오늘의 계산 54 - Some problems

Problem 1. Show that[각주:1] the product

\begin{equation*} P = \left( \frac{2}{1} \right)^{\frac{1}{8}} \left( \frac{2 \cdot 2}{1 \cdot 3} \right)^{\frac{3}{16}} \left( \frac{2 \cdot 2 \cdot 2 \cdot 4}{1 \cdot 3 \cdot 3 \cdot 3} \right)^{\frac{6}{32}} \left( \frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 4 \cdot 4 \cdot 4 \cdot 4}{1 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 5} \right)^{\frac{10}{64}} \cdots \tag{1} \end{equation*}

is equal to $P = \exp \left( \frac{7 \zeta(3)}{24 \zeta(2)} \right)$.

 

Solution. Note that

\begin{align*} \log P &= - \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \log (k+1). \end{align*}

Using the Frullani integral[각주:2]

\begin{align*} \int_{0}^{\infty} \frac{e^{-a x} - e^{-b x}}{x} \, \mathrm{d}x &= \log \left( \frac{b}{a} \right), \end{align*}

we find that

\begin{align*} \log P &= \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{\infty} \frac{e^{-(k+1)x} - e^{-x}}{x} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \frac{n(n+1)}{2^{n+3}} \int_{0}^{\infty} \frac{e^{-x} (1 - e^{-x})^{n}}{x} \, \mathrm{d}x \\ &= \frac{1}{4} \int_{0}^{\infty} \frac{e^{-x}}{x} \sum_{n=1}^{\infty} \binom{n+1}{2} \left( \frac{1 - e^{-x}}{2} \right)^{n} \, \mathrm{d}x \\ &= \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-x})}{x (1 + e^{-x})^{3}} \, \mathrm{d}x . \end{align*}

Now we introduce the function

\begin{equation*} I(s) = \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, dx. \tag{2} \end{equation*}

It is clear that $\text{(2)}$ is analytic for $\Re s > -1$ and $ \log P = \lim_{s\to 0^{+}} I(s) $. But for large real $s$, we have

\begin{align*} I(s) &= \int_{0}^{\infty} x^{s-1} \frac{e^{-x}(1 - e^{-x})}{(1+e^{-x})^{3}} \, \mathrm{d}x \\ &= \int_{0}^{\infty} x^{s-1} \sum_{n=1}^{\infty} (-1)^{n-1} n^2 e^{-n x} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} (-1)^{n-1} n^2 \int_{0}^{\infty} x^{s-1} e^{-n x} \, \mathrm{d}x \\ &= \Gamma(s) \eta(s-2) = \Gamma(s) (1 - 2^{3-s}) \zeta(s-2). \end{align*}

By the Identity Theorem, this remains valid on any open subset of $\Bbb{C}$ where $\text{(2)}$ is analytic. In particular, this holds for $\Re s > -1$. Thus taking $s \to 0^{+} $, we obtain

\begin{align*} \log P = -7 \zeta'(-2). \end{align*}

Finally, differentiating both sides of the functional equation

\begin{align*} \zeta(s) =& 2^{s} \pi^{s-1} \sin \left( \frac{\pi s}{2} \right) \Gamma (1-s) \zeta(1-s) \end{align*}

and plugging $s = -2$, we obtain

\begin{align*} \zeta'(-2) = -\frac{\zeta(3)}{4\pi^2} = -\frac{\zeta(3)}{24\zeta(2)}. \end{align*}

This completes the proof.

 

Problem 2. Show that[각주:3]

\begin{equation*} \lim_{t\to\infty} \int_{0}^{\infty} \frac{\sin ( t \arctan x ) }{(1+x^{2})^{t/2} (e^{2\pi x}-1)} \, \mathrm{d}x = \frac{1}{4}. \tag{3}\end{equation*}

 

Solution. Note that for $x > 0$, we have

\begin{align*} \operatorname{Im} (1 - ix)^{-t} = \frac{\sin (t \arctan (x)) }{(1 + x^{2})^{t/2}} \end{align*}

So if we denote the integral inside the limit by $I(t)$, then we have

\begin{align*} I(t) = \int_{0}^{\infty} \frac{\operatorname{Im} (1 - ix)^{-t}}{e^{2\pi x} - 1} \, \mathrm{d}x = \frac{1}{2} \int_{0}^{\infty} \operatorname{Im} (1 - ix)^{-t} ( \coth \pi x - 1) \, \mathrm{d}x. \end{align*}

It is easy to show that

\begin{align*} -\frac{1}{2} \int_{0}^{\infty} \operatorname{Im} (1 - ix)^{-t} \, \mathrm{d}x = \frac{1}{2(1-t)}. \end{align*}

Also, $ \operatorname{Im} (1 + ix)^{-t} = \operatorname{Im} \overline{(1 - ix)^{-t}} = - \operatorname{Im} (1 - ix)^{-t} $ and we have

\begin{align*} I(t) = \frac{1}{2(1-t)} + \frac{1}{4} \int_{-\infty}^{\infty} \mathrm{Im} (1 - ix)^{-t} \coth \pi x \, \mathrm{d}x. \end{align*}

Plugging $z = 1 - ix$, we find that

\begin{align*} I(t) = \frac{1}{2(1-t)} + \frac{1}{4} \operatorname{Im} \operatorname{PV} \! \! \int_{1+i \infty}^{1 - i \infty} z^{-t} \cot \pi z \, \mathrm{d}z. \end{align*}

Recalling that $\pi \cot \pi z = \sum_{n=-\infty}^{\infty} \frac{1}{z - n}$, considering an appropriate contour and taking limit, it is easy to see that

\begin{align*} \operatorname{PV} \! \! \int_{1+i \infty}^{1 - i \infty} z^{-t} \cot \pi z \, \mathrm{d}z = 2 i \left( \zeta(t) - \frac{1}{2} \right). \end{align*}

Therefore

\begin{align*} I(t) = \frac{1}{2(1-t)} + \frac{1}{2} \left( \zeta(t) - \frac{1}{2} \right), \end{align*}

and taking $t \to \infty$ gives the desired limit.

Footnotes

1. galactus, Tough, but fun-looking, infinite product - Integrals and Series [본문으로]
2. The idea of using the Frullani integral is attributed to galactus. I also derived the same formula but my derivation required a few more lines. [본문으로]
3. tom-nowy, A limit of an intricate integral - Art of Problem Solving [본문으로]