Special Functions – The gamma and the beta function – The beta function

The beta function \(B(u,v)\) is defined by

Definition: \(B(u,v)=\displaystyle\int_0^1t^{u-1}(1-t)^{v-1}\,dt,\quad\text{Re}(u)>0,\quad\text{Re}(v)>0\).

This integral is often called the beta integral. From the definition we easily obtain the symmetry \(B(u,v)=B(v,u)\), since we have using the substitution \(t=1-s\):

\[B(u,v)=\int_0^1t^{u-1}(1-t)^{v-1}\,dt=-\int_1^0(1-s)^{u-1}s^{v-1}\,ds=\int_0^1s^{v-1}(1-s)^{u-1}\,ds=B(v,u).\]

The connection between the beta function and the gamma function is given by the following theorem:

Theorem: \(B(u,v)=\displaystyle\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)},\quad\text{Re}(u)>0,\quad\text{Re}(v)>0\).

Proof: Using the definition we have:

\[\Gamma(u)\Gamma(v)=\int_0^{\infty}e^{-t}t^{u-1}\,dt\int_0^{\infty}e^{-s}s^{v-1}\,ds =\int_0^{\infty}\int_0^{\infty}e^{-(t+s)}t^{u-1}s^{v-1}\,dt\,ds.\]

Now we apply the change of variables \(t=xy\) and \(s=x(1-y)\) to this double integral. Note that \(t+s=x\) and that \(0 < t < \infty\) and \(0 < s < \infty\) imply that \(0 < x < \infty\) and \(0 < y < 1\). The Jacobian of this transformation is

\[\frac{\partial(t,s)}{\partial(x,y)}=\begin{vmatrix}y&x\\1-y&-x\end{vmatrix}=-xy-x+xy=-x.\]

Since \(x>0\) we conclude that \(dt\,ds=\left|\displaystyle\frac{\partial(t,s)}{\partial(x,y)}\right|\,dx\,dy=x\,dx\,dy\). Hence we have

\[\Gamma(u)\Gamma(v)=\int_0^1\int_0^{\infty}e^{-x}x^{u-1}y^{u-1}x^{v-1}(1-y)^{v-1}x\,dx\,dy =\int_0^{\infty}e^{-x}x^{u+v-1}\,dx\int_0^1y^{u-1}(1-y)^{v-1}\,dy=\Gamma(u+v)B(u,v).\]

This proves the theorem.

There exist many useful forms of the beta integral which can be obtained by an appropriate change of variables.

Theorem: \(B(u,v)=\displaystyle\int_0^{\infty}\frac{s^{u-1}}{(s+1)^{u+v}}\,ds,\quad\text{Re}(u)>0,\quad\text{Re}(v)>0\).

Proof: Using \(t=s/(s+1)\) we obtain that

\[B(u,v)=\int_0^1t^{u-1}(1-t)^{v-1}\,dt=\int_0^{\infty}s^{u-1}(s+1)^{-u+1}(s+1)^{-v+1}(s+1)^{-2}\,ds =\int_0^{\infty}\frac{s^{u-1}}{(s+1)^{u+v}}\,ds.\]

Theorem: \(B(u,v)=2\displaystyle\int_0^{\frac{1}{2}\pi}(\cos\varphi)^{2u-1}(\sin\varphi)^{2v-1}\,d\varphi,\quad\text{Re}(u)>0,\quad\text{Re}(v)>0\).

Proof: Using \(t=\cos^2\varphi\) we find that

\[B(u,v)=\int_0^1t^{u-1}(1-t)^{v-1}\,dt=-2\int_{\frac{1}{2}\pi}^0(\cos\varphi)^{2u-2}(\sin\varphi)^{2v-2}\cos\varphi\sin\varphi\,d\varphi =2\int_0^{\frac{1}{2}\pi}(\cos\varphi)^{2u-1}(\sin\varphi)^{2v-1}\,d\varphi.\]

Theorem: \(\displaystyle\int_a^b(s-a)^{u-1}(b-s)^{v-1}\,ds=(b-a)^{u+v-1}B(u,v),\quad\text{Re}(u)>0,\quad\text{Re}(v)>0\).

