오늘 계산은 지금까지 MathLinks에 답변했던 계산들 중 2개를 추려서 올려본 것입니다.

 


이번에 계산할 적분은 아래 적분입니다. 특별한 설명 없이 쭉 이어나가겠습니다.

 

\begin{eqnarray} I & = & \int_{0}^{1}\frac{\sqrt{x-x^{2}}}{(1+x^{2})^{2}}\, dx \nonumber \ & = & \int_{0}^{\infty}\frac{\sqrt{y}(1+y)}{(1+(1+y)^{2})^{2}}\, dt\qquad (x = 1/(1+y)) \nonumber \ & = & \frac{1}{4} \int_{0}^{\infty} \frac{dy}{\sqrt{y}(1+(1+y)^{2})} \qquad \text{(int. by parts)} \nonumber \ & = & \frac{1}{2} \int_{0}^{\infty}\frac{ds}{1+(1+s^{2})^{2}}\qquad (y = s^{2}) \nonumber \ & = & \frac{1}{2}\int_{0}^{\infty} \int_{0}^{\infty} \sin t e^{-(1+s^{2})t} \, dt ds \nonumber \ & = & \frac{1}{2}\int_{0}^{\infty} \sin t e^{-t} \left[\int_{0}^{\infty}e^{-ts^{2}}\, ds\right] dt \nonumber \ & = & \frac{\sqrt{\pi}}{4}\int_{0}^{\infty}\frac{\sin t}{\sqrt{t}}e^{-t}\, dt \nonumber \ & = & \frac{\sqrt{\pi}}{2}\int_{0}^{\infty}\sin (z^{2}) e^{-z^{2}}\, dz\qquad (t = z^{2})\nonumber \ & = & \frac{\sqrt{\pi}}{2}\Im \left[ \int_{0}^{\infty}e^{-(1-i)z^{2}}\, dz \right] \nonumber \end{eqnarray}

 

이제 복소적분 테크닉을 이용하면

 

\begin{eqnarray} I & = & \frac{\sqrt{\pi}}{2}\Im \left[ e^{\frac{i\pi}{8}} \int_{0}^{\exp(-\frac{i\pi}{8}) \infty}e^{-\sqrt{2}w^{2}}\, dw \right] \qquad (z = e^{\frac{i\pi}{8}} w) \nonumber \ & = & \frac{\sqrt{\pi}}{2}\Im \left( e^{\frac{i\pi}{8}} \lim_{R\to\infty} \left[ \int_{0}^{R} e^{-\sqrt{2}w^{2}}\, dw + \int_{R}^{\exp(-\frac{i\pi}{8})R} e^{-\sqrt{2}w^{2}}\, dw \right]\right) \nonumber \ & = & \frac{\sqrt{\pi}}{2}\Im \left( e^{\frac{i\pi}{8}} \int_{0}^{\infty} e^{-\sqrt{2}w^{2}}\, dw \right) \nonumber \ & = & \frac{\pi}{4\sqrt[4]{2}}\Im e^{\frac{i\pi}{8}} \nonumber \ & = & \frac{\pi}{4\sqrt[4]{2}} \sin \frac{\pi}{8} \nonumber \ & = & \frac{\pi}{8} \sqrt{\sqrt{2}-1} \nonumber \end{eqnarray} 

 


이번에 계산할 적분은

I =\int_{0}^{\infty}\left(\frac{1}{x}-\frac{e^{-x}}{2}-\frac{1}{e^{x}-1}\right)\,\frac{dx}{x} 

 

입니다. 위 적분이 절대수렴한다는 것은 원점에서의 행동을 조사해보면 쉽게 알 수 있습니다. 이제 위 적분을 계산하기 위해, 좀 더 일반적인 다음 식을 증명해보도록 하겠습니다.

\begin{eqnarray}F(s) & = &\int_{0}^{\infty}\left(\frac{1}{x}-\frac{e^{-x}}{2}-\frac{1}{e^{x}-1}\right)\,\frac{e^{-sx}}{x} \, dx \nonumber \ & = & s\log s-s+\frac{1}{2}\log (1+s)-\log\Gamma (1+s)+\log\sqrt{2\pi} \nonumber \end{eqnarray} 

 

즉, F(s)는 Stirling formula를 적분으로 표현한 버전이라 할 수 있겠습니다. 증명은 간단합니다. F''(s)를 계산한 다음, F(∞) = F'(∞) = 0 이라는 사실을 이용하여 두 번 적분해주면 됩니다. F''(s)는 Leibniz's integral rule에 의해

