Let $$I_n = \int_1^e x^{19}(\log|x|)^n dx$$, where $$n \in N$$. If $$(20)I_{10} = \alpha I_9 + \beta I_8$$, for natural numbers $$\alpha$$ and $$\beta$$, then $$\alpha - \beta$$ is equal to ________.

We have $$I_n = \int_1^e x^{19}(\log|x|)^n\,dx$$. Since $$x \in [1, e]$$, $$|x| = x$$, so $$I_n = \int_1^e x^{19}(\log x)^n\,dx$$.

Using integration by parts with $$u = (\log x)^n$$ and $$dv = x^{19}\,dx$$, we get $$du = \frac{n(\log x)^{n-1}}{x}\,dx$$ and $$v = \frac{x^{20}}{20}$$.

So $$I_n = \left[\frac{x^{20}}{20}(\log x)^n\right]_1^e - \int_1^e \frac{x^{20}}{20} \cdot \frac{n(\log x)^{n-1}}{x}\,dx = \frac{e^{20}}{20}(1)^n - 0 - \frac{n}{20}\int_1^e x^{19}(\log x)^{n-1}\,dx$$.

This gives the recurrence $$I_n = \frac{e^{20}}{20} - \frac{n}{20}I_{n-1}$$, or equivalently $$20 I_n = e^{20} - n I_{n-1}$$.

For $$n = 10$$: $$20 I_{10} = e^{20} - 10 I_9$$. We need to express this in the form $$20 I_{10} = \alpha I_9 + \beta I_8$$.

From the recurrence for $$n = 9$$: $$20 I_9 = e^{20} - 9 I_8$$, so $$e^{20} = 20 I_9 + 9 I_8$$.

Substituting into the equation for $$I_{10}$$: $$20 I_{10} = (20 I_9 + 9 I_8) - 10 I_9 = 10 I_9 + 9 I_8$$.

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