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Question

For α,β>0\alpha, \beta>0, show that

B(α,β)=2∫0∞t2α−1(1+t2)−(α+β)dt.B(\alpha, \beta)=2 \int_0^{\infty} t^{2 \alpha-1}\left(1+t^2\right)^{-(\alpha+\beta)} d t .

B(α,β)=∫01xα−1(1−x)β−1dx.B(\alpha, \beta)=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1} d x .

Solution

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Let α,β>0\alpha, \beta >0. We know that B(α,β)=∫01xα−1(1−x)β−1dxB(\alpha, \beta)=\int\limits_0^1 x^{\alpha-1}(1-x)^{\beta-1} dx. If we use substitution

x=t21+t2⇒dx=2t(1+t2)2dt ;x=\dfrac{t^2}{1+t^2} \Rightarrow dx=\dfrac{2t}{(1+t^2)^2}dt\,;

x=0⇒t=0;x=1⇒t=∞(lim⁡t→∞t21+t2=1) ,x=0 \Rightarrow t=0; \quad x=1 \Rightarrow t= \infty\left(\lim\limits_{t \to \infty }\dfrac{t^2}{1+t^2}=1 \right)\,,

we get:

B(α,β)=∫01xα−1(1−x)β−1dx=B(\alpha, \beta)=\int\limits_0^1 x^{\alpha-1}(1-x)^{\beta-1} dx=

∫0∞(t21+t2)α−1(1−t21+t2)β−12t(1+t2)2dt=\int\limits_0^{\infty} \left( \dfrac{t^2}{1+t^2}\right)^{\alpha-1} \left( 1-\dfrac{t^2}{1+t^2}\right)^{\beta-1} \dfrac{2t}{(1+t^2)^2}dt=

2∫0∞(t2α−2(1+t2)α−1)(1(1+t2)β−1)t(1+t2)2dt=2 \int\limits_0^{\infty} \left(\dfrac{t^{2\alpha-2}}{(1+t^2)^{\alpha-1}}\right)\left( \dfrac{1}{(1+t^2)^{\beta-1}}\right)\dfrac{t}{(1+t^2)^2} dt=

2∫0∞t(2α−2)+1(1+t2)(α−1)+(β−1)+2dt=2 \int\limits_0^{\infty} \dfrac{t^{(2\alpha-2)+1}}{(1+t^2)^{(\alpha-1)+(\beta-1)+2}}dt=

2∫0∞t2α−1(1+t2)α+βdt=2∫0∞t2α−1(1+t2)−(α+β)dt .✓2 \int\limits_0^{\infty} \dfrac{t^{2\alpha-1}}{(1+t^2)^{\alpha+\beta}}dt=2 \int\limits_0^{\infty} t^{2\alpha-1}(1+t^2)^{-(\alpha+\beta)}dt \,. \quad \checkmark

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