Borwein–Borwein integrals and sumsDirect calculation

1 Direct calculation

Turns out it is possible to calculate the integrals above by pure brute force, and this gives explicit formulas for the integrals as we see above.

In general, let a0,a1,a2,,ana_0, a_1, a_2, \ldots , a_n be a sequence of positive real numbers, and consider the integral

k=0nsinc(akx)  dx. \int _{-\infty }^\infty \prod _{k = 0}^n \operatorname{sinc}(a_k x) \; \mathrm{d}x.

Let us put aside the 1x\frac{1}{x} factors for a moment, and expand out k=0nsin(akx)\prod _{k = 0}^n \sin (a_k x):

k=0nsin(akx)=1(2i)n+1(eia0xeia0x)k=1n(eiakxeiakx)=1(2i)n+1γ{1,1}nεγ(eibγx(1)neibγx), \begin{aligned} \prod _{k = 0}^n \sin (a_k x) & = \frac{1}{(2i)^{n + 1}} (e^{i a_0 x} - e^{-i a_0 x}) \prod _{k = 1}^n (e^{i a_k x} - e^{-i a_k x})\\ & = \frac{1}{(2i)^{n + 1}} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma (e^{i b_\gamma x} - (-1)^n e^{-i b_\gamma x}), \end{aligned}


bγ=a0+k=1nγkak,εγ=k=1nγk. b_\gamma = a_0 + \sum _{k = 1}^n \gamma _k a_k,\quad \varepsilon _\gamma = \prod _{k = 1}^n \gamma _k.

Note that each of the terms in the right-hand sum is some sort of trigonometric function, depending on the value of nn mod 22.

The original integral was

xn1k=0nsin(akx)  dx. \int _{-\infty }^\infty x^{-n - 1} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x.

Since sin(akx)\sin (a_k x) has a simple zero at x=0x = 0, we know from this expression that we can integrate this by parts nn times and have vanishing boundary terms:

xn1k=0nsin(akx)  dx=1n!1xdndxnk=0nsin(akx)  dx. \int _{-\infty }^\infty x^{-n - 1} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x = \frac{1}{n!} \int _{-\infty }^\infty \frac{1}{x} \frac{\mathrm{d}^n}{\mathrm{d}x^n} \prod _{k = 0}^n \sin (a_k x)\; \mathrm{d}x.

We now use the expression above to compute this nn-fold derivative, and get

k=0nsin(akx)x  dx=1n!1x12nγ{1,1}nεγbγnsin(bγx)  dx=π2nn!γ{1,1}nεγbγnsign(bγ). \begin{aligned} \int _{-\infty }^\infty \prod _{k = 0}^n \frac{\sin (a_k x)}{x} \; \mathrm{d}x & = \frac{1}{n!}\int _{-\infty }^\infty \frac{1}{x} \frac{1}{2^n} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma b_\gamma ^n \sin (b_\gamma x) \; \mathrm{d}x \\ & = \frac{\pi }{2^n n!} \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma b_\gamma ^n \operatorname{sign}(b_\gamma ). \end{aligned}

We claim this is equal to πa0\frac{\pi }{a_0} when a0k=1naka_0 \geq \sum _{k = 1}^n a_k, and smaller otherwise. Indeed, this is exactly the condition that all the bγb_\gamma are positive, so that the sign term disappears. The remaining claim is then that

γ{1,1}εγbγn=2nn!k=1nak. \sum _{\gamma \in \{ -1, 1\} } \varepsilon _\gamma b_\gamma ^n = 2^n n! \prod _{k = 1}^n a_k.

Indeed, this follows by considering the nnth Taylor coefficient of the equality

ea0tk=1n(eakteakt)=γ{1,1}nεγebγt, e^{a_0 t} \prod _{k = 1}^n (e^{a_k t} - e^{- a_k t}) = \sum _{\gamma \in \{ -1, 1\} ^n} \varepsilon _\gamma e^{b_\gamma t},

where on the left we use that eakteakt=2akte^{a_k t} - e^{-a_k t} = 2 a_k t.