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$\int^{\infty}_{0}\frac{x^{p}}{a+x}dx=a^p\int^{\infty}_{0}\frac{y^{p}}{1+y}dy=\frac{\pi a^p}{\sin(p+1)\pi}$
(令$y=x/a$)
80.
$\int^{\infty}_{0}\frac{x^{-p}}{a+x}dx=a^{-p}\int^{\infty}_{0}\frac{y^{-p}}{1+y}dy$
(令$y=x/a$)
$a^{-p}\int^{\infty}_{0}\frac{y^{-p}}{1+y}dy=a^{-p}\int^{\infty}_{0}\frac{t^{p-1}}{1+t}dt=\frac{\pi a^{-p}}{\sin p\pi}$
$\int^{\infty}_{0}\frac{x}{a^n+x^n}dx=\frac{1}{a^{n-2}}\int^{\infty}_{0}\frac{y}{1+y^n}dy=\frac{1}{na^{n-2}}\int^{\infty}_{0}\frac{z^{\frac{1}{n}}z^{\frac{1}{n}-1}}{1+z}$
(令$y=\frac{x}{a}$)
$\frac{1}{na^{n-2}}\int^{\infty}_{0}\frac{z^{\frac{1}{n}}z^{\frac{1}{n}-1}}{1+z}=\frac{1}{na^{n-2}}\int^{\infty}_{0}\frac{z^{\frac{2}{n}-1}}{1+z}dz=\frac{1}{na^{n-2}}\frac{\pi}{\sin\left(\frac{n-2}{n}\pi\right)}=\frac{1}{na^{n-2}}\frac{\pi}{\sin\frac{2\pi}{n}}$
$\int^{\infty}_{0}\frac{x^m}{(a+bx)^n}dx=\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{0}\frac{y^m}{(1+y)^n}dy$
(令$y=bx/a$)
$\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{0}\frac{y^m}{(1+y)^n}dy=\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{1}\frac{(z-1)^m}{z^n}dz$
(令$z=y+1$)
$\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{1}\frac{(z-1)^m}{z^n}dz=\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{1}\frac{(
\frac{1}{t}-1)^m}{t^{-n}t^2}dt=\frac{1}{b^{m+1}a^{n-m-1}}\int^{\infty}_{1}t^{n-m-2}(1-t)^m dt=\frac{B(m+1,n-m-1)}{b^{m+1}a^{n-m-1}}$
$\int^{\infty}_{0}\frac{dx}{(a^2+x^2)^n}=\frac{1}{2}\int^{\infty}_{0}\frac{2xdx}{x(a^2+x^2)^n}=\frac{1}{2}\int^{\infty}_{0}\frac{y^{-1/2}dy}{(a^2+y)^n}$
(令$y=x^2$)
$\frac{1}{2}\int^{\infty}_{0}\frac{y^{-1/2}dy}{(a^2+y)^n}=\frac{B(\frac{1}{2},n-\frac{1}{2})}{a^{2n-1}}$
$=\frac{\Gamma(1/2)(n-3/2)...(1/2)\Gamma(1/2)}{(n-1)...1\Gamma(1)a^{2n-1}}$
$=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{a^{2n-1}}$
$(1-e^{-i2\pi p})\int^{\infty}_{0}\frac{x^{-p}dx}{1-x}=-\pi i(1+e^{-i2\pi p})$
所以
$\int^{\infty}_{0}\frac{x^{-p}dx}{1-x}=-\frac{\pi i(e^{i\pi p}+e^{-i\pi p})}{e^{i\pi p}-e^{-i\pi p}}=-\pi\cot p\pi$
105.
$\int^{1}_{0}\frac{x^p}{(1-x)^p}dx=\int^{\infty}_{1}\frac{1}{(t-1)^p t^2}dt$
$\int(\sqrt{x^2+1}-x)^n dx=(\sqrt{x^2+1}-x)^{n}x +n\int\frac{x(\sqrt{x^2+1} -x)^n}{\sqrt{x^2+1}}dx=(\sqrt{x^2+1}-x)^{n}x +n\int(\sqrt{x^2+1} -x)^nd\sqrt{x^2+1}$
$=(\sqrt{x^2+1}-x)^{n}x +n\int(\sqrt{x^2+1} -x)^nd(\sqrt{x^2+1}-x)+n\int(\sqrt{x^2+1} -x)^ndx=(\sqrt{x^2+1}-x)^{n}x+\frac{n}{n+1}(\sqrt{x^2+1} -x)^{n+1}+n\int(\sqrt{x^2+1} -x)^ndx$
所以
$\int(\sqrt{x^2+1} -x)^ndx=-\frac{1}{n-1}(\sqrt{x^2+1}-x)^{n}x-\frac{n}{n^2-1}(\sqrt{x^2+1} -x)^{n+1}$
故
$\int^{\infty}_{b}(\sqrt{x^2+1} -x)^ndx=\frac{1}{n-1}(\sqrt{x^2+1}-x)^{n}x\vert^b_{\infty}+\frac{n}{n^2-1}(\sqrt{x^2+1} -x)^{n+1}\vert^b_{\infty}=\frac{1}{n-1}(\sqrt{b^2+1}-b)^{n}b+\frac{n}{n^2-1}(\sqrt{b^2+1} -b)^{n+1}$
137.
$\int^{1}_{0}\left(\frac{1}{1-x}-\frac{px^{p-1}}{1-x^p}\right)dx=\ln(1-x)\vert^0_1-\ln(1-x^p)\vert^0_1=\lim_{x\to 1}\ln\frac{1-x^p}{1-x}=\ln p$
$(1-e^{-i(1/2)2\pi})\int^{1}_{0}\frac{dz}{(q-pz)\sqrt{z(1-z)}}=\frac{2\pi i}{p\sqrt{\frac{q}{p}(1-\frac{q}{p})}}$
所以
$\int^{1}_{0}\frac{dx}{(q-px)\sqrt{x(1-x)}}=\frac{\pi}{p\sqrt{\frac{q}{p}(\frac{q}{p}-1)}}=\frac{\pi}{\sqrt{q(q-p)}}$
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