {{x^{19}}{{(\log \left| x \right|)}^n}dx} \], so we can write \[\int\limits_1^e {{x^{19}}{{(\log x)}^{n - 1}}dx} = {I_{n - 1}}\]
On substituting this, we get
\[{I_n} = \dfrac{{{e^{20}}}}{{20}} - \dfrac{n}{{20}}{I_{n - 1}}\]
On multiplying both side of the equations with \[20\], we get
\[20{I_n} = {e^{20}} - n{I_{n - 1}}\]
On substituting \[n = 10\], we get
\[20{I_{10}} = {e^{20}} - 10{I_9}\] …… (1)
On substituting \[n = 9\], we get
\[20{I_9} = {e^{20}} - 9{I_8}\] …… (2)

On subtracting equation (2) from (1), we get
\[
  20{I_{10}} - 20{I_9} = {e^{20}} - 10{I_9} - ({e^{20}} - 9{I_8}) \\
  20{I_{10}} = 20{I_9} - 10{I_9} + 9{I_8} \\
  20{I_{10}} = 10{I_9} + 9{I_8} \\
 \]
On comparing this with \[20{I_{10}} = \alpha {I_9} + \beta {I_8}\], we get value of \[\alpha ,\beta \], as \[10,9\] respectively.
Thus \[
  \alpha - \beta = 10 - 9 \\
   = 1 \\
 \]
Therefore the value of \[\alpha - \beta \] is \[1\]

Note : Here, students may make mistake while using the integration by parts formula \[\int {uv = u\int {v - \int {u'v} } } \]. Here, often they get confused about what to assume as the values of \[u,v\]. While choosing this value, they should consider \[u\] as such a term which can be easily differentiable. This will ease their solving of integration and thus they can find the required answer easily. This is the reason why in this question we chose \[u = \ln x\],\[v = {x^{19}}\] and not \[u = {x^{19}}\],\[v = \ln x\].