# The convergence of $\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}$

I need to evaluate the following integral

$$\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}dx$$ where $$\alpha\in (0,1)$$ and $$[x]$$ is the floor function. Without $$x-[x]$$, we can evaluate the integral easily using analytical continuation. $$\int_0^{\infty} \sin(x)/x^{\alpha}dx=\Im \int_0^{\infty} e^{ix}/x^{\alpha}dx=\Im [\Gamma(1+\alpha)/i^{\alpha-1}]$$

In the presence of $$x-[x]$$, my idea is that one could use inequality to upper bound the integral if it converges or lower bound the integral if it diverges. But I'm not sure what kind of inequality I should use.

I tried Hölder's inequality, but it will square the sin function in the upper bound, which makes it diverge.

• You have checked that the integral converges, I guess. – user436658 Jan 3 at 21:06
• $x-\lfloor x\rfloor=\text{frac}(x)$, the fractional part of $x$ – Raffaele Jan 3 at 21:46
• The Fourier series of $x-\lfloor x\rfloor$ gives $\sin(x)(x-\lfloor x\rfloor)=\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n}$, integrating termwise gives a Dirichlet series – reuns Jan 3 at 22:09
• @reuns Integrating this series termwise can be justified? I don't think it's absolutely convergent. – user436658 Jan 4 at 7:57

Let's settle (as in the title of the OP) just the convergence of the integral. The obvious choice is Dirichlet's test: we only have to show that $$\int^N_0\sin(x)(x-[x])\,dx$$ is uniformly bounded. Since $$\sin(x)(x-[x])$$ is bounded, it's sufficient to show that for integer $$N$$. But then, $$\int^N_0\sin(x)(x-[x])\,dx=\sum^{N-1}_{n=0}\int^{n+1}_n\sin(x)(x-[x])\,dx=\int^1_0y\,\sum^{N-1}_{n=0}\sin(n+y)\,dy,$$ and $$\sum^{N-1}_{n=0}\sin(n+y)=\cos y\,\sum^{N-1}_{n=0}\sin n+\sin y\,\sum^{N-1}_{n=0}\cos n.$$ Since $$\left|\sum^{N-1}_{n=0}\sin n\right|=\left|\frac{\cos(1/2)-\cos(N-1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)}$$ and $$\left|\sum^{N-1}_{n=0}\cos n\right|=\left|\frac{\sin(N-1/2)+\sin(1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)},$$ this is uniformly bounded. $$x^{-\alpha}\to0$$ as $$x\to\infty$$ guarantees the convergence of the integral.

BONUS: If $$\alpha>0$$, $$e^{-bx}\,x^{-\alpha}\to0$$ as $$x\to\infty$$ uniformly in $$b\ge0$$, so $$\displaystyle\int_0^{\infty} e^{-bx}\,\sin(x)(x-[x])\,x^{-\alpha}\,dx$$ converges uniformly in $$b\ge0$$.

For $$a\in (1,2)$$, $$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx= \lim_{b\to 0} \int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $$\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ is analytic for $$a < 2$$.

For $$a \in (0,1/2)$$

• $$e^{-bx/2}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})$$ converges in $$L^2(0,\infty)$$

• and $$x^{-a}e^{-bx/2}$$ is $$L^2(0,\infty)$$

Thus no problem to switch $$\int,\sum$$ $$\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $$= \int_0^{\infty} e^{-bx}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})x^{-a}dx$$ $$= \Gamma(1-a)\Re(\frac{(b+i)^{a-1}}{2i}-\sum_{n\ge 1} \frac{(b+i(2\pi n-1))^{a-1}-( b+i(2\pi n+1))^{a-1}}{2\pi n})$$

The latter series extends analytically to $$\Re(a)\in (0,2)$$ so it must stay equal to $$\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$.

Thus for $$a\in (1,2)$$ we can let $$b\to 0$$ to obtain

$$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx = \Gamma(1-a)\Re(\frac{i^{a-1}}{2i}-\sum_{n\ge 1} \frac{(i(2\pi n-1))^{a-1}-(i(2\pi n+1))^{a-1}}{2\pi n})$$

• $\int^\infty_0e^{-bx}\sin x\,dx=\frac1{1+b^2}$, now letting $b\to 0$ gives...? – user436658 Jan 5 at 8:02
• I don't see what you mean. – reuns Jan 5 at 16:00
• You do, thus the edit. $b\to0$ is justified for $a\in(1,2)$, but the OP has $\alpha\in(0,1)$. – user436658 Jan 5 at 16:10
• No I don't. I thought it was $(1,2)$. I think my answer proves it converges for $(0,1)$ but it is not fully obvious. – reuns Jan 5 at 16:12
• It converges just fine, and with the additional factor $e^{-bx}$, even uniformly in $b$, but that needs another line or two. – user436658 Jan 5 at 16:14