Study the improper integral for convergence and absolute convergence

Question

$$\int_1^\infty \cos( x^2\ln(x) )\ dx $$ here is an integration which I have to study for convergence and absolute convergence.

I cannot find f(x) and g(x) to use Dirichlet test.

(Hint: Use Dirichlet test+negation of Cauchy condition)

Answer 1

Following the discussions in comments with @OliverDiaz $$x^2_n\log(x_n)=\frac{(2n+1)\pi}{2} \implies x_n=\sqrt{ \frac{\pi (2 n+1)}{W(\pi (2 n+1))}}$$ where $W(.)$ is Lambert function. So, the length of the interval
$$\Delta_n=x_{n+1}-x_n=\sqrt{ \frac{\pi (2 n+3)}{W(\pi (2 n+3))}}-\sqrt{ \frac{\pi (2 n+1)}{W(\pi (2 n+1))}}\tag1$$ is a decreasing function of $n$. The expansion for large values of $n$ is quite difficult. So, assuming $W(y)\sim \log(y)$,we have the approximation $$\Delta_n\sim\sqrt{ \frac{\pi }{2 n}}\log \left(\frac{2 \pi n}{e}\right) \big[\log(2\pi n)\big]^{-3/2}\tag 2$$ CHecking for a few values $$\left( \begin{array}{ccc} n & (2) & (1) \\ 5 & 0.214310 & 0.228359 \\ 10 & 0.147734 & 0.162283 \\ 15 & 0.118389 & 0.131434 \\ 20 & 0.101099 & 0.112808 \end{array} \right)$$ which is not fantastic but, in my opinion, acceptable for the analysis.

Using a quick and dirty nonlinear regression for the model $$\Delta_n=a\,n^b+c$$ using the values of $(1)$ tabulated for $1 \leq n \leq 100$, we have with $R^2=0.99936$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +0.481053 & 0.002469 & \{+0.476153,+0.485954\} \\ b & -0.372854 & 0.006015 & \{-0.384793,-0.360915\} \\ c & -0.042170 & 0.002965 & \{-0.048056,-0.036285\} \\ \end{array}$$

Edit (after @OliverDiaz's answer)

Computing

$$I_k=\int_1^{10^k}\cos( x^2\ln(x) )\, dx$$ we can see a quite fast convergence $$\left( \begin{array}{cc} k & I_k \\ 1 & 0.2559212210 \\ 2 & 0.2708833458 \\ 3 & 0.2701534539 \\ 4 & 0.2710161375 \\ \cdots & \cdots \\ \infty & 0.2701136093 \end{array} \right)$$

Answer 2

This is not a rigorous answer, but it seems that the improper integral in the OP converges based on numerical calculation done by @ClaudeLeibovici.

Define the function $$F(x)=\int^x_1\cos(t^2\log(t))\,dt$$ and consider the intervals $[x_n,x_{n+1}]$ where $x^2_n\log(x_n)=\frac{(2n+1)\pi}{2}$. The function $\phi:t\mapsto\cos(t^2\log(t))$ vanishes at each $x_n$, it has sign $(-1)^{n+1}$ in $[x_n,x_{n+1}]$, and $|\phi(x'_n)|=1$ for some point $x_n<x'_n<x_{n+1}$. Since $g(t)=t^2\log(t)$ is monotone increasing, $x_n\xrightarrow{n\rightarrow\infty}\infty$. The calculations of Claude's suggests that the length $\Delta_n$ of $[x_n,x_{n+1}]$ decays as $$\Delta_n\sim\sqrt{ \frac{\pi }{2 n}}\log \left(\frac{2 \pi n}{e}\right) \big[\log(2\pi n)\big]^{-3/2}=O\big(n^{−0.372854}\big)$$ Notice that $$\begin{align} F(x_n)&=\int^{x_0}_1\phi(t)\,dt +\sum^n_{j=1}\int^{x_j}_{x_{j-1}}\phi(t)\,dt\\ \end{align}$$ and $$ \frac12\Delta_n\leq \Big|\int^{x_j}_{x_{j-1}}\phi(t)\,dt\Big|\leq\Delta_n$$ Then, the series $\sum^\infty_{n=1}\int^{x_n}_{x_{n-1}}\phi(t)\,dt$ would be a convergent alternating series. This would suggest that improper integral $\int^\infty_1\phi(t)\,dt$ converges. Still some additional works need to be done to address existence of $\lim_{x\rightarrow\infty}F(x)$ not only along the sequence $x_n$.