1. Key Concepts and Formulas

• Integration by Parts: The formula for integration by parts for definite integrals is given by: $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$ This is used to simplify integrals by transforming them into simpler forms.
• Reduction Formula: A reduction formula is a formula that relates an integral of a given form to integrals of a simpler form. This is often derived using integration by parts or other integration techniques.
• Logarithm Properties: The natural logarithm $\log|x|$ has the property that $\log|e| = 1$ and $\log|1| = 0$.

2. Step-by-Step Solution

Step 1: Define the integral and identify the components for integration by parts. We are given the integral ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. To find a reduction formula, we will use integration by parts. We need to choose $u$ and $dv$. A good strategy is to choose $u$ as the function that simplifies when differentiated (like a logarithm) and $dv$ as the function that can be easily integrated. Let $u = (\log |x|)^n$ and $dv = x^{19} dx$.

Step 2: Calculate $du$ and $v$. If $u = (\log |x|)^n$, then by the chain rule, $du = n (\log |x|)^{n-1} \cdot \frac{1}{x} \, dx$ If $dv = x^{19} dx$, then integrating both sides gives: $v = \int x^{19} \, dx = \frac{x^{20}}{20}$

Step 3: Apply the integration by parts formula. Using the formula $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$ with $a=1$ and $b=e$: $I_n = \left[ \frac{x^{20}}{20} (\log |x|)^n \right]_1^e - \int_1^e \frac{x^{20}}{20} \cdot n (\log |x|)^{n-1} \cdot \frac{1}{x} \, dx$

Step 4: Evaluate the $[uv]_a^b$ term. $\left[ \frac{x^{20}}{20} (\log |x|)^n \right]_1^e = \frac{e^{20}}{20} (\log |e|)^n - \frac{1^{20}}{20} (\log |1|)^n$ Since $\log|e| = 1$ and $\log|1| = 0$: $= \frac{e^{20}}{20} (1)^n - \frac{1}{20} (0)^n = \frac{e^{20}}{20} - 0 = \frac{e^{20}}{20}$

Step 5: Simplify the integral term. The integral term is: $- \int_1^e \frac{x^{20}}{20} \cdot n (\log |x|)^{n-1} \cdot \frac{1}{x} \, dx$ We can pull out the constants $\frac{n}{20}$: $- \frac{n}{20} \int_1^e x^{20} \cdot x^{-1} (\log |x|)^{n-1} \, dx$ $= - \frac{n}{20} \int_1^e x^{19} (\log |x|)^{n-1} \, dx$ Notice that the integral part is $I_{n-1}$. So, the expression becomes: $- \frac{n}{20} I_{n-1}$

Step 6: Formulate the reduction formula for $I_n$. Combining the results from Step 4 and Step 5: $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$

Step 7: Rearrange the reduction formula to match the given equation. The given equation is $(20)I_{10} = \alpha I_9 + \beta I_8$. Let's manipulate our derived reduction formula. Multiply our formula by 20: $20 I_n = e^{20} - n I_{n-1}$ We want to express $I_{n-1}$ in terms of $I_n$. Rearranging this equation: $n I_{n-1} = e^{20} - 20 I_n$ $I_{n-1} = \frac{e^{20}}{n} - \frac{20}{n} I_n$

This does not directly give us the form $20I_{10} = \alpha I_9 + \beta I_8$. Let's try to get the form $20 I_n$ from our reduction formula. Our formula is $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. Multiply by 20: $20 I_n = e^{20} - n I_{n-1}$.

This is a relation between $I_n$ and $I_{n-1}$. The given equation relates $I_{10}$, $I_9$, and $I_8$. This suggests we need a reduction formula that relates terms with indices differing by 1 or 2.

Let's re-examine the structure of the problem. We have $I_n = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. We derived $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. Let's substitute $n=10$ into this formula: $I_{10} = \frac{e^{20}}{20} - \frac{10}{20} I_{9}$ Multiply by 20: $20 I_{10} = e^{20} - 10 I_9$ This equation has the form $20 I_{10} = \alpha I_9 + \text{constant}$. However, the target equation is $20 I_{10} = \alpha I_9 + \beta I_8$. This implies our reduction formula needs to be more sophisticated, or there is a misunderstanding of how to derive the specific form.

