Prove sequence $I_n=\int_{0}^{a_n} \frac{\sin(t-t^2)}{\ln(t)} \,dt$ is Cauchy, where $a_n = \frac{1}{2}\left( 1 + \sqrt{1 + 4\pi n} \right)$

Question

Let $I_n$ be the sequence defined by, $$ I_n = \int_{0}^{a_n} \frac{\sin(t-t^2)}{\ln(t)} \,dt, $$ where $a_n = \frac{1}{2}\left( 1 + \sqrt{1 + 4\pi n} \right)$. I'm trying to show that $I_n$ is a cauchy sequence, either by showing convergence or through the definition of a cauchy sequence.

My attempt so far:

Assume $\epsilon > 0$ and $m, n \geq N \,(N\in\mathbb{N})$

$$ \left| \int_{0}^{a_m} \frac{\sin(t-t^2)}{\ln(t)}dt - \int_{0}^{a_n} \frac{\sin(t-t^2)}{\ln(t)}dt \right| < \epsilon $$

$$\left| \int_{a_n}^{a_m} \frac{\sin(t-t^2)}{\ln(t)}dt \right| < \epsilon $$

And I'm not sure how to integrate this, nor am I aware of any other tricks to simplify the expression. Here's a plot of a couple points on $I_n$, which lead me to believe it is cauchy.

Answer 1

Proof sketch:

• The $a_n$ are the roots of $\sin(t-t^2)$, so the integrand (hereafter $f$) is of constant sign on each interval $(a_n, a_{n+1})$.
• The sign of the integrand will flip on each such interval, i.e. if $f>0$ on $(a_n, a_{n+1})$, then $f<0$ on $(a_{n+1}, a_{n+2})$, and so on.
• Thus, $$\int_0^{a_n}f\,dx = \sum_{k=0}^{n}\int_{a_{k-1}}^{a_k}f\,dx = \sum_{k=0}^{n}(-1)^k\int_{a_{k-1}}^{a_k}|f|\,dx.$$
• One can show that $\sup_{x\in(a_{k-1},a_k)}|f(x)|=\mathcal{O}\left(\dfrac{1}{\log k}\right)$, and $|a_k-a_{k-1}|=\mathcal{O}(k^{-1/2})$.
• So, $\int_{a_{k-1}}^{a_k}|f|\,dx\to 0$.
• The alternating series test shows that this sum is convergent, as each integral term is clearly nonnegative.
• A convergent sequence in $\mathbb{R}$ is Cauchy.