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 , 1008 阅读

计算$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx}.$$

Let $u = \ln \tan x$, so that $\frac{\pi}{4} < x< \frac{\pi}{2}$ is mapped to $u>0$, $x=\arctan(\exp(u))$ and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \cosh(u)}$. Then

$$\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\ln (u)}{\cosh(u)} \mathrm{d}u = \frac{1}{2} \lim_{s \to 0^+}\frac{\mathrm{d}}{\mathrm{d} s} \int_0^\infty \frac{u^s}{\cosh(u)} \mathrm{d}u$$

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.

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

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 $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$.