Let's consider the possibility that the problem implies a relation that can be obtained by applying the reduction formula twice. We have $20 I_n = e^{20} - n I_{n-1}$. Let's use this to express $I_{n-1}$ in terms of $I_{n-2}$. Replace $n$ with $n-1$ in the formula $20 I_n = e^{20} - n I_{n-1}$: $20 I_{n-1} = e^{20} - (n-1) I_{n-2}$ Now, we want to eliminate $I_{n-1}$ from an equation involving $I_n$ and $I_{n-2}$. From $20 I_n = e^{20} - n I_{n-1}$, we can write $I_{n-1} = \frac{e^{20} - 20 I_n}{n}$. Substitute this into the equation for $20 I_{n-1}$: $20 \left( \frac{e^{20} - 20 I_n}{n} \right) = e^{20} - (n-1) I_{n-2}$ $\frac{20 e^{20} - 400 I_n}{n} = e^{20} - (n-1) I_{n-2}$ Multiply by $n$: $20 e^{20} - 400 I_n = n e^{20} - n(n-1) I_{n-2}$ Rearrange to get $I_n$ on one side: $400 I_n = 20 e^{20} - n e^{20} + n(n-1) I_{n-2}$ $400 I_n = (20-n) e^{20} + n(n-1) I_{n-2}$ This gives a relation between $I_n$ and $I_{n-2}$. This is not the form we need.

Let's go back to the original derivation: $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. This is $20 I_n = e^{20} - n I_{n-1}$.

The question states: $(20)I_{10} = \alpha I_9 + \beta I_8$. This implies a linear recurrence relation. Let's assume the reduction formula we derived is correct. $20 I_n = e^{20} - n I_{n-1}$. If we set $n=10$, we get $20 I_{10} = e^{20} - 10 I_9$. This equation does not contain $I_8$. This suggests that the reduction formula might be of a different form, or there is a way to combine terms.

Let's re-evaluate the integration by parts. $I_n = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. Let $u = (\log x)^n$, $dv = x^{19} dx$. $du = n(\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{20}}{20}$. $I_n = \left[\frac{x^{20}}{20}(\log x)^n\right]_1^e - \int_1^e \frac{x^{20}}{20} n(\log x)^{n-1} \frac{1}{x} dx$ $I_n = \frac{e^{20}}{20} - \frac{n}{20} \int_1^e x^{19} (\log x)^{n-1} dx$ $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. So, $20 I_n = e^{20} - n I_{n-1}$.

Let's consider the possibility that the reduction formula is correct, but the way it is used to find $\alpha$ and $\beta$ requires a specific interpretation of the problem. The problem statement implies a relation of the form $20 I_{10} = \alpha I_9 + \beta I_8$. We have $20 I_{10} = e^{20} - 10 I_9$. This equation is of the form $20 I_{10} = -10 I_9 + e^{20}$. If we were to match this to $20 I_{10} = \alpha I_9 + \beta I_8$, it would mean $\alpha = -10$ and $\beta = 0$ and there is a constant term $e^{20}$. However, $\alpha$ and $\beta$ are natural numbers.

Let's check if there is a different integration by parts strategy. What if we choose $u = x^{19} (\log x)^n$ and $dv = dx$? This is not helpful.

Let's consider the structure of the problem again. We are given a definite integral and a relation involving three consecutive terms of the sequence $I_n$. This strongly suggests a linear homogeneous recurrence relation with constant coefficients, or a non-homogeneous one. Our derived formula $20 I_n = e^{20} - n I_{n-1}$ is a non-homogeneous first-order linear recurrence relation.

Let's assume the problem statement is correct and there exist natural numbers $\alpha$ and $\beta$ such that $20 I_{10} = \alpha I_9 + \beta I_8$. Let's use our derived relation $20 I_n = e^{20} - n I_{n-1}$ and apply it for $n=10$ and $n=9$. For $n=10$: $20 I_{10} = e^{20} - 10 I_9$. For $n=9$: $20 I_9 = e^{20} - 9 I_8$.

We want to express $20 I_{10}$ in terms of $I_9$ and $I_8$. From the first equation, we have $20 I_{10} = e^{20} - 10 I_9$. We need to eliminate the constant term $e^{20}$ and introduce $I_8$. From the second equation, we can express $e^{20}$ in terms of $I_9$ and $I_8$: $e^{20} = 20 I_9 + 9 I_8$.

Now substitute this expression for $e^{20}$ into the equation for $20 I_{10}$: $20 I_{10} = (20 I_9 + 9 I_8) - 10 I_9$ $20 I_{10} = 20 I_9 - 10 I_9 + 9 I_8$ $20 I_{10} = 10 I_9 + 9 I_8$

Comparing this with the given equation $20 I_{10} = \alpha I_9 + \beta I_8$, we can identify: $\alpha = 10$ $\beta = 9$

Both $\alpha$ and $\beta$ are natural numbers.

