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151页至156页
1.
$\int^{\infty}_{0}x^n p^{-x} dx = \int^{\infty}_{0}x^n e^{-(\ln p) x} dx= \frac{1}{(\ln p)^{n+1}} \int^{\infty}_{0}y^n e^{-y} dy$
2.
3.
4.
$\int^{1}_{0}z^{m-1}\left(1-z\right)^{n-1}dz=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$
29.
$\int^{1}_{0}x^{p-1}(1-x)^{q-1}(1+ax)^{-p-q}dx=\int^{1}_{0}\left(\frac{x}{1+ax}\right)^{p-1}\left(1-\frac{(a+1)x}{1+ax}\right)^{q-1}\frac{dx}{(1+ax)^2}$
(令$y=\frac{x}{1+ax}$,所以$\frac{1}{y}=\frac{1}{x}+a$,故$x=\frac{y}{1-ay}$,$dx=\frac{dy}{(1-ay)^2}$)
$\int^{\frac{1}{1+a}}_{0}y^{p-1}\left(1-(a+1)y\right)^{q-1}dy=\frac{1}{(a+1)^p}\int^{1}_{0}z^{p-1}(1-z)^{q-1}dz$
(令$z=(a+1)y$)
$\frac{1}{(a+1)^p}\int^{1}_{0}z^{p-1}(1-z)^{q-1}dz=(a+1)^{-p}B(p,q)$
$\int^{b}_{a}(x-a)^{p-1}(b-x)^{q-1}(x-c)^{-p-q}dx=\int^{b-a}_{0}y^{p-1}(b-a-y)^{q-1}(y+a-c)^{-p-q}dy$
(令$y=x-a$)
$\int^{b-a}_{0}y^{p-1}(b-a-y)^{q-1}(y+a-c)^{-p-q}dy=\frac{1}{b-a}\int^{1}_{0}z^{p-1}(1-z)^{q-1}\left(z+\frac{a-c}{b-a}\right)^{-p-q}dz$
(令$z=\frac{y}{b-a}$)
$\frac{1}{b-a}\int^{1}_{0}z^{p-1}(1-z)^{q-1}\left(z+\frac{a-c}{b-a}\right)^{-p-q}dz=\frac{(a-c)^{-p-q}}{(b-a)^{-p-q+1}}\left(1+\frac{b-a}{a-c}\right)^{-p}B(p,q)=(a-c)^{-q}(b-c)^{-p}(b-a)^{p+q-1}B(p,q)$
$\int^{\infty}_{1}\frac{x^{p-1}+x^{q-1}}{(1+x)^{p+q}}dx=\int^{1}_{0}\frac{\frac{1}{t^{p-1}}+\frac{1}{t^{q-1}}}{\left(1+\frac{1}{t}\right)^{p+q}t^2}dt$
(令$t=\frac{1}{x}$)
$\int^{1}_{0}\frac{\frac{1}{t^{p-1}}+\frac{1}{t^{q-1}}}{\left(1+\frac{1}{t}\right)^{p+q}t^2}dt=\int^{1}_{0}\frac{t^{p-1}+t^{q-1}}{(1+t)^{p+q}}dt$
$=\int^{1/2}_{0}\frac{\left(\frac{y}{1-y}\right)^{p-1}+\left(\frac{y}{1-y}\right)^{q-1}}{\left(\frac{1}{1-y}\right)^{p+q}}\frac{dy}{(1-y)^2}$
(令$y=\frac{t}{1+t}$,有$x=\frac{y}{1-y}$、$dx=\frac{dy}{(1-y)^2}$)
$\int^{1/2}_{0}\frac{\left(\frac{y}{1-y}\right)^{p-1}+\left(\frac{y}{1-y}\right)^{q-1}}{\left(\frac{1}{1-y}\right)^{p+q}}\frac{dy}{(1-y)^2}$
$=\int^{1/2}_{0}[y^{p-1}(1-y)^{q-1}+y^{q-1}(1-y)^{p-1}]dy=\int^{1/2}_{0}y^{p-1}(1-y){q-1}dy+\int^{1}_{\frac{1}{2}}z^{p-1}(1-z)dz$
(令$z=1-y$)
$\int^{\frac{1}{2}}_{0}y^{p-1}(1-y){q-1}dy+\int^{1}_{\frac{1}{2}}z^{p-1}(1-z)dz=\int^{1}_{0}y^{p-1}(1-y)^{q-1}=B(p,q)$
$\int^{1}_{0}\frac{x^{p}-x^{-p}}{x+1}dx=\int^{1}_{0}\frac{x^{p}}{x+1}dx-\int^{1}_{0}\frac{x^{-p}}{x+1}dx=\int^{1}_{0}\frac{x^{p}}{x+1}dx-\int^{\infty}_{1}\frac{t^{p}}{(t+1)t}dt$
(令$t=1/x$)
$\int^{1}_{0}\frac{x^{p}}{x+1}dx-\int^{\infty}_{1}\frac{t^{p}}{(t+1)t}dt=\int^{\infty}_{0}\frac{x^{p}}{x+1}dx-\int^{\infty}_{1}\frac{t^{p}}{t+1}\left(\frac{1}{t}+1\right)dt=-\frac{\pi}{\sin p\pi}-\int^{\infty}_{1}t^{p-1}dt=\frac{1}{p}-\frac{\pi}{\sin p\pi}$
$\int^{a}_{0}\frac{dx}{a^2+ax+x^2}=\frac{1}{a}\int^{1}_{0}\frac{dy}{1+y+y^2}$
(令$y=x/a$)
$\frac{1}{a}\int^{1}_{0}\frac{dy}{1+y+y^2}=\frac{1}{a}\int^{1}_{0}\frac{dy}{(y+\frac{1}{2})^2+\frac{3}{4}}=\frac{2}{\sqrt{3}a}\int^{\sqrt{3}}_{\frac{1}{\sqrt{3}}}\frac{dz}{z^2+1}$
(令$z=(y+\frac{1}{2})/(\frac{\sqrt{3}}{2})$)
$\frac{2}{\sqrt{3}a}\int^{\sqrt{3}}_{\frac{1}{\sqrt{3}}}\frac{dz}{z^2+1}=\frac{2}{\sqrt{3}a}\arctan z\vert^{\sqrt{3}}_{\frac{1}{\sqrt{3}}}=\frac{2\sqrt{3}}{3a}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{3a\sqrt{3}}$
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