1. Since the interval of integration is $[1,e]$, we have $x>0$, so $\log|x|=\log x$.

   Thus $I_n=\int_1^e x^{19}(\log x)^n\,dx.$

2. We need a relation involving $I_{10}, I_9, I_8$. Use integration by parts on $I_n=\int_1^e x^{19}(\log x)^n\,dx.$

   Take $u=(\log x)^n, \qquad dv=x^{19}dx.$ Then $du=\frac{n(\log x)^{n-1}}{x}dx, \qquad v=\frac{x^{20}}{20}.$

   So, $I_n=\left[\frac{x^{20}}{20}(\log x)^n\right]_1^e-\frac{n}{20}\int_1^e x^{19}(\log x)^{n-1}dx.$

   Hence, $I_n=\left[\frac{x^{20}}{20}(\log x)^n\right]_1^e-\frac{n}{20}I_{n-1}.$

3. Evaluate the boundary term:

   • At $x=e$, $\log e=1$, so contribution is $\dfrac{e^{20}}{20}$.
   • At $x=1$, $\log 1=0$, so contribution is $0$.

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

   Multiply by $20$: $20I_n=e^{20}-nI_{n-1}.$

4. Apply this for $n=10$: $20I_{10}=e^{20}-10I_9. \qquad (1)$

   Apply it for $n=9$: $20I_9=e^{20}-9I_8.$ So, $e^{20}=20I_9+9I_8. \qquad (2)$

5. Substitute (2) into (1): $20I_{10}=(20I_9+9I_8)-10I_9=10I_9+9I_8.$

   Comparing with $20I_{10}=\alpha I_9+\beta I_8,$ we get $\alpha=10, \qquad \beta=9.$

6. Therefore, $\alpha-\beta=10-9=1.$

So the required integer is $1$.