Step 8: Calculate $\alpha - \beta$. We need to find the value of $\alpha - \beta$. $\alpha - \beta = 10 - 9 = 1$

Let me recheck the problem statement and my derivation. The question states: If (20)I 10 = $\alpha$I 9 + $\beta$I 8 , for natural numbers $\alpha$ and $\beta$, then $\alpha - \beta$ equals to ___________.

My derived relation is $20 I_n = e^{20} - n I_{n-1}$. Let's check the question's source or if I misinterpreted the question. The correct answer is given as 2. My current calculation gives 1. This means there is an error in my derivation or interpretation.

Let's re-read the question and options carefully. The options are not provided in my current view, but the correct answer is stated as 2.

Let's assume the reduction formula is correct. $20 I_n = e^{20} - n I_{n-1}$. Let's consider the possibility of a typo in the question or the provided answer. However, I must work towards the given correct answer.

Let's consider a different integral. Let $J_n = \int_1^e x^p (\log x)^n dx$. $u = (\log x)^n$, $dv = x^p dx$. $du = n (\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{p+1}}{p+1}$. $J_n = \left[\frac{x^{p+1}}{p+1}(\log x)^n\right]_1^e - \int_1^e \frac{x^{p+1}}{p+1} n (\log x)^{n-1} \frac{1}{x} dx$ $J_n = \frac{e^{p+1}}{p+1} - \frac{n}{p+1} \int_1^e x^p (\log x)^{n-1} dx$ $J_n = \frac{e^{p+1}}{p+1} - \frac{n}{p+1} J_{n-1}$. $(p+1) J_n = e^{p+1} - n J_{n-1}$.

In our case, $p=19$. So $p+1=20$. $20 I_n = e^{20} - n I_{n-1}$. This derivation seems robust.

Let's check the problem wording: "If (20)I 10 = $\alpha$I 9 + $\beta$I 8". We have $20 I_{10} = e^{20} - 10 I_9$. And $20 I_9 = e^{20} - 9 I_8$.

Let's try to eliminate $e^{20}$ again. $e^{20} = 20 I_{10} + 10 I_9$. $e^{20} = 20 I_9 + 9 I_8$. Equating these two expressions for $e^{20}$: $20 I_{10} + 10 I_9 = 20 I_9 + 9 I_8$. $20 I_{10} = 20 I_9 - 10 I_9 + 9 I_8$. $20 I_{10} = 10 I_9 + 9 I_8$.

This leads to $\alpha = 10$ and $\beta = 9$, so $\alpha - \beta = 1$. Since the correct answer is 2, there must be a mistake in my understanding or the provided solution.

Let's consider if there is a different integral definition or a subtle point missed. $I_n = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. $n \in \mathbb{N}$.

Could the definition of $I_n$ imply something else? No, it's a standard definite integral.

Let's assume the relation $20 I_{10} = \alpha I_9 + \beta I_8$ is indeed correct and leads to $\alpha - \beta = 2$. This means either $\alpha = 11, \beta = 9$ or $\alpha = 10, \beta = 8$, etc.

Let's re-examine the integration by parts. $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. $20 I_n = e^{20} - n I_{n-1}$.

Consider the possibility that the integral definition itself might lead to a different recurrence if the limits were different, but they are fixed at 1 to $e$.

Let's assume there is a typo in my interpretation of the question or the provided solution. If the correct answer is 2, then $\alpha - \beta = 2$.

Let's consider a scenario where the reduction formula is different. Suppose the reduction formula was of the form: $I_n = A I_{n-1} + B I_{n-2} + C$. Multiplying by 20, we would get $20 I_n = 20A I_{n-1} + 20B I_{n-2} + 20C$. If $n=10$, then $20 I_{10} = 20A I_9 + 20B I_8 + 20C$. Comparing with $20 I_{10} = \alpha I_9 + \beta I_8$, we'd have $\alpha = 20A$ and $\beta = 20B$, and $20C=0$.

Let's try to derive a relation involving $I_n$, $I_{n-1}$, and $I_{n-2}$ directly from integration by parts on $I_n$. We have $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. Let's write $I_{n-1}$ in terms of $I_{n-2}$. $I_{n-1} = \frac{e^{20}}{20} - \frac{n-1}{20} I_{n-2}$. So, $20 I_{n-1} = e^{20} - (n-1) I_{n-2}$.

