Let I_n=_1^e x^19(|x|)^n,dx, where n N.

Question

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

Answer: 1

Accepted Answer

$I_n=\displaystyle\int_1^e x^{19}(\ln|x|)^n\,dx$. Integrating by parts (treating $x^{19}$ as the part to integrate):
$$I_n=\left[\frac{x^{20}}{20}(\ln|x|)^n\right]_1^e-\int_1^e\frac{x^{20}}{20}\cdot n(\ln|x|)^{n-1}\cdot\frac{1}{x}dx=\frac{e^{20}}{20}-\frac{n}{20}\int_1^e(\ln|x|)^{n-1}x^{19}dx$$
$$\Rightarrow I_n=\frac{e^{20}}{20}-\frac{n}{20}I_{n-1}\ \Rightarrow\ 20I_n+nI_{n-1}=e^{20}$$

Put $n=10$: $20I_{10}+10I_9=e^{20}\ \ldots(i)$
Put $n=9$: $20I_9+9I_8=e^{20}\ \ldots(ii)$

Subtracting: $20I_{10}-10I_9-9I_8=0\Rightarrow 20I_{10}=10I_9+9I_8$

Comparing with $20I_{10}=\alpha I_9+\beta I_8$: $\alpha=10,\ \beta=9\ \Rightarrow\ \alpha-\beta=1$

Solutions · 1