 

\begin{eqnarray}F''(s) & = &\int_{0}^{\infty}\frac{\partial^{2}}{\partial s^{2}}\left(\frac{1}{x}-\frac{e^{-x}}{2}-\frac{1}{e^{x}-1}\right)\,\frac{e^{-sx}}{x}\, dx\nonumber \ & = &\int_{0}^{\infty}\left( 1-\frac{x e^{-x}}{2}-\frac{x}{e^{x}-1}\right) e^{-sx}\, dx\nonumber \\ & = &\int_{0}^{\infty}e^{-sx}\, dx-\frac{1}{2}\int_{0}^{\infty}x e^{-(s+1)x}\, dx-\int_{0}^{\infty}\frac{x e^{-sx}}{e^{x}-1}\, dx\nonumber \\ & = &\frac{1}{s}-\frac{1}{2}\frac{1}{(s+1)^{2}}-\sum_{n=1}^{\infty}\int_{0}^{\infty}x e^{-(n+s)x}\, dx\nonumber \\ & = &\frac{1}{s}-\frac{1}{2}\frac{1}{(s+1)^{2}}-\sum_{n=1}^{\infty}\frac{1}{(n+s)^{2}}\nonumber \\ & = &\frac{1}{s}-\frac{1}{2}\frac{1}{(s+1)^{2}}-\psi_{1}(s+1). \nonumber \end{eqnarray} 

 

로 주어집니다. 이제 위 식을 두 번 적분하고 Stirling formula를 적용하든가 다른 방법을 이용하면 원하는 식을 보일 수 있습니다. 따라서

 

I = F(0) = \log \sqrt{2\pi} 

 

입니다.

이 글은 스프링노트에서 작성되었습니다.

Posted by aficionado

댓글을 달아 주세요

  1. faye 2011.07.10 23:41  댓글주소  수정/삭제  댓글쓰기

    Hi,

    I have just read your solution and I have a question to ask.
    I don't know what is going on when you write the first and the second steps of your second picture...

    We integrate the function from z=0 to z=+infinity, but when you put z=exp(i*pi/8)w, what limit should we take when z=+infinity?

    Thank you for being so kind as to give me a reply.

    • aficionado 2011.07.11 06:26 신고  댓글주소  수정/삭제

      What I did at the first line of the second image is just the substitution \( z = e^{\frac{i\pi}{8}} w \). As a result, as you can see, the upper limit is changed from \( +\infty \) to \( e^{-\frac{i\pi}{8}} \infty \), which denotes that the integration is taken along the half-line which starts from the origin and extends toward the direction of \( e^{-\frac{i\pi}{8}} \).

      Then, at the next line, I changed the contour of integration. Since $$ \int_{0}^{e^{-\frac{i\pi}{8}} \infty} = \lim_{R \to +\infty} \int_{0}^{e^{-i\pi / 8} R},$$ we may choose the contour to be the union of a line segment from 0 to \( R \) and a circular arc from \( R \) to \( e^{-\frac{i\pi}{8}} \). So we have $$ \int_{0}^{e^{-i\pi / 8} \infty} = \lim_{R \to +\infty} \left( \int_{0}^{R} + \int_{R}^{e^{-i\pi / 8} R} \right).$$

  2. faye 2011.07.12 07:57  댓글주소  수정/삭제  댓글쓰기

    Hi,

    So, you have apply the Cauchy's integral theorem to change the contour line segment?


    I have another idea about this type of integral, which is not so rigor, I have uploaded it for you to take a look:

    http://upload.trend.hk/images/1310424630.jpg
    http://upload.trend.hk/images/1310424639.jpg
    http://upload.trend.hk/images/1310424651.jpg
    http://upload.trend.hk/images/1310424660.jpg

    Is my idea above correct?
    I am a secondary school student only so I am asking for your advice.

    Thank you and looking for your reply.

    • aficionado 2011.07.13 19:16 신고  댓글주소  수정/삭제

      I skimmed your arguments, but was unable to find any logical flaw. Moreover, the method you adopted in the 3rd page is impressive in that it requires barely no preliminary knowledges except for calculus. Finally, your conclusion is valid.

      But I think I have to mention a more general approach which is also applicable to other situations. If you know complex analysis, then you will find that your question is essentially a particular case of the problem of finding analytic continuation.

      Theory of complex analysis tells us that, if two analytic functions f and g defined on the same open set coincide on a sufficiently thick subset, then f and g are in fact identical. We can give a detailed condition for the 'thicnkess', but it is sufficient that the subset should contain a line segment or an arc.

      Now we know that

      $$\int_{0}^{\infty} e^{-z x^2} \; dx = \frac{1}{2} \sqrt{\frac{\pi}{z}}$$

      holds for z > 0, and both sides are defined and analytic for Re(z) > 0. Thus they must coincide on Re(z) > 0. So we only have to check for the case where z is nonzero and purely imaginary.