Does the integral of sin(xlogx) from 0 to infinity converge? I saw an interesting post today: Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge? 
I was wondering if anyone could explain the answer given in the post further. Specifically, why are each of the In's decreasing in their absolute value?
Any tips would be greatly appreciated!
 A: In the explanation of why $\displaystyle \int_1^\infty \sin(x \log x) \mathrm{d} x$ converges, Johnathan Y. says that if you call $I_n$ the integral between the $n$th and $n+1$st sign changes of $\sin(x \log x)$, then the values of $I_n$ are flipping signs and decreasing in absolute value, towards $0$. 
To see this, first note that $x\log x$ is a nonnegative, strictly increasing function on $[1, \infty]$. Further, it is concave up, meaning that it has positive second derivative: so the rate at which it is increasing is itself increasing.
The zeroes of the sine function are on multiples of $\pi$, and $\sin (x)$ flips sign after each zero. So it's clear that $I_n$ are flipping signs, because they are integrals over regions where $\sin$ is either entirely positive or entirely negative. 
Why is it that $|I_{n+1}| < |I_n|$? The idea is that since $x\log x$ is concave up, successive zeroes occur more rapidly. Equivalently, $\sin(x\log x)$ oscillates more rapidly as $x$ increases. The graph of $\sin (x\log x)$ looks roughly the same on each interval, except that it gets "compressed" more and more, yielding smaller areas.
Why is $|I_n| \to 0$? As $|\sin(x)|$ is bounded by $1$, we see that $|I_n|$ is bounded by the width of the interval, i.e. the distance between successive zeroes of $\sin(x\log x)$. This again boils down to examining the idea that since $x \log x$ is concave up, successive zeroes occur more and more rapidly. In fact, since the zeroes of $\sin x$ are spaced linearly, but the growth of $x \log x$ is more than linear, the distance between zeroes of $\sin (x \log x)$ goes to $0$, making $|I_n| \to 0$. 
