CONTOUR INTEGRAL RECIPROCAL O BETA FUNCTION

Today we will show the following result, a integral representation for the reciprocal of the Beta Function


\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta=\frac{\pi}{2^{b+1}}\cdot \frac{\Gamma\left( b+1\right)}{\Gamma\left( \frac{2-a+b}{2}\right)\Gamma\left( \frac{a+b+2}{2}\right)}


For a,b \in C, \operatorname{Re}(a)>\operatorname{Re}(b), and \operatorname{Re}(b)>−1


To this end We will use a contour integral. The function that we will integrate is:


F(z)=z^{a-b-1}(1+z^2)^b=z^{a-b-1}(z+i)^b(z-i)^b


Note that this function pocesses three branch points. We will integrate along the contour C_r below.






The dot lines represent the branch cuts. Since F(z) is analytic inside this contour, by Cauchy´s theorem the  contour integral equals zero, i.e.



\oint_{C_r} z^{a-b-1}(1+z^2)^b \,dz=\int_{A_0}+\int_{A_1}+\int_{I_1}+\int_{A_2}+\int_{I_2}+\int_{A_3}\left( z^{a-b-1}(1+z^2)^b \,dz\right)=0(1)


Lets first show that the integral along A_2 vanishes as r \to 0


\lim_{r \to 0}\oint_{A_2} z^{a-b-1}(1+z^2)^b \,dz \to 0


Choosing z=re^{i \theta} the integral over A_2 becomes


\begin{aligned} I_{A_2}&=\int_{\pi/2}^{-\pi/2} \left(re^{i \theta} \right)^{a-b-1}\left(1+r^2e^{2i \theta} \right)^b ire^{i \theta}\,d \theta\\ &=-i r^{a-b}\int_{-\pi/2}^{\pi/2} e^{i \theta(a-b)}e^{b \ln\left(1+r^2e^{2i \theta} \right)}\,d \theta\\ &=-i r^{a-b}\int_{-\pi/2}^{\pi/2} e^{i \theta(a-b)}e^{\frac{b}{2} \ln\left(\left(1+r^2\cos(2 \theta)\right)^2+r^4\sin(2 \theta)\right)+ib\arctan\left( \frac{r^2\sin(2 \theta)}{1+r^2 \cos(2 \theta)}\right) }\,d \theta\\ &\leq \Bigg|-i r^{a-b}\int_{-\pi/2}^{\pi/2} e^{i \theta(a-b)}e^{\frac{b}{2} \ln\left(\left(1+r^2\cos(2 \theta)\right)^2+r^4\sin(2 \theta)\right)+ib\arctan\left( \frac{r^2\sin(2 \theta)}{1+r^2 \cos(2 \theta)}\right) }\,d \theta \Bigg|\\ & \leq r^{a-b}\int_{-\pi/2}^{\pi/2} e^{\frac{b}{2} \ln\left(\left(1+r^2\cos(2 \theta)\right)^2+r^4\sin(2 \theta)\right) }\,d \theta \end{aligned}


Taking the limit as r \to 0 in the last expression


\begin{aligned} \lim_{r \rightarrow 0} \,I_{A_2}& \leq \lim_{r \rightarrow 0} r^{a-b}\int_{-\pi/2}^{\pi/2} e^{\frac{b}{2} \ln\left(\left(1+r^2\cos(2 \theta)\right)^2+r^4\sin(2 \theta)\right) }\,d \theta\\ & = \lim_{r \rightarrow 0} r^{a-b}\int_{-\pi/2}^{\pi/2} \lim_{r \rightarrow 0} e^{\frac{b}{2} \ln\left(\left(1+r^2\cos(2 \theta)\right)^2+r^4\sin(2 \theta)\right) }\,d \theta\\ & = \lim_{r \rightarrow 0} r^{a-b}\int_{-\pi/2}^{\pi/2} e^{\frac{b}{2} \ln\left(1\right) }\,d \theta\\ & = \lim_{r \rightarrow 0} r^{a-b}\int_{-\pi/2}^{\pi/2} \,d \theta\\ & = \pi \lim_{r \rightarrow 0} r^{a-b} \to 0\\ \end{aligned}


The other two integrals along A_1 and A_3 also vanishes as \oepratorname{O}\left( r^{\operatorname{Re}(a)-\operatorname{Re}(b)}\right) \to 0 for \operatorname{Re}(a)>\operatorname{Re}(b), and \operatorname{Re}(b)>−1

