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Numerical result shows that

$$ \lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{\ln x}\,dx=\frac\pi2,\qquad \lim_{n\to\infty}\int_0^1\frac{\sin\left((2n+1)\pi x\right)}{\ln x}\,dx=-\frac\pi2. $$

I want to know if it's true or my computer is just lying to me.

What I've tried:

  • Something like here and here. Maybe it can work but I don't know how to deal with the limit. It just ruins everything.

  • Residue theorem for $$f:z\mapsto\frac{\exp(i\cdot2n\pi z)}{\log z}$$ using a curve around the $1/4$ unit disk without center and outer edge: $(\epsilon_1, 0)\to(1-\epsilon_2, 0)\to(0, 1-\epsilon_2)\to(0,\epsilon_1)\to(\epsilon_1, 0)$.

    It didn't work. Maybe using directly $z\mapsto \sin (2n\pi z)/\log z$ can lead to something but I didn't try.

  • Substitution using $t = 2nx$, then dividing the integration interval into $2n$ parts and trying to get a Riemann sum. I failed.

Any other ideas?

Aforest
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  • It makes good sense that the two limits are negatives of each other. The large contribution comes from the $x=1$ end. Both sines are approaching $0$, cancelling the $0$ in the denominator, but the even $n$ case is negative while the odd $n$ case is positive. As $n$ gets large the wiggles of the sine will be so fast the log doesn't have time to change, so the integral over a range of $\frac 1n$ will go to zero – Ross Millikan Dec 13 '20 at 03:34

1 Answers1

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Both results follow from the well known Dirichlet integral $\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$

We first notice that $g(x)=\frac{1}{\log x}+\frac{1}{1-x}$ is continuous on $[0,1]$ (where we define by continuity $g(0)=1, g(1)=1/2$ since $1-x+\log x=-(1-x)^2/2+O((1-x)^3), \log x=-(1-x)(1+O(1-x))$ near $1^-$)

Hence by the Riemann-Lebesgue lemma:

$\lim_{m\to\infty}\int_0^1\sin(m\pi x)g(x)dx=0$ so the required limits are $-\lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{1-x}dx$ and $-\lim_{n\to\infty}\int_0^1\frac{\sin((2n+1)\pi x)}{1-x}dx$ respectively once we show that those exists.

Changing variables $x \to 1-x$ and using $\sin(2n\pi (1-x))=-\sin(2n\pi x)$ the first integral becomes

$\int_0^1\frac{\sin(2n\pi x)}{x}dx=\int_0^{2n\pi}\frac{\sin y}{y}dy$ and the limit follows from the Dirichlet integral as noted.

Obviously $\sin((2n+1)\pi (1-x))=\sin((2n+1)\pi x)$ so the second integral preserves the minus sign and the limit is then $-\pi/2$

Conrad
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