Let $I_n={\int }_1^e{x}^{19}\left(\log \right|x|{)}^{n}dx,$ where $n\in N$. If  $\left(20\right)I_{10}=\alpha I_9+\beta I_8,$ for natural numbers $\alpha$ and $\beta$, then $\alpha -\beta$ equal to _______.

Given $I_n={\int }_1^e{x}^{19}\left(\log \right|x|{)}^{n}dx, n\in N$

$\Rightarrow I_n={\int }_1^e{x}^{19}{\left(\log x\right)}^{n}dx, \because$ for $1<x<e, x>0$

Integrating by using by parts rule, ${\int }_a^b\left(u·v\right)dx={\left[u\int vdx\right]}_a^b-{\int }_a^b\left[\frac{d}{dx}\left(u\right)\int vdx\right]dx,$ taking $u=\log x \& v=x^{19},$ we get

$I_n={\left[{\left(\log x\right)}^n\int x^{19}dx\right]}_1^e-{\int }_1^e\left[\frac{d}{dx}{\left(\log x\right)}^n\int x^{19}dx\right]dx$

$\Rightarrow I_n={\left[{\left(\log x\right)}^n\left(\frac{x^{20}}{20}\right)\right]}_1^e-{\int }_1^e\left[n{\left(\log x\right)}^{n-1}\times \frac{1}{x}\times \left(\frac{x^{20}}{20}\right)\right]dx$

$\Rightarrow I_n=\left[{\left(\log e\right)}^n\left(\frac{e^{20}}{20}\right)-0\right]-{\int }_1^e\left[n{\left(\log x\right)}^{n-1}\times \left(\frac{x^{19}}{20}\right)\right]dx$

$\Rightarrow I_n=\left(\frac{e^{20}}{20}\right)-\frac{n}{20}{\int }_1^e{x}^{19}{\left(\log x\right)}^{n-1}dx$

$\Rightarrow 20I_n=e^{20}-n{I}_{n-1}$

Hence, we get $20I_{10}=e^{20}-10I_9 ...\left(i\right)$ and

$20I_9=e^{20}-9I_8 ...\left(ii\right)$

On subtracting the above two equations, we get

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

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

Given $20I_{10}=\alpha I_9+\beta I_8$

$\Rightarrow \alpha =10, \beta =9$ and hence $\alpha -\beta =1.$