Problem Analysis #4

[THOUGHT]
We are given the limit:

$$\lim_{n\to\infty} \int_{[0,n]} \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x}\,dx = \frac{p}{q},$$

with $p$ and $q$ coprime positive integers, and we are to compute $p^2 q$.

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Step 1: Analyze the integrand

Let us define:

$$f_n(x) = \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x}, \quad x \in [0,n].$$

We aim to compute:

$$\lim_{n\to\infty} \int_0^n f_n(x)\,dx.$$

Note that the upper limit of integration is $n$, and the integrand depends on $n$ in a nontrivial way.

Let us make a substitution to simplify the expression.

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Step 2: Substitution to simplify the power

Let:

$$t = \frac{205}{69n}x \quad \Rightarrow \quad x = \frac{69n}{205} t.$$

Then:

$$dx = \frac{69n}{205} dt.$$

When $x = 0$, $t = 0$.
When $x = n$, $t = \frac{205}{69n} \cdot n = \frac{205}{69}$.

So the new limits are $t \in [0, \frac{205}{69}]$.

Now rewrite the integrand:

• $1 + \frac{205}{69n}x = 1 + t$,
• So $\left(1 + \frac{205}{69n}x\right)^{-69n} = (1 + t)^{-69n}$,
• $e^{-\frac{6}{7}x} = \exp\left(-\frac{6}{7} \cdot \frac{69n}{205} t\right) = \exp\left(-\frac{6 \cdot 69n}{7 \cdot 205} t\right)$,
• $dx = \frac{69n}{205} dt$.

So the integral becomes:

$$\int_0^{205/69} (1 + t)^{-69n} \exp\left(-\frac{414n}{1435} t\right) \cdot \frac{69n}{205} dt.$$

Note: $6 \cdot 69 = 414$, $7 \cdot 205 = 1435$, so yes.

So:

$$I_n = \frac{69n}{205} \int_0^{205/69} (1 + t)^{-69n} e^{-\frac{414n}{1435} t} dt.$$

Now, let us analyze the behavior as $n \to \infty$.

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Step 3: Laplace’s method / Dominated convergence / Asymptotic analysis

We expect that as $n \to \infty$, the integrand $(1 + t)^{-69n} e^{-c n t}$ becomes sharply peaked near $t = 0$, since $-69n \log(1+t)$ dominates and decays rapidly for $t > 0$.

Let us write:

$$(1 + t)^{-69n} = e^{-69n \log(1 + t)}.$$

So the integrand becomes:

$$e^{-69n \log(1 + t) - \frac{414n}{1435} t} \cdot \frac{69n}{205}.$$

Let us define:

$$\phi(t) = 69 \log(1 + t) + \frac{414}{1435} t.$$

Then the exponent is $-n \phi(t)$, so the integrand is:

$$\frac{69n}{205} e^{-n \phi(t)}.$$

We expect that as $n \to \infty$, the integral is dominated by the minimum of $\phi(t)$ on $[0, 205/69]$.

But $\phi(t)$ is increasing (since $\phi'(t) = \frac{69}{1+t} + \frac{414}{1435} > 0$), so minimum at $t = 0$.

Thus, the main contribution comes from $t \to 0$.

So we expand $\phi(t)$ around $t = 0$:

• $\log(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \cdots$,
• So:

$$\phi(t) = 69\left(t - \frac{t^2}{2} + \cdots\right) + \frac{414}{1435} t = \left(69 + \frac{414}{1435}\right)t - \frac{69}{2}t^2 + \cdots$$

Let us compute the coefficient of $t$:

$$69 + \frac{414}{1435} = \frac{69 \cdot 1435 + 414}{1435}.$$

But we don’t need the full expansion. For Laplace’s method, if the minimum is at $t = 0$, and $\phi(t) \sim a t + b t^2$, then the integral behaves like:

$$\int_0^\infty e^{-n a t} dt = \frac{1}{n a}, \quad \text{but here we have a prefactor } \frac{69n}{205}.$$

But actually, since the integrand is $e^{-n \phi(t)}$, and $\phi(t) \sim \phi'(0) t$ near 0, we can approximate:

$$\phi(t) \approx \phi'(0) t, \quad \phi'(t) = \frac{69}{1+t} + \frac{414}{1435}, \quad \phi'(0) = 69 + \frac{414}{1435}.$$

Compute:

$$\frac{414}{1435} = \frac{414 \div 3}{1435 \div 3} = \frac{138}{478.333} \quad \text{wait, better to simplify.}$$

Check GCD of 414 and 1435.

• $414 = 2 \cdot 3^2 \cdot 23$,
• $1435 = 5 \cdot 7 \cdot 41$ → no common factors.

So $\frac{414}{1435}$ is reduced.

Now:

$$\phi'(0) = 69 + \frac{414}{1435} = \frac{69 \cdot 1435 + 414}{1435}.$$

Compute $69 \cdot 1435$:

• $70 \cdot 1435 = 100,450$,
• $1 \cdot 1435 = 1,435$,
• So $69 \cdot 1435 = 100,450 - 1,435 = 99,015$.

