# Finding $\text{PV}\int_0^\infty\frac{\sec\left(\pi B(xt-\lfloor xt+\frac12 \rfloor\right)-\sec\left(\pi B(x-\lfloor x+\frac12 \rfloor\right)}{x}dx$

The following problem is proposed by a friend

$$\text{PV}\int_0^\infty\frac{\sec\left(\pi B(xt-\lfloor xt+\frac12 \rfloor\right)-\sec\left(\pi B(x-\lfloor x+\frac12 \rfloor\right)}{x}\mathrm{d}x,\quad t>0,\quad |B|<1$$

where the $$\text{PV}$$ is the Cauchy principal value and $$\lfloor x \rfloor$$ is the floor function.

First I wrote $$\int_0^\infty=\lim_{N\to \infty}\int_0^N$$, then I separated the integrand and let $$xt\to x$$ in the first integral. I found the two integrals cancel out but my friend claims that the integral is not zero. Any idea?

• The integrals don't quite "cancel"; you get $\lim\limits_{N\to\infty}\left(\int_0^{\color{red}{tN}}(\ldots)-\int_0^N(\ldots)\right)$. Apr 27, 2022 at 6:33

Let $$f$$ be any $$1$$-periodic function such that (Lebesgue's) $$\int_0^1\big(f(x)/x\big)\,dx$$ exists. Put $$F(y):=\int_0^y f(x)\,dx,\qquad G(y):=\int_0^y\frac{f(x)}{x}\,dx.$$ Then $$f(x+1)=f(x)$$ implies $$F(y+1)=F(y)+F(1)$$, i.e. $$g(y):=F(y)-yF(1)$$ is (absolutely) continuous and $$1$$-periodic, hence bounded; thus, $$\Delta(y):=G(y)-F(1)\log y=G(1)+\frac{g(y)}{y}+\int_1^y\frac{g(x)}{x^2}\,dx$$ has a finite limit as $$y\to\infty$$ (i.e., the $$\int_1^\infty$$ converges). Therefore, for any $$t>0$$, $$\lim_{y\to\infty}\int_0^y\frac{f(tx)-f(x)}{x}\,dx=\lim_{y\to\infty}\big(G(ty)-G(y)\big)\\{}=F(1)\log t+\lim_{y\to\infty}\big(\Delta(ty)-\Delta(y)\big)=\color{red}{F(1)\log t}.$$
Your integral is obtained at $$f(x)=\sec\pi B(x-\lfloor x+1/2\rfloor)-1$$, for which $$F(1)=\frac2{\pi B}\log\tan\frac{\pi(1+B)}{4}-1.$$