Let I_{n} = int_{1}^{e} x^{19}(log |x|)^{n} d | vedclass

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(B) Given $I_{n} = \int_{1}^{e} x^{19}(\log |x|)^{n} dx$.Using integration by parts,let $u = (\log |x|)^{n}$ and $dv = x^{19} dx$. Then $du = n(\log |

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(B) Given $I_{n} = \int_{1}^{e} x^{19}(\log |x|)^{n} dx$.Using integration by parts,let $u = (\log |x|)^{n}$ and $dv = x^{19} dx$. Then $du = n(\log |x|)^{n-1} \cdot \frac{1}{x} dx$ and $v = \frac{x^{20}}{20}$.$I_{n} = \left[ \frac{x^{20}}{20} (\log |x|)^{n} \right]_{1}^{e} - \int_{1}^{e} \frac{x^{20}}{20} \cdot n(\log |x|)^{n-1} \cdot \frac{1}{x} dx$.$I_{n} = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$.Multiplying by $20$,we get $20 I_{n} = e^{20} - n I_{n-1}$.For $n=10$,$20 I_{10} = e^{20} - 10 I_{9}$.For $n=9$,$20 I_{9} = e^{20} - 9 I_{8}$,which implies $e^{20} = 20 I_{9} + 9 I_{8}$.Substituting $e^{20}$ into the first equation: $20 I_{10} = (20 I_{9} + 9 I_{8}) - 10 I_{9} = 10 I_{9} + 9 I_{8}$.Comparing with $20 I_{10} = \alpha I_{9} + \beta I_{8}$,we get $\alpha = 10$ and $\beta = 9$.Therefore,$\alpha - \beta = 10 - 9 = 1$.

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 ..... .

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