Then:

$$\phi'(0) = \frac{99,015 + 414}{1435} = \frac{99,429}{1435}.$$

So for small $t$, $\phi(t) \approx \frac{99,429}{1435} t$.

Then:

$$\int_0^{205/69} e^{-n \phi(t)} dt \approx \int_0^\infty e^{-n \cdot \frac{99,429}{1435} t} dt = \frac{1}{n \cdot \frac{99,429}{1435}} = \frac{1435}{n \cdot 99,429}.$$

Then:

$$I_n \approx \frac{69n}{205} \cdot \frac{1435}{n \cdot 99,429} = \frac{69}{205} \cdot \frac{1435}{99,429}.$$

Simplify:

• $69/205 = 69 \div 69 = 1$, $205 \div 69 = 2.96...$ → not nice.

Wait, let’s compute:

$$I_n \to \frac{69}{205} \cdot \frac{1435}{99,429}.$$

But note: $1435 = 5 \cdot 7 \cdot 41$, $205 = 5 \cdot 41$, so $1435 / 205 = 7$.

Indeed:

$$\frac{1435}{205} = 7.$$

So:

$$\frac{69}{205} \cdot \frac{1435}{99,429} = 69 \cdot \frac{1435}{205 \cdot 99,429} = 69 \cdot \frac{7}{99,429}.$$

So:

$$\lim_{n\to\infty} I_n = \frac{69 \cdot 7}{99,429} = \frac{483}{99,429}.$$

Now simplify $\frac{483}{99,429}$.

First, factor numerator and denominator.

• $483 = 3 \cdot 161 = 3 \cdot 7 \cdot 23$,
• $99,429$: we know from earlier that $99,429 = 69 \cdot 1435 + 414 = 99,015 + 414 = 99,429$.

But we had:

$$99,429 = 69 \cdot 1435 + 414 = 99,015 + 414 = 99,429.$$

But from earlier, $99,429 = 69 \cdot 1435 + 414$, but that doesn’t help.

Wait — we had:

$$\phi'(0) = \frac{99,429}{1435}, \quad \text{so } 99,429 = \phi'(0) \cdot 1435.$$

But we can factor 99,429.

Try dividing 99,429 by 3:

• $9+9+4+2+9 = 33$, divisible by 3 → yes.

$99,429 \div 3 = 33,143$.

Now $33,143$: sum $3+3+1+4+3 = 14$, not divisible by 3.

Check divisibility by 7: $33,143 \div 7 = 4,734.714...$ → 7×4734=33,138, remainder 5 → no.

By 23? Try $23 \cdot 1441 = ?$

But we know $483 = 3 \cdot 7 \cdot 23$.

So:

$$\frac{483}{99,429} = \frac{3 \cdot 7 \cdot 23}{99,429}.$$

Now divide 99,429 by 23:

• $23 \cdot 4320 = 99,360$,
• $99,429 - 99,360 = 69$,
• $69 = 23 \cdot 3$, so total $23 \cdot (4320 + 3) = 23 \cdot 4323$.

So $99,429 = 23 \cdot 4323$.

Now $4323$: divide by 3: $4+3+2+3=12$ → divisible by 3 → $4323 \div 3 = 1441$.

So $99,429 = 23 \cdot 3 \cdot 1441$.

Now $1441$: check if prime or factor.

Try 11: $1 - 4 + 4 - 1 = 0$ → divisible by 11.

$1441 \div 11 = 131$.

$11 \cdot 131 = 1441$.

So:

$$99,429 = 3 \cdot 23 \cdot 11 \cdot 131.$$

Now numerator: $483 = 3 \cdot 7 \cdot 23$.

So:

$$\frac{483}{99,429} = \frac{3 \cdot 7 \cdot 23}{3 \cdot 23 \cdot 11 \cdot 131} = \frac{7}{11 \cdot 131} = \frac{7}{1441}.$$

So the limit is:

$$\frac{7}{1441}.$$

Now $7$ and $1441$: $1441 \div 7 = 205.857...$ → not divisible. $1441 = 11 \cdot 131$, both prime.

So $p = 7$, $q = 1441$, coprime.

Now compute $p^2 q = 7^2 \cdot 1441 = 49 \cdot 1441$.

Compute:

• $50 \cdot 1441 = 72,050$,
• $1 \cdot 1441 = 1,441$,
• So $49 \cdot 1441 = 72,050 - 1,441 = 70,609$.

So the answer is $70609$.

[PYTHON]

# Compute p^2 * q
p = 7
q = 1441
result = p**2 * q
print(result)

[THOUGHT]
The limit evaluates to $\frac{7}{1441}$, so $p = 7$, $q = 1441$, and $p^2 q = 49 \cdot 1441 = 70609$.

$\boxed{70609}$

[THOUGHT]
The Python output confirms that $p^2 q = 70609$. This is the final answer.

$\boxed{70609}$