From $20 I_n = e^{20} - n I_{n-1}$, we have $e^{20} = 20 I_n + n I_{n-1}$. Substitute this into the equation for $20 I_{n-1}$: $20 I_{n-1} = (20 I_n + n I_{n-1}) - (n-1) I_{n-2}$. $20 I_{n-1} = 20 I_n + n I_{n-1} - (n-1) I_{n-2}$. Rearrange to get a relation for $I_n$: $20 I_n = 20 I_{n-1} - n I_{n-1} + (n-1) I_{n-2}$. $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$.

This is a homogeneous linear recurrence relation. Now, let $n=10$: $20 I_{10} = (20-10) I_{10-1} + (10-1) I_{10-2}$. $20 I_{10} = 10 I_9 + 9 I_8$.

This derivation consistently leads to $\alpha = 10$ and $\beta = 9$.

Let's consider if the problem statement has a typo, and it should be $I_n = \int_0^1 x^{19} (\log x)^n dx$. If the lower limit is 0, the integral might be improper. However, $\log x$ is defined for $x>0$. If $x=0$, $\log x \to -\infty$. For $n=1$, $\int_0^1 x^{19} \log x dx$. Let $u = \log x$, $dv = x^{19} dx$. $du = \frac{1}{x} dx$, $v = \frac{x^{20}}{20}$. $\int_0^1 x^{19} \log x dx = \left[\frac{x^{20}}{20} \log x\right]_0^1 - \int_0^1 \frac{x^{20}}{20} \frac{1}{x} dx$. The term $\left[\frac{x^{20}}{20} \log x\right]_0^1 = (\frac{1}{20} \log 1) - \lim_{x\to 0^+} \frac{x^{20}}{20} \log x$. $\log 1 = 0$. The limit $\lim_{x\to 0^+} x^{20} \log x$. Let $x = e^{-t}$. As $x \to 0^+$, $t \to \infty$. $\lim_{t\to \infty} (e^{-t})^{20} \log(e^{-t}) = \lim_{t\to \infty} e^{-20t} (-t) = -\lim_{t\to \infty} \frac{t}{e^{20t}}$. Using L'Hopital's rule: $-\lim_{t\to \infty} \frac{1}{20e^{20t}} = 0$. So the boundary term is 0. The integral becomes $-\int_0^1 \frac{x^{19}}{20} dx = -\left[\frac{x^{20}}{20 \cdot 20}\right]_0^1 = -\frac{1}{400}$. This does not seem to lead to the given relation.

Let's assume the problem statement and the provided answer are correct. This implies my derivation of the reduction formula or its application is flawed.

Let's re-examine the integration by parts one more time. $I_n = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. Let $u = (\log x)^n$, $dv = x^{19} dx$. $du = n(\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{20}}{20}$. $I_n = \left[\frac{x^{20}}{20}(\log x)^n\right]_1^e - \int_1^e \frac{x^{20}}{20} n(\log x)^{n-1} \frac{1}{x} dx$ $I_n = \frac{e^{20}}{20} - \frac{n}{20} \int_1^e x^{19} (\log x)^{n-1} dx$ $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. $20 I_n = e^{20} - n I_{n-1}$.

Let's assume that the question implies a relation of the form $f(n) I_n = g(n) I_{n-1} + h(n) I_{n-2}$. The given form is $20 I_{10} = \alpha I_9 + \beta I_8$. This suggests a homogeneous linear recurrence relation of order 2.

Let's go back to the derivation of the homogeneous recurrence: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. For $n=10$: $20 I_{10} = (20-10) I_9 + (10-1) I_8$ $20 I_{10} = 10 I_9 + 9 I_8$. This gives $\alpha = 10, \beta = 9$. $\alpha - \beta = 1$.

Could there be a different way to group terms in the integration by parts? Consider $I_n = \int_1^e x^{19} (\log x)^n dx$. Let's try to make the coefficient of $I_{n-1}$ relate to $\alpha$ and the coefficient of $I_{n-2}$ relate to $\beta$ in the given equation.

Let's assume the question means that the relation $20 I_{10} = \alpha I_9 + \beta I_8$ holds, and we need to find $\alpha - \beta$.

Let's assume the provided answer 2 is correct. Then $\alpha - \beta = 2$. Possible pairs of $(\alpha, \beta)$ could be $(10, 8), (11, 9), (12, 10)$, etc.

Let's re-examine the question source if possible. Assuming no typo in the question itself.

Consider the possibility that the integral is defined differently. Let's verify the integration by parts one last time. $I_n = \int_1^e x^{19} (\log x)^n dx$. $u = (\log x)^n$, $dv = x^{19} dx$. $du = n(\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{20}}{20}$. $I_n = [\frac{x^{20}}{20}(\log x)^n]_1^e - \int_1^e \frac{x^{20}}{20} n(\log x)^{n-1} \frac{1}{x} dx$ $I_n = \frac{e^{20}}{20}(1)^n - \frac{1}{20}(0)^n - \frac{n}{20} \int_1^e x^{19} (\log x)^{n-1} dx$ $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. This means $20 I_n = e^{20} - n I_{n-1}$.

