标签：frac 函数 int oint 2n pi 复变 Gamma
C-R
\[f=u+iv \\
\Re \left(\oint_{C}\overline{f(z)}f'(z)dz\right) \\
=\Re \left(\oint_{C}(u-iv)(u'_x+v'_x)(dx+dy)\right) \\
=\oint_{C}(uu'_x+vv'_y)dx+(vu'_x-uv'_x)dy \\
=\oint_{C}(uu'_x+vv'_y)dx+(vv'_y+uu'_y)dy \\
=\oint_C{} d \frac{u^2+v^2}{2}=0
\]
Arg
\[\oint_{C}\frac{1}{z}dz=\oint_{C}\frac{z'}{z}dz=2\pi i(N(z,C)-P(z,C)) \\
\int_{L}\frac{1}{z}dz=\ln(2)-\ln(2e^{i\pi})+2\pi i(0-1+0)=-3\pi i
\]
\(\Gamma\)
\[\psi(z) =\frac{d}{dz} \ln \Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)} \\
(\int_{L}+\oint_{C^+})\psi(z)dz
=\ln \Gamma(z_1)-\ln \Gamma(z_0)+2\pi i(N(\Gamma,C)-P(\Gamma,C)) \\
=\ln \Gamma(z_1)-\ln \Gamma(z_0)-2\pi i(P(\Gamma,C))
\]
\[\Gamma(z+1)=z\Gamma (z) \\
(z\Gamma(z))'_{z=0}=\Gamma'(z+1)|_{z=0}=\Gamma'(1)=-\gamma
\]
\[\lim_{b \to 0} (\Gamma(b)-B(a,b))=\Gamma(b)\left(1-\frac{\Gamma(a)}{\Gamma(a+b)}\right)=b\Gamma(b) \frac{\Gamma(a+b)-\Gamma(b)}{b\Gamma(a+b)}=\Gamma(1+b)\psi(a)=\psi(a)
\]
\[\int_0^{1}\frac{1-t^{a-1}}{1-t}dt=\lim_{b \to 0}\int_0^{1}(t^0(1-t)^{b-1}-t^{a-1}(1-t)^{b-1})dt=B(1,b)-B(a,b) \\
=\Gamma(b)\left(\frac{1}{\Gamma(1+b)}-\frac{\Gamma(a)}{\Gamma(a+b)}\right) \\
=\Gamma(b) \frac{\Gamma(a+b)-\Gamma(a)+\Gamma(a)(1-\Gamma(1+b))}{\Gamma(1+b)\Gamma(a+b)} \\
=\Gamma(1+b)\frac{1}{\Gamma(1+b)}\left(\psi(a)+\frac{-\psi(1)\Gamma(a)}{\Gamma(a+b)}\right) \\
=\psi(a)+\gamma
\]
\[x=\frac{t}{1+t},t=\frac{x}{1-x},1+t=\frac{1}{1-x},dt=\frac{1}{(1-x)^2}dx \\
\int_0^{\infty} \frac{t^p}{1+t}dt=\int_0^{1}x^p(1-x)^{-1-p}dt=B(p+1,-p)=\frac{\Gamma(p+1)\Gamma(-p)}{\Gamma(1)} \\
=\frac{\pi}{\sin (-p\pi)}
\]
\[t=x^{2n},dt=2nx^{2n-1}dx=2nt^{\frac{2n-1}{2n}}dx \\
\int_{-\infty}^{\infty} \frac{x^n}{1+x^{2n}}dx
=2\int_0^{\infty} \frac{t^{1/2}}{1+t}\frac{1}{2n}t^{\frac{1-2n}{2n}}dt \\
=\frac{1}{n}\int_0^{\infty} \frac{t^{\frac{1-n}{2n}}}{1+t}dt \\
=\frac{1}{n} \frac{\pi}{\sin(\pi \frac{n-1}{2n})} \\
=\frac{1}{n} \frac{\pi}{\cos(\frac{\pi}{2n})} \\
=\frac{\pi}{n\cos \frac{\pi}{2n}}
\]
标签：frac,
函数,
int,
oint,
2n,
pi,
复变,
Gamma
From： https://www.cnblogs.com/nekko/p/16873116.html