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Step-by-Step Solution
Step 1: Define the integral
Let
I_n = \int_{1}^{e} x^{19} \bigl(\ln x\bigr)^{n} \, dx,
where n is a natural number. Our goal is to derive a relationship involving I_{n} , I_{n-1} , and then apply it specifically for n=10 and n=9 .
Step 2: Apply integration by parts
Recall the integration by parts formula:
\int u \, dv = uv - \int v \, du.
For this integral, choose
u = (\ln x)^{n}
and
dv = x^{19}\,dx.
Then we have:
du = n \bigl(\ln x\bigr)^{n-1} \cdot \frac{1}{x} \, dx.
v = \frac{x^{20}}{20}.
Thus,
I_n \;=\; \int_{1}^{e} x^{19} \bigl(\ln x\bigr)^n \, dx
\;=\; \left[ \bigl(\ln x\bigr)^n \cdot \frac{x^{20}}{20} \right]_{1}^{e}
\;-\; \int_{1}^{e} \frac{x^{20}}{20} \cdot \Bigl(n \bigl(\ln x\bigr)^{n-1} \cdot \frac{1}{x}\Bigr) \, dx.
Step 3: Evaluate the boundary term
At x=e, we have \ln(e)=1, so
\Bigl(\ln(e)\Bigr)^n \cdot \frac{e^{20}}{20} = \frac{e^{20}}{20}.
At x=1, we have \ln(1)=0, so that boundary contribution is 0. Hence,
\left[ \bigl(\ln x\bigr)^n \cdot \frac{x^{20}}{20} \right]_{1}^{e}
= \frac{e^{20}}{20} - 0 = \frac{e^{20}}{20}.
Step 4: Simplify the remaining integral
The subtracted integral becomes:
\int_{1}^{e} \frac{x^{20}}{20} \cdot n \bigl(\ln x\bigr)^{n-1} \cdot \frac{1}{x} \, dx
= \frac{n}{20} \int_{1}^{e} x^{19} \bigl(\ln x\bigr)^{n-1} \, dx
= \frac{n}{20}\, I_{n-1}.
Therefore,
I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}.
Step 5: Multiply both sides by 20
Multiply the entire equation by 20 to get a clean relationship:
20\,I_n = e^{20} - n \, I_{n-1}.
This is the key recurrence relation for I_n .
Step 6: Substitute n=10
Using n=10 in the relation
20\,I_n = e^{20} - n\,I_{n-1},
we get:
20\,I_{10} = e^{20} - 10\,I_{9}.
\quad (1)
Step 7: Substitute n=9
Similarly, for n=9, the relation becomes
20\,I_{9} = e^{20} - 9\,I_{8}.
\quad (2)
Step 8: Express 20\,I_{10} in terms of I_9 and I_8
Subtract equation (2) from equation (1):
From (1): 20\,I_{10} = e^{20} - 10\,I_{9}.
From (2): 20\,I_{9} = e^{20} - 9\,I_{8}.
Subtracting (2) from (1) effectively eliminates e^{20} :
20\,I_{10} - 20\,I_{9} = \bigl(e^{20} - 10\,I_{9}\bigr) - \bigl(e^{20} - 9\,I_{8}\bigr),
20\,I_{10} = 10\,I_{9} + 9\,I_{8}.
So we can write
20\,I_{10} = \alpha \, I_{9} + \beta \, I_{8},
where \alpha = 10 and \beta = 9.
Step 9: Find \alpha - \beta
From \alpha = 10 and \beta = 9, we conclude:
\alpha - \beta = 10 - 9 = 1.
Final Answer
The value of \alpha - \beta is \boxed{1}.