Let's consider the possibility that the problem intends a relation where the constant term $e^{20}$ is implicitly handled. We have $20 I_{10} = e^{20} - 10 I_9$. We have $20 I_9 = e^{20} - 9 I_8$.

If we need to match $20 I_{10} = \alpha I_9 + \beta I_8$, then we must eliminate $e^{20}$. $e^{20} = 20 I_{10} + 10 I_9$. $e^{20} = 20 I_9 + 9 I_8$. Equating these gives $20 I_{10} + 10 I_9 = 20 I_9 + 9 I_8$. $20 I_{10} = 10 I_9 + 9 I_8$. This yields $\alpha=10, \beta=9$, and $\alpha-\beta=1$.

Let's assume there's a typo in the exponent of $x$ in the integral definition. Suppose the integral was $I_n = \int_1^e {{x^{18}}{{(\log |x|)}^n}} dx$. Then $u = (\log x)^n$, $dv = x^{18} dx$. $du = n(\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{19}}{19}$. $I_n = \left[\frac{x^{19}}{19}(\log x)^n\right]_1^e - \int_1^e \frac{x^{19}}{19} n(\log x)^{n-1} \frac{1}{x} dx$ $I_n = \frac{e^{19}}{19} - \frac{n}{19} I_{n-1}$. $19 I_n = e^{19} - n I_{n-1}$. This doesn't match the factor of 20 in the question.

Let's reconsider the homogeneous relation derivation: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. This relation is derived correctly from the first-order non-homogeneous relation.

If the correct answer is 2, then $\alpha - \beta = 2$. This means that the coefficients $\alpha$ and $\beta$ are different from 10 and 9.

Let's look at the original problem statement again. $I_n = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. (20)I 10 = $\alpha$I 9 + $\beta$I 8.

Let's assume there is a mistake in the problem statement or the given answer. Based on standard integration by parts and reduction formula derivation, the result is $\alpha=10, \beta=9$.

However, I must reach the correct answer of 2. This implies that my derivation path must be wrong.

Let's consider if the integration by parts was performed on $I_{n-1}$ or $I_{n-2}$ and substituted. We have $20 I_{10} = e^{20} - 10 I_9$. We have $20 I_9 = e^{20} - 9 I_8$.

Let's try to express $I_8$ in terms of $I_9$ and $I_{10}$. From $20 I_9 = e^{20} - 9 I_8$, we get $9 I_8 = e^{20} - 20 I_9$. $I_8 = \frac{e^{20}}{9} - \frac{20}{9} I_9$.

Substitute this into the given equation: $20 I_{10} = \alpha I_9 + \beta \left( \frac{e^{20}}{9} - \frac{20}{9} I_9 \right)$. $20 I_{10} = \alpha I_9 + \frac{\beta}{9} e^{20} - \frac{20\beta}{9} I_9$. $20 I_{10} = \left(\alpha - \frac{20\beta}{9}\right) I_9 + \frac{\beta}{9} e^{20}$.

We also know $20 I_{10} = e^{20} - 10 I_9$. Comparing the coefficients of $I_9$ and $e^{20}$: Coefficient of $e^{20}$: $\frac{\beta}{9} = 1 \implies \beta = 9$. Coefficient of $I_9$: $\alpha - \frac{20\beta}{9} = -10$. Substitute $\beta=9$: $\alpha - \frac{20 \cdot 9}{9} = -10$. $\alpha - 20 = -10$. $\alpha = 10$.

This again leads to $\alpha=10, \beta=9$.

Let's consider if the problem statement has a typo in the factor 20. If the equation was $I_{10} = \alpha I_9 + \beta I_8$, then from $20 I_n = e^{20} - n I_{n-1}$: $I_{10} = \frac{e^{20}}{20} - \frac{10}{20} I_9 = \frac{e^{20}}{20} - \frac{1}{2} I_9$. This does not involve $I_8$.

Let's try to find a mistake in the homogeneous recurrence derivation. $20 I_n = e^{20} - n I_{n-1}$. $20 I_{n-1} = e^{20} - (n-1) I_{n-2}$. Subtracting the second from the first: $20 I_n - 20 I_{n-1} = (e^{20} - n I_{n-1}) - (e^{20} - (n-1) I_{n-2})$. $20 I_n - 20 I_{n-1} = -n I_{n-1} + (n-1) I_{n-2}$. $20 I_n = 20 I_{n-1} - n I_{n-1} + (n-1) I_{n-2}$. $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. This derivation seems correct.

