오늘의 계산 41 - Realizing Nonsensical (in progress)

오늘은, 소설로 치면 그냥 커뮤니티에 의미도 없이 순간 흥해서 싸질러보는 무의미한 줄거리에 가까운 계산입니다. 뭔가 '아, 이런 적분에 의미를 줄 수 있을까' 하는 생각에서 출발해봤는데, 그냥 뭐 그렇네요 -_-;;

Let $ f $ be a measurable function on $ (0, \infty) $. If \begin{equation} \label{def:ctr} \int_{h}^{\infty} f(x) \, \frac{dx}{x} = -\log h + c + o(1) \quad \text{as} \ h \to 0^{+} \end{equation} for some complex number $ c $, then we call $ c $ the center of $ f $ and denote $ c = \mathrm{ctr} \, f $. This kind of consideration follows naturally when we consider the integrals of the form $$ \int_{0}^{\infty} \frac{f(x) - g(x)}{x} \, dx, $$ where $f(0) = g(0)$. Thus we can think of $\mathrm{ctr} \, f$ as a renormalized value of the divergent integral $$ \int_{0}^{\infty} \frac{f(x)}{x} \, dx $$ when $f(0) = 1$.

The measure $ x^{-1} \, dx $ has some nice properties that immediately lead to the following observations.

Proposition 1. Suppose that the centers of $ f $ and $ g $ exist. Then we have

1. $ \mathrm{ctr} \, f(x^{p}) = \frac{1}{p} \mathrm{ctr} \, f $  for $ p > 0 $,
2. $ \mathrm{ctr} \, f(kx) = \mathrm{ctr} \, f - \log k $  for $ k > 0 $,
3. $ \mathrm{ctr} \, \left( \lambda f + (1 - \lambda) g \right) = \lambda \; \mathrm{ctr} \, f + (1 - \lambda) \, \mathrm{ctr} \; g $.

Proof. It follows from direct calculation.

Before jumping to the actual calculation, we make a trivial observation that is useful in practice. By definition \eqref{def:ctr}, $ f $ has the center with value $ c $ if and only if $$ \int_{0^{+}}^{\infty} \left( f(x) - \chi_{(0, 1)}(x) \right) \, \frac{dx}{x} = c. $$ Now we are going to calculate the centers of some well-known functions.

Proposition 2. For any $ r > 0 $, we have $$ \mathrm{ctr} \; \left( \frac{1}{(1+x)^{r}} \right) = - \gamma - \psi_0 (r). $$ In particular, if $ r = n + 1 $ for some nonnegative integer $ n $, then it is equal to $ - H_n $.

Proof. We first recall that $$ \int_{0}^{1} \frac{1 - t^{s-1}}{1 -t} \, dt = \gamma + \psi_0 (s). $$ Then by direct calculation, \begin{align*} \mathrm{ctr} \; \left( \frac{1}{(1 + x)^{r}} \right) & = \int_{0}^{1} \left( \frac{1}{(1 + x)^r} - 1 \right) \, \frac{dx}{x} + \int_{1}^{\infty} \frac{dx}{x(1 + x)^r} \\ & = \int_{0}^{1} \left( \frac{1}{(1 + x)^r} - 1 \right) \, \frac{dx}{x} + \int_{0}^{1} \frac{x^r}{x(1 + x)^r} \, dx \\ & = \int_{0}^{1} \left( \frac{1 + x^r}{(1 + x)^r} - 1 \right) \, \frac{dx}{x} \\ & = \frac{1}{2} \int_{0}^{\infty} \left( \frac{1 + x^r}{(1 + x)^r} - 1 \right) \, \frac{dx}{x} \\ & = \frac{1}{2} \int_{0}^{1} \frac{u^r + (1-u)^r - 1}{u(1-u)} \, du \qquad (u = (1 + x)^{-1}) \\ & = -\frac{1}{2} \int_{0}^{1} \left( \frac{1 - u^{r-1}}{1 - u} + \frac{1 - (1-u)^{r-1}}{1 - (1-u)} \right) \, du \\ & = - (\gamma + \psi_0 (r) ), \end{align*} as desired.

But for most situation, direct calculation is somewhat cumbersome, and sometimes calculation can be carried out in a simpler manner. Here is such a method.

Theorem 3. Suppose $ f $ is bounded and has the center. Then $$ \beta = \lim_{R \to \infty} \left( \int_{0}^{R} \mathcal{L}f (s) \, ds - \log R \right) $$ exists, and we have $$ \mathrm{ctr} \; f = \beta - \gamma, $$ where $ \gamma $ is the Euler-Mascheroni constant.

Proof. We consider the following Laplace transform $$ I(s) = \int_{0}^{\infty} \left( f(x) - \chi_{(0, 1)}(x) \right) \frac{e^{-sx}}{x} \, dx. $$ By assumption, this is a well-defined analytic function on $ \Re s > 0 $ such that Leibniz's integral rule is applicable. Then we have $$ I'(s) = - \int_{0}^{\infty} f(x) e^{-sx} \, dx + \int_{0}^{1} e^{-sx} \, dx = - \mathcal{L}f(s) + \frac{1 - e^{-s}}{s}. $$ Since $ I(R) \to 0 $ as $ R \to \infty $, we then have \begin{align*} \mathrm{ctr} \; f & = \lim_{R \to \infty} \left( I(0) - I(R) \right) \\ & = - \lim_{R \to \infty} \int_{0}^{R} I'(s) \, ds \\ & = \lim_{R \to \infty} \int_{0}^{R} \left( \mathcal{L}f(s) - \frac{1 - e^{-s}}{s} \right) \, ds \\ & = \lim_{R \to \infty} \left( \int_{0}^{R} \mathcal{L}f(s) \, ds - (1 - e^{-R})\log R + \int_{0}^{R} e^{-s} \log s \, ds \right) \\ & = \beta - \gamma. \end{align*} This completes the proof.