For the integral along the segment I_1 we have


\begin{aligned} \oint_{I_1} z^{a-b-1}(1+z^2)^b \,dz&=\int_{i-ir}^{ir} z^{a-b-1}(1+z^2)^b \,dz\\ &=\int_{1-r}^{r} (ix)^{a-b-1}(1+(ix)^2)^b \,idx & (z \to ix)\\ &=-i^{a-b}\int_0^1 x^{a-b}(1-x^2)^b\,dx &(\text{taking the limit }\,r \to 0)\\ &=-\frac{e^{\frac{i \pi}{2}(a-b)}}{2}\int_0^1 x^{(a-b)/2-1}\left(1-x \right)^b\,dx\\ &=-\frac{e^{-\frac{i \pi}{2}(a-b)}}{2}\int_0^1 x^{(a-b)/2-1}\left(1-x \right)^b\,dx\\ &=-\frac{e^{-\frac{i \pi}{2}(a-b)}}{2} \cdot \frac{\Gamma\left( \frac{a-b}{2}\right)\Gamma\left( b+1\right)}{\Gamma\left( \frac{a+b}{2}+1\right)} \end{aligned}


\begin{aligned} \oint_{I_2} z^{a-b-1}(1+z^2)^b \,dz&=\int_{-ir}^{-i+ir} z^{a-b-1}(1+z^2)^b \,dz\\ &=\int_{r}^{1-r} (-ix)^{a-b-1}(1+(-ix)^2)^b \,(-idx) & (z \to -ix)\\ &=(-i)^{a-b}\int_0^1 x^{a-b}(1-x^2)^b\,dx &(\text{taking the limit }\,r \to 0)\\ &=\frac{e^{-\frac{i \pi}{2}(a-b)}}{2}\int_0^1 x^{(a-b)/2-1}\left(1-x \right)^b\,dx\\ &=\frac{e^{\frac{i \pi}{2}(b-a)}}{2}\cdot \frac{\Gamma\left( \frac{a-b}{2}\right)\Gamma\left( b+1\right)}{\Gamma\left( \frac{a+b}{2}+1\right)} \end{aligned}

For the integral along let z=e^{i \theta} since the radius of the semi-circle is 1


\begin{aligned} \oint_{A_0} z^{a-b-1}(1+z^2)^b \,dz&=\int_{-\pi/2}^{\pi/2}(e^{i \theta})^{a-b-1}\left(1+e^{2 i \theta} \right)^b i e^{i \theta}\,d\theta \\ &=i \int_{-\pi/2}^{\pi/2}e^{i a\theta}\left(e^{ i \theta}+e^{- i \theta} \right)^b \,d\theta \\ &=i \int_{-\pi/2}^{0}e^{i a\theta}\left(e^{ i \theta}+e^{- i \theta} \right)^b \,d\theta+i \int_{0}^{\pi/2}e^{i a\theta}\left(e^{ i \theta}+e^{- i \theta} \right)^b \,d\theta \\ &=i \int_{\pi/2}^{0}e^{-i a\theta}\left(e^{ i \theta}+e^{- i \theta} \right)^b \,(-d\theta)+i \int_{0}^{\pi/2}e^{i a\theta}\left(e^{ i \theta}+e^{- i \theta} \right)^b \,d\theta \\ &=i \int_0^{\pi/2}\left(e^{i a\theta}+e^{-i a\theta}\right)\left(e^{ i \theta}+e^{- i \theta} \right)^b \,d\theta\\ &=i 2^{b+1}\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta\\ \end{aligned}

Putting all together back in (1) we obtain:


\begin{aligned} i 2^{b+1}\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta+\left(\frac{e^{\frac{i \pi}{2}(b-a)}}{2}-\frac{e^{-\frac{i \pi}{2}(a-b)}}{2} \right)\cdot \frac{\Gamma\left( \frac{a-b}{2}\right)\Gamma\left( b+1\right)}{\Gamma\left( \frac{a+b}{2}+1\right)}=0 \end{aligned}

\begin{aligned} i 2^{b+1}\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta &=-i\sin\left(\frac{\pi}{2}(b-a) \right)\cdot \frac{\Gamma\left( \frac{a-b}{2}\right)\Gamma\left( b+1\right)}{\Gamma\left( \frac{a+b}{2}+1\right)}\end{aligned}

And


\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta=-\frac{\sin\left(\frac{\pi}{2}(b-a) \right)}{2^{b+1}}\cdot \frac{\Gamma\left( \frac{a-b}{2}\right)\Gamma\left( b+1\right)}{\Gamma\left( \frac{a+b}{2}+1\right)}(2)


Now recall the reflection formual of the gamma function


\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}(3)

From (3) we have that


\sin\left(\frac{\pi}{2}(a-b) \right)=\frac{\pi}{\Gamma\left(\frac{a-b}{2}\right)\Gamma\left(1-\frac{a-b}{2}\right)}(4)


Plugging (4) in (2) we Obtain


\int_0^{\pi/2}\cos\left( a\theta}\right)\cos^b\left( \theta\right) \,d\theta=\frac{\pi}{2^{b+1}}\cdot \frac{\Gamma\left( b+1\right)}{\Gamma\left( \frac{2-a+b}{2}\right)\Gamma\left( \frac{a+b+2}{2}\right)}

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