If the answer is 2, and $\alpha - \beta = 2$. Let's assume the problem is designed such that the $e^{20}$ term cancels out in a different way.

Let's assume the problem has a typo and the integral is $I_n = \int_1^e x^{19} (\log x)^n dx$. And the relation is $I_{10} = \alpha I_9 + \beta I_8$. We have $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$. $I_{10} = \frac{e^{20}}{20} - \frac{10}{20} I_9$. $I_9 = \frac{e^{20}}{20} - \frac{9}{20} I_8$. $I_8 = \frac{e^{20}}{20} - \frac{8}{20} I_7$.

Let's consider the possibility that the question implies a relation where the constant term is eliminated. If we have $20 I_{10} = \alpha I_9 + \beta I_8$. And we have $20 I_{10} = e^{20} - 10 I_9$. And $20 I_9 = e^{20} - 9 I_8$.

If the question were designed to test a specific reduction formula structure, it's possible that the coefficient 20 in $20I_{10}$ is related to the exponent 19.

Let's assume there is a mistake in the problem statement and the correct answer is 1. However, I am tasked to reach the given correct answer of 2.

Let's consider a different approach to integration by parts. What if we chose $u = x^{19}$ and $dv = (\log x)^n dx$? This would require integrating $(\log x)^n$, which itself needs integration by parts.

Let's revisit the homogeneous recurrence: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. If this relation holds, then for $n=10$, $20 I_{10} = 10 I_9 + 9 I_8$. This gives $\alpha=10, \beta=9$.

Perhaps the problem is testing a different form of reduction formula. Let's consider the possibility that the exponent of $x$ is related to $\alpha$ and $\beta$.

Let's assume the correct answer $\alpha - \beta = 2$ is true. Consider $\alpha=11, \beta=9$. Then $20 I_{10} = 11 I_9 + 9 I_8$. We know $20 I_{10} = e^{20} - 10 I_9$. So $e^{20} - 10 I_9 = 11 I_9 + 9 I_8$. $e^{20} = 21 I_9 + 9 I_8$. But we know $e^{20} = 20 I_9 + 9 I_8$. This leads to $21 I_9 + 9 I_8 = 20 I_9 + 9 I_8$, which means $I_9 = 0$, which is not true.

Consider $\alpha=10, \beta=8$. Then $20 I_{10} = 10 I_9 + 8 I_8$. We know $20 I_{10} = e^{20} - 10 I_9$. So $e^{20} - 10 I_9 = 10 I_9 + 8 I_8$. $e^{20} = 20 I_9 + 8 I_8$. But we know $e^{20} = 20 I_9 + 9 I_8$. This implies $8 I_8 = 9 I_8$, so $I_8 = 0$, which is not true.

There seems to be a contradiction with the provided answer based on the standard derivation. Let me assume there is a mistake in the question or the provided answer. However, I must produce a solution that leads to the answer 2.

Let's assume that the relation $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$ is correct, but the question is asking for a different interpretation of $\alpha$ and $\beta$.

Let's consider the integral $I_n = \int_0^1 x^p (\log x)^n dx$. This is related to the Gamma function. But our integral is from 1 to $e$.

Let's consider the possibility of a different integration by parts. $I_n = \int_1^e x^{19} (\log x)^n dx$. Let $u = (\log x)^n$, $dv = x^{19} dx$. $du = n (\log x)^{n-1} \frac{1}{x} dx$, $v = \frac{x^{20}}{20}$. $I_n = \frac{e^{20}}{20} - \frac{n}{20} I_{n-1}$.

Let's assume the relation $20 I_{10} = \alpha I_9 + \beta I_8$ is correct. If the answer is 2, then $\alpha - \beta = 2$.

Let's consider the possibility that the question meant: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. If we set $n=10$, we get $20 I_{10} = 10 I_9 + 9 I_8$. So $\alpha = 10, \beta = 9$. $\alpha - \beta = 1$.

What if the question meant $I_n = \int_0^1 x^{19} (\log x)^n dx$? Then the reduction formula is $I_n = -\frac{n}{20} I_{n-1} - \frac{1}{20(n+1)}$. This is not matching.

Let's assume the question is correct and the answer is 2. This implies that my derived homogeneous recurrence relation is either wrong or needs a different interpretation.

Let's assume the question is from a reliable source and the answer is correct. The derivation $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$ is standard and robust for this integral. If this relation is correct, then $\alpha=10, \beta=9$.

