Calculus and Patterns

Finding an Integral expression for the area:

We first need to find a way to write this area as an integral. For the bounds of integration solve $\log^n x = \log^{n-1} x$ . Let $u = \log x$ to get the polynomial equation $u^n - u^{n-1} = 0 \implies u^{n-1} (u - 1) = 0 \implies u=0,1$ for $n>1$ . If $n=1$ no area is bound between the curves. So we have to solve $\log x = 0,1 \implies x = 1, e$ . Also, as the function ranges from $0$ to $1$ we will have that $\log^{n-1} x > \log^n x$ so an expression for this area is:

$\large \int_1^e \log^{n-1} x - \log^{n} x dx$ .

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Introducing a new sequence:

Computing this integral ended up being remarkably difficult but as the problem hints, it is possible to write it in terms of two arithmetic functions. My procedure is as follows: First, let $u = \log x$ . Then $dx = e^u du$ and thus the integral is equal to $\large \int_0^1 x^{n-1} e^x - x^{n} e^x dx$ . To solve this, I will introduce the following sequence. Let $s_n = \int_0^1 x^n e^x dx$ . This sequence has the property that the integral we want to compute is just $s_{n-1} - s_n$ so if we can find a general formula for $s_n$ we can tackle the original problem.

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Recurrence relation for $s_n$

We have that $s_n = \int_0^1 x^n e^x dx$ . Let $u = x^n, dv = e^x$ so $du = nx^{n-1}, dv = e^x$ to get, by integration by parts, that this is equal to $e - n \int_0^1 x^{n-1} e^x dx = e - n s_{n-1}$ . So the recurrence relation is $s_n = e - n s_{n-1}$ . We can completely define the sequence in with this recurrence relation if we only compute $s_1 = \int_0 ^1 x e^x dx = 1$ .

I now show some computed values: $s_2 = e - 2 \\ s_3 = -2e + 6 \\ s_4 = 9e - 24 \\ s_5 = -44e + 120$ . At a quick glance one can immediately notice that the integer term is $(-1)^{n+1} n!$ . But the $e$ term is much harder to find by just looking, some extra work has to be put in. If one writes out what the $e$ term's coefficient is they will find the following expression: $1 - n + n(n-1) - n(n-1)(n-2) + ... \pm \frac{n!}{2!} = 1 - (-1)^{n+1}\sum_{k=2}^{n-1} \frac{n!}{k!} (-1)^k$

With this we have finally solved $\large s_n = (-1)^{n+1} n! + e \left( 1 - (-1)^{n+1}\sum_{k=2}^{n-1} \frac{n!}{k!} (-1)^k \right )$

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General formula for the area

With the previous result finding the area is just an exercise in algebra but we end up with: $\Large (-1)^{n+1} \left( \left( \sum_{k=2}^{n-1} \frac{n!}{k!} (-1)^k + \sum_{k=2}^{n-2} \frac{(n-1)!}{k!} (-1)^k \right)e - (n! + (n-1)!) \right)$ with, specifically:

$a_n = \sum_{k=2}^{n-1} \frac{n!}{k!} (-1)^k + \sum_{k=2}^{n-2} \frac{(n-1)!}{k!} (-1)^k$

$b_n = n! + (n-1)! = (n-1)!(n+1)$

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Solving for $\alpha$

For the fraction limit, I will consider separately both of the sums that make up $a_n$ . Diving the first part of $a_n$ by $b_n$ yields $\frac{n}{n+1} \sum_{k=2}^{n-1} \frac{(-1)^k}{k!}$ . As $n \to \infty$ the fraction outside goes to 1 and the sum goes to $e^{-1}$ so this part of the limit equals $\frac{1}{e}$ . Now notice that for the second part of $a_n$ we just get $\frac{1}{n+1} \sum_{k=2}^{n-2} \frac{(-1)^k}{k!}$ . The sum approaches the constant $e^{-1}$ but the fraction outside goes to 0 so this part vanishes. Leaving us with the limit being $\frac{1}{e}$ and thus $\alpha = 1$ .

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Solving for $\gamma$

Notice that $b_{20} = 19! (21)$ . 21 is not a prime number, and $19!$ will contain all the prime numbers from 1 to 19 so $\gamma = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77$ .

Solving for $\beta$

Here I did not figure out any clever way to compute this in an easy way. I used a computer program to compute this using the formula I gave for $a_n$ . The result is $\beta = 112$ .