Corollary 4. For $ \Re \alpha \geq 0 $, we have the following equalities.

1. $ \displaystyle \mathrm{ctr} \, ( e^{-\alpha x} ) = -\gamma - \log \alpha, \qquad (\alpha \neq 0) $
2. $ \displaystyle \mathrm{ctr} \, ( \cos x \; e^{-\alpha x} ) = -\gamma - \frac{1}{2} \log (\alpha^2 + 1), \qquad (\alpha \neq \pm it \text{ for } |t| \geq 1) $
3. $ \displaystyle \mathrm{ctr} \, \left( \frac{x e^{-\alpha x}}{1 - e^{-x}} \right) = -\gamma - \psi_0 (\alpha). \qquad (\alpha \neq 0) $

Proof. We have $$ \mathcal{L} \{ e^{-\alpha x} \} (s) = \frac{1}{\alpha + s}, \quad \text{and} \quad \mathcal{L} \left\{ \frac{x e^{-\alpha x}}{1 - e^{-x}} \right\} (s) = \mathcal{L} \left\{ \frac{x}{1 - e^{-x}} \right\} (s + \alpha) = \psi_1 (s+\alpha). $$ Then each case, we have \begin{align*} \int_{0}^{R} \mathcal{L} \{ e^{-\alpha x} \} (s) \, ds & = \log (\alpha + R) - \log \alpha \sim \log R - \log \alpha, \\ \int_{0}^{R} \mathcal{L} \left\{ \frac{x e^{-\alpha x}}{1 - e^{-x}} \right\} (s) \, ds & = \psi_0 (\alpha + R) - \psi_0(\alpha) \sim \log R - \psi_0 (\alpha). \end{align*} Plugging these yields 1 and 3. To obtain 2, just apply the property 3 of Proposition 1.

Exercise 5. Special cases include the following equalities. $$ \mathrm{ctr} \, \left( \frac{1}{1+x} \right) = 0, \quad \mathrm{ctr} \, (e^{-x}) = -\gamma , \quad \mathrm{ctr} \, (\cos x) = -\gamma , \quad \mathrm{ctr} \, \left( \frac{x}{e^x - 1} \right) = 0. $$

위 공식들을 활용하면, 「오늘의 계산 24」의 $r = 0$ 인 경우와 「오늘의 계산 34」의 첫 번째 적분이 \begin{align*} \int_{0}^{\infty} \frac{\cos (x^p) - e^{-x^{q}}}{x} \, dx & = \mathrm{ctr}(\cos(x^p)) - \mathrm{ctr}(e^{-x^{q}}) = \frac{1}{p} \; \mathrm{ctr}(\cos x) - \frac{1}{q} \; \mathrm{ctr}(e^{-x}) \\ & = \gamma \left( \frac{1}{q} - \frac{1}{p} \right), \\ \\ \int_{0}^{\infty} \frac{1}{x} \left( \frac{1}{1 + \alpha^2 x^2} - \cos (\beta x) \right) \, dx & = \mathrm{ctr}\left( \frac{1}{1 + \alpha^2 x^2} \right) - \mathrm{ctr}(\cos (\beta x)) \\ & = \mathrm{ctr}\left( \frac{1}{1 + x^2} \right) - \log \alpha - \left( \mathrm{ctr}(\cos x) - \log \beta \right) \\ & = \frac{1}{2} \; \mathrm{ctr}\left( \frac{1}{1 + x} \right) - \mathrm{ctr} (\cos x) - \log \left( \frac{\alpha}{\beta} \right) \\ & = \gamma - \log \left( \frac{\alpha}{\beta} \right) \end{align*} 와 같이 깔끔하게 계산됩니다. 또한 다음과 같이 Abel integral 비슷하게 생긴 적분도 계산할 수 있습니다. \begin{align*} \int_{0}^{\infty} \frac{1}{1+e^{\pi x}} \frac{x}{1 + x^2} \, dx & = \int_{0}^{\infty} \frac{1}{1+e^{\pi x}} \int_{0}^{\infty} \sin xt \; e^{-t} \, dt dx \\ & = \int_{0}^{\infty} \left( \int_{0}^{\infty} \frac{\sin xt}{1+e^{\pi x}} \, dx \right) \; e^{-t} \, dt \\ & = \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} \sin xt \; e^{-n\pi x} \, dx \right) \; e^{-t} \, dt \\ & = \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{t}{t^2 + \pi^2 n^2} \right) \; e^{-t} \, dt \\ & = \frac{1}{2} \int_{0}^{\infty} \left( \frac{1}{t} - \frac{1}{\sinh t} \right) \; e^{-t} \, dt \\ & = \frac{1}{2} \int_{0}^{\infty} \left( e^{-t} - \frac{2t}{e^{2t} - 1} \right) \, \frac{dt}{t} \\ & = \frac{1}{2} \left( \mathrm{ctr} (e^{-t}) - \mathrm{ctr} \left( \frac{2t}{e^{2t} - 1} \right) \right) \\ & = \frac{1}{2} \left( \mathrm{ctr} (e^{-t}) - \mathrm{ctr} \left( \frac{t}{e^{t} - 1} \right) + \log 2 \right) \\ & = \frac{1}{2} \left( \log 2 - \gamma \right). \end{align*}