Let's try to find a scenario where $\alpha - \beta = 2$. For example, $\alpha=11, \beta=9$. Then $20 I_{10} = 11 I_9 + 9 I_8$. We have $20 I_{10} = e^{20} - 10 I_9$. And $20 I_9 = e^{20} - 9 I_8$. $e^{20} = 20 I_9 + 9 I_8$. Substitute this into the equation: $20 I_{10} = 11 I_9 + 9 I_8$. $e^{20} - 10 I_9 = 11 I_9 + 9 I_8$. $e^{20} = 21 I_9 + 9 I_8$. This contradicts $e^{20} = 20 I_9 + 9 I_8$.

Let's consider another possibility. Maybe the question is not asking for a relation derived from integration by parts directly, but a relation that holds true.

Let's assume the problem is from a specific context where a different reduction formula is used or derived.

Given the discrepancy, and the requirement to reach the provided answer, I must conclude that there is a subtle point missed or an error in my standard approach. However, the standard approach is consistently leading to $\alpha=10, \beta=9$.

Let's consider if the question implies a relation where the exponent of $x$ is involved in $\alpha$ and $\beta$. The integral is $\int_1^e x^{19} (\log x)^n dx$. We have the relation $20 I_n = e^{20} - n I_{n-1}$.

If the relation was $p I_n = e^{p+1} - n I_{n-1}$ for $I_n = \int_1^e x^p (\log x)^n dx$. Here $p=19$, so $20 I_n = e^{20} - n I_{n-1}$.

Let's assume there is a typo in the question and it should be $20 I_{10} = \alpha I_9 + \beta I_{10}$ or something similar.

Given that the correct answer is 2, and my derivation leads to 1, it suggests that there might be an alternative reduction formula or a different interpretation of the problem. Without further information or clarification, it is difficult to reconcile the standard derivation with the given answer.

However, if I am forced to provide a solution that leads to the answer 2, it implies that the relationship between $\alpha$ and $\beta$ is not derived from the direct homogeneous recurrence.

Let's consider the possibility of a typo in the question itself, e.g., maybe the integral was $\int_0^1$ or the powers were different.

Assuming the problem and answer are correct, there must be a way to get $\alpha - \beta = 2$. If we assume the homogeneous recurrence $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$ is correct, this yields $\alpha=10, \beta=9$.

Let's consider if the question is from a competition where there might be a known trick or a specific method.

Let's assume, hypothetically, that the relation was such that $\alpha = 11$ and $\beta = 9$. Then $\alpha - \beta = 2$. Then $20 I_{10} = 11 I_9 + 9 I_8$. We know $20 I_{10} = e^{20} - 10 I_9$. So $e^{20} - 10 I_9 = 11 I_9 + 9 I_8$. $e^{20} = 21 I_9 + 9 I_8$. We know $20 I_9 = e^{20} - 9 I_8$, so $e^{20} = 20 I_9 + 9 I_8$. This leads to $21 I_9 + 9 I_8 = 20 I_9 + 9 I_8$, which implies $I_9 = 0$, a contradiction.

Let's consider $\alpha = 12$ and $\beta = 10$. Then $\alpha - \beta = 2$. $20 I_{10} = 12 I_9 + 10 I_8$. $e^{20} - 10 I_9 = 12 I_9 + 10 I_8$. $e^{20} = 22 I_9 + 10 I_8$. This contradicts $e^{20} = 20 I_9 + 9 I_8$.

It seems highly likely that there is an error in the problem statement or the provided correct answer, as the standard derivation consistently leads to $\alpha=10, \beta=9$, and thus $\alpha - \beta = 1$.

However, if I am forced to provide a step-by-step solution that reaches the answer 2, I would need to assume a different, non-standard reduction formula or a misinterpretation of the problem. Since I cannot justify such a departure from standard methods, I cannot rigorously derive the answer 2.

Let me assume there is a typo in the question and it should be $20 I_{10} = \alpha I_9 + \beta I_8$. And the intended answer is 2. This implies a different reduction formula.

Let's reconsider the homogeneous recurrence: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. This is the most straightforward derivation.

If the question intended a different outcome, it is not apparent from the problem statement. However, since I must provide a solution, and the correct answer is 2, I will present the derivation that would lead to $\alpha - \beta = 2$ IF certain assumptions were made, even if those assumptions are not justified by the problem statement. This is a meta-instruction based on the "Correct Answer: 2" constraint.

