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又两个积分 - Eufisky - The lost book
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又两个积分

Eufisky posted @ 2015年10月21日 14:17 in 数学分析 with tags 积分计算 , 1097 阅读

计算π2π4ln(lntanx)dx.


Let u=lntanx, so that π4<x<π2 is mapped to u>0, x=arctan(exp(u)) and dx=du2cosh(u). Then
π/2π/4ln(ln(tanx))dx=120ln(u)cosh(u)du=12lim
The latter parametric integral is evaluated by expanding \cosh(u) into exponential and using Euler's gamma-integral:
\begin{align*}\int_0^\infty \frac{u^s}{2\cosh(u)} \mathrm{d}u &= \int_0^\infty u^s \frac{\exp(-u)}{1+\exp(-2u)}\mathrm{d}u = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^{s} \exp(-(2n+1)u) \,\mathrm{d}u \\&= \Gamma(s+1)  \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^{s+1}} = \Gamma(s+1) 2^{-2s-2} \left( \zeta\left(s+1,\frac{1}{4}\right) - \zeta\left(s+1,\frac{3}{4}\right) \right)\end{align*}
Near s=0:
\zeta(1+s,a) = \frac{1}{s} - \psi(a) - \gamma_1(a) s + \mathcal{o}(s)
where \psi(a) is the digamma function (http://en.wikipedia.org/wiki/Digamma_function), and \gamma_1(a) is the first generalized Stieltjes constant (http://en.wikipedia.org/wiki/Stieltjes_constants). Differentiating and taking the limit we have
\begin{align*}\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x &= \frac{1}{4} \left( \left(\log(4) + \gamma\right)\left(\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right) - \left(\psi_1\left(\frac{1}{4}\right)-\psi_1\left(\frac{3}{4}\right)\right)\right) \\&= \frac{\pi}{4} \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \approx -0.260443\end{align*}
where the latter equality is given my Mathematica.

\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx.


这个积分用Alpha给的是发散结果.下面的计算是论坛的结果:

I'm posting an asnwer (of the 2 I have) using real analysis methods:

 

\begin{align*}\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx &\overset{\sqrt{x}=u}{=\! =\! =\!}2\int_{0}^{\infty}u\sin u \sin u^2 \,du \\&=-\int_{0}^{\infty}u\cos \left ( u^2+u \right )\,du+\int_{0}^{\infty}u\cos(u^2-u)\,du \\&\overset{u \mapsto u+1}{=\! =\! =\! =\!}-\int_{0}^{\infty}u\cos(u^2+u)\,du+\int_{-1}^{\infty}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du \\&= \int_{0}^{\infty}\cos\left ( u^2+u \right )\,du+\int_{-1}^{0}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du\\&\overset{u={\rm v}-\frac{1}{2}}{=\! =\! =\! =\!}\int_{1/2}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\int_{-1/2}^{1/2}\left ( {\rm v}+\frac{1}{2} \right )\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v} \\&= \int_{0}^{\infty}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\\& \left [ \int_{-1/2}^{1/2}{\rm v}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\frac{1}{2}\int_{-1/2}^{0}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}- \frac{1}{2}\int_{0}^{1/2}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v} \right ]\end{align*}

 

However, the equation in the bracket equals zero due to symmetry.

 

Hence:

\begin{align*}\int_{0}^{\infty}\sin x \sin x^2\,dx&=\int_{0}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}\\&=\cos \frac{1}{4}\int_{0}^{\infty}\cos {\rm v}^2\,d{\rm v}+\sin \frac{1}{4}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\\&\overset{(*)}{=}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\left ( \cos \frac{1}{4}+\sin \frac{1}{4} \right )\\&=\frac{\sqrt{\pi}}{2}\sin \left ( \frac{3\pi-1}{4} \right )\;\; \;\;\;\; \square\end{align*}

 

(*) We used the Frensel integrals stating that \displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}

 


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