与双重对数函数有关的积分
求∫π20x21+cos2xdx.
对于|b|<a,注意到
令a=2+√22和b=−2+√22,我们有
求∫π0x21+sin2xdx.
令t=x−π2,我们有
求∫10∫10∫10∫10(1−x2y2z2t2)dxdydzdt√(1−x2)(1−y2)(1−z2)(1−t2)(1+x2y2z2t2).
解.原积分等于I=∫π20∫π20∫π20∫π201−cos2αcos2βcos2θcos2γ√1+cos2αcos2βcos2θcos2γdαdβdθdγ.
由于\frac{1}{\sqrt{1+x}}=\sum_{n=0}^{\infty}{\binom{-1/2}{n}x^n}=1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^n},
因此
由于\int_0^{\frac{\pi}{2}}{\cos ^{2n}xdx}=\frac{\sqrt{\pi}\Gamma \left( n+\frac{1}{2} \right)}{2\Gamma \left( n+1 \right)}=\frac{\sqrt{\pi}}{2n!}\cdot \frac{\left( 2n-1 \right) !!}{2^n}\sqrt{\pi}=\frac{\pi}{2}\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!},我们有
\begin{align*}\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}&=\int_0^{\infty}{\frac{2\sqrt{t}}{1+t^2}\ln^2 \left( \frac{2t}{1+t^2} \right) \text{d}t}\hspace{0.5cm}t=\tan \frac{\theta}{2}\\&=\int_0^{\infty}{\frac{2\sqrt{1/t}}{1+t^2}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_0^{\infty}{\frac{\sqrt{1/t}+\sqrt{1/t^3}}{t+1/t}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_{-\infty}^{\infty}{\frac{2}{x^2+2}\ln^2 \left( \frac{2}{x^2+2} \right) \text{d}x}\\&=2\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln^2 \left( \cos ^2u \right) \text{d}u}~~x=\sqrt2\tan u\\&=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln ^2\sin u\text{d}u}=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\left( -\ln 2-\sum_{k=1}^{\infty}{\frac{\cos \left( 2kx \right)}{k}} \right)^2 \text{d}u}\\&=8\sqrt{2}\left( \int_0^{\frac{\pi}{2}}{\ln ^22\text{d}u}+\sum_{n=1}^{\infty}{\frac{1}{k}\int_0^{\frac{\pi}{2}}{\frac{1+\cos 4kx}{2}\text{d}x}} \right)\\&=4\sqrt{2}\pi \ln ^22+2\sqrt{2}\pi \zeta \left( 2 \right) =\frac{\sqrt{2}}{3}\pi ^3+4\sqrt{2}\ln^2 2.\end{align*}
f是[0,1]上严格单增的凸实值连续函数,满足f(0)=0,f(1)=1,且g(x)满足g(f(x))=x对任意x\in[0,1]成立,证明
\large{\textbf{\textcolor{blue}{证明}}} 首先我们有凸函数的等价定义:
注意到g(x)=f^{-1}(x)是f(x)的反函数,只需要证明f(x)f^{-1}(x)\leqslant x^2即可,即
求最大的常数b>0,使得对任意a>0和一切(1,+\infty)上连续可导且单增的实值函数f(x)满足f(x)\leqslant x^{2a}\ln^bx,x\in(1,+\infty)就有积分\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x=+\infty}.
\large{\textbf{\textcolor{blue}{解}}} 首先如果b>1,我们取f(x)=x^{2a}\ln^{b}x,则求导后很容易得到积分\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x<+\infty}
\begin{split}& (1-\frac{f}{M})(1-\frac{m}{f}) \geq 0 \Longrightarrow 1+\frac{m}{M} \geq \frac{f}{M}+\frac{m}{f}\\\Longrightarrow & 1+\frac{m}{M} \geq \frac1{M}\int^{1}_{0}{f}dx +m\int^{1}_{0}{\frac1{f}}dx \geq 2\sqrt{\frac{m}{M}\int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx}\\\Longrightarrow & \int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx\leq \frac{(m+M)^2}{4mM}.\end{split}
2022年9月14日 21:21
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2023年1月22日 20:22
K2 challan also referred to as Khajane 2, an integrated financial management system from the Government of Karnataka. The K2 has been brought into working with an aim to manage the financial business of the government. K2 Challan Generation It works to simplify the process of remittance of departments under government by bringing an option of anywhere-anytime payment options. Firstly every department under the government of Karnataka will have access to Khajane 2 which allows their customers to remit to the government through the easy payment links provided.