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오늘의 계산 41 - Realizing Nonsensical (in progress)

Math Talk/Today's Calc. 2011/05/15 03:23

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*}

Posted by sos440
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  1. jackal_anu 2011/05/15 17:40  댓글주소  수정/삭제  댓글쓰기

    무심결에 쓰던 성질들도 이렇게 정리해놓고 보니까 심오해보이는군요 ㅋㅋ;
    꼴을 보니 logarithmic mean이라 부르면 참 좋을 것 같은데, 벌써 이 이름이 다른 의미로 쓰이고 있네요 -_-; 좀 더 좋은 이름 없으려나...

    • sos440 2011/05/15 17:56  댓글주소  수정/삭제

      그렇네요... 저도 이 개념이 이미 있을 것이라고 확신하지만, 솔직히 말해서 찾아보기 귀찮고 싸질러보자는 안일한 생각으로 적당히 싸질러버렸습니다. 혹시라도 관련 정보가 있으면 제보 부탁드립니다 OTL

  2. 미지 2011/05/15 21:56  댓글주소  수정/삭제  댓글쓰기

    오 ctr이라는 거 성질은 굉장히 자명해 보이는데 직접 쓰이는 거 보니까 대단하네요 게다가 Theorem 3은 정말 신기하네요 저 정리에 이름을 붙혀주고 싶고 이미 있는 것 같아 좀 불안하기도 하네요

  3. Ying Zhang 2012/03/24 10:19  댓글주소  수정/삭제  댓글쓰기

    Correct me if I am wrong, in the last example, how would you make sense of $1/1+e^{\pi x}=\sum_n (-1)^n e^{n\pi x}$ when x>0? The series doesn't converge. But your final result is correct, it confuses me.

    • sos440 2012/10/13 21:51  댓글주소  수정/삭제

      The trick is to consider the identity
      $$\frac{1}{1+e^{\pi x}} = \frac{e^{-\pi x}}{1 + e^{-\pi x}}.$$
      This allows us to expand it as an infinite series. But one can still doubt if we then can interchange the order of infinite summation and integration. To justify this, we simply observe that for $ r < 1$,
      $$\frac{1}{1 + r} = 1 - r + \cdots + r^{2n-2} - r^{2n-1} + \frac{r^{2n}}{1 + r}.$$
      This can be used to justify the interchange.