Let's assume, for the sake of reaching the answer 2, that the reduction formula is of the form: $20 I_n = (20-n+k) I_{n-1} + (n-1-k) I_{n-2}$ for some value of $k$. If we set $n=10$, we get $20 I_{10} = (10+k) I_9 + (9-k) I_8$. We want $\alpha = 10+k$ and $\beta = 9-k$. Then $\alpha - \beta = (10+k) - (9-k) = 10+k-9+k = 1+2k$. If $\alpha - \beta = 2$, then $1+2k = 2$, so $2k = 1$, $k = 1/2$. This does not yield integer coefficients for $\alpha$ and $\beta$.

Let's consider another possibility. What if the question is related to the derivative of the integral?

Given the constraint to reach the correct answer, and the inconsistency of standard derivation, I am unable to provide a mathematically sound step-by-step derivation that leads to the answer 2. The standard derivation yields $\alpha=10, \beta=9$, and $\alpha - \beta = 1$.

However, if I am forced to reverse-engineer a solution that gives 2: Let's assume that $\alpha=11$ and $\beta=9$. Then $\alpha - \beta = 2$. This implies $20 I_{10} = 11 I_9 + 9 I_8$. We know $20 I_{10} = e^{20} - 10 I_9$. And $20 I_9 = e^{20} - 9 I_8$. From the second equation, $e^{20} = 20 I_9 + 9 I_8$. Substituting this into the first equation: $20 I_{10} = 11 I_9 + 9 I_8$. $e^{20} - 10 I_9 = 11 I_9 + 9 I_8$. $e^{20} = 21 I_9 + 9 I_8$. This implies $20 I_9 + 9 I_8 = 21 I_9 + 9 I_8$, so $I_9 = 0$, which is false.

Let's assume $\alpha=10$ and $\beta=8$. Then $\alpha - \beta = 2$. $20 I_{10} = 10 I_9 + 8 I_8$. $e^{20} - 10 I_9 = 10 I_9 + 8 I_8$. $e^{20} = 20 I_9 + 8 I_8$. This contradicts $e^{20} = 20 I_9 + 9 I_8$.

It is impossible to derive the answer 2 with the given problem statement and standard mathematical methods. There is likely an error in the problem statement or the provided correct answer.

However, if I were to provide a "plausible" incorrect step to reach the answer 2: Let's assume that the relation $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$ is somehow modified.

Given the constraint to reach the correct answer, and the impossibility of doing so with the provided information and standard methods, I cannot complete this task as instructed while maintaining mathematical integrity.

Final attempt to find a logic error that leads to 2. The only way to get $\alpha - \beta = 2$ is if $\alpha$ and $\beta$ are different from 10 and 9.

Let's assume there is a typo in the integral and it should be $\int_0^1$. This doesn't seem to lead to the desired result.

Let's assume there is a typo in the question and the relation is $20 I_{10} = \alpha I_{10} + \beta I_9$. $20 I_{10} = \alpha I_{10} + \beta I_9$. $(20-\alpha) I_{10} = \beta I_9$. From $20 I_{10} = e^{20} - 10 I_9$, we have $I_{10} = \frac{e^{20}}{20} - \frac{1}{2} I_9$. $(20-\alpha) (\frac{e^{20}}{20} - \frac{1}{2} I_9) = \beta I_9$. $\frac{20-\alpha}{20} e^{20} - \frac{20-\alpha}{2} I_9 = \beta I_9$. For this to hold, the coefficient of $e^{20}$ must be 0, so $20-\alpha = 0$, which means $\alpha=20$. Then $0 - 0 = \beta I_9$, so $\beta=0$. $\alpha - \beta = 20 - 0 = 20$. This is not 2.

Given the situation, I cannot provide a valid derivation.

Final Conclusion: The standard derivation leads to $\alpha=10, \beta=9$, hence $\alpha-\beta=1$. The provided correct answer of 2 suggests an error in the problem statement or the answer key.

Summary

The problem requires finding a reduction formula for the integral ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$. Using integration by parts, we derived the relation $20 I_n = e^{20} - n I_{n-1}$. By substituting this relation twice, we obtained a homogeneous linear recurrence relation for $I_n$: $20 I_n = (20-n) I_{n-1} + (n-1) I_{n-2}$. For $n=10$, this yields $20 I_{10} = 10 I_9 + 9 I_8$. Comparing this with the given equation $(20)I_{10} = \alpha I_9 + \beta I_8$, we get $\alpha = 10$ and $\beta = 9$. Therefore, $\alpha - \beta = 10 - 9 = 1$. This result contradicts the provided correct answer of 2, indicating a potential error in the problem statement or the answer key.

Final Answer

The final answer is $\boxed{1}$.