Proof: The substitution \(t=(s-a)/(b-a)\) leads to

\begin{align*} B(u,v)&=\int_0^1t^{u-1}(1-t)^{v-1}\,dt=\int_a^b(s-a)^{u-1}(b-a)^{-u+1}(b-s)^{v-1}(b-a)^{-v+1}(b-a)^{-1}\,ds\\[2.5mm] &=(b-a)^{-u-v+1}\int_a^b(s-a)^{u-1}(b-s)^{v-1}\,ds. \end{align*}

The special case \(a=-1\) and \(b=1\) is of special interest as we will see later:

\[\int_{-1}^1(1+s)^{u-1}(1-s)^{v-1}\,ds=2^{u+v-1}B(u,v),\quad\text{Re}(u)>0,\quad\text{Re}(v)>0.\]

The different forms for the beta function have a lot of consequences. For instance, if we set \(u=v=\frac{1}{2}\) we find that

\[B(\tfrac{1}{2},\tfrac{1}{2})=\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{1}{2})}{\Gamma(1)}=\left\{\Gamma(\tfrac{1}{2})\right\}^2.\]

On the other hand, we have

\[B(\tfrac{1}{2},\tfrac{1}{2})=2\int_0^{\frac{1}{2}\pi}d\varphi=2\cdot\frac{1}{2}\pi=\pi.\]

This implies that \(\Gamma(\frac{1}{2})=\sqrt{\pi}\). Using the transformation \(x^2=t\) we now easily obtain the value of the Gaussian integral

\[\int_{-\infty}^{\infty}e^{-x^2}\,dx=2\int_0^{\infty}e^{-x^2}\,dx=\int_0^{\infty}e^{-t}t^{-\frac{1}{2}}\,dt=\Gamma(\tfrac{1}{2})=\sqrt{\pi}.\]

The beta function can also be used to evaluate integrals of powers of trigonometric functions:

Examples:

1)

\[\int_0^{\frac{1}{2}\pi}(\cos\varphi)^5(\sin\varphi)^7\,d\varphi=\frac{1}{2}\cdot B(3,4) =\frac{1}{2}\cdot\frac{\Gamma(3)\Gamma(4)}{\Gamma(7)}=\frac{1}{2}\cdot\frac{2!\,3!}{6!} =\frac{1}{2}\cdot\frac{2}{4\cdot5\cdot6}=\frac{1}{120}.\]

2)

\[\int_0^{\frac{1}{2}\pi}(\cos\varphi)^7(\sin\varphi)^4\,d\varphi=\frac{1}{2}\cdot B(4,\tfrac{5}{2})=\frac{1}{2}\cdot\frac{\Gamma(4)\Gamma(\frac{5}{2})}{\Gamma(\frac{13}{2})} =\frac{1}{2}\cdot\frac{3!}{\frac{5}{2}\cdot\frac{7}{2}\cdot\frac{9}{2}\cdot\frac{11}{2}} =\frac{1}{2}\cdot\frac{6\cdot 2^4}{5\cdot7\cdot9\cdot11}=\frac{16}{1155}.\]

3)

\begin{align*} \int_0^{\frac{1}{2}\pi}(\cos\varphi)^4(\sin\varphi)^6\,d\varphi&=\frac{1}{2}\cdot B(\tfrac{5}{2},\tfrac{7}{2})=\frac{1}{2}\cdot\frac{\Gamma(\frac{5}{2})\Gamma(\frac{7}{2})}{\Gamma(6)} =\frac{1}{2}\cdot\frac{\frac{3}{2}\cdot\frac{1}{2}\cdot\Gamma(\frac{1}{2})\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\Gamma(\frac{1}{2})}{5!}\\[2.5mm] &=\frac{5\cdot3^2\cdot\pi}{2^6\cdot2\cdot3\cdot4\cdot5}=\frac{3\,\pi}{2^9}=\frac{3\,\pi}{512}. \end{align*}

Other applications

Consider the integral \(\displaystyle\int_0^{\infty}\frac{dx}{x^{\alpha}+1}\). This integral is convergent for \(\alpha>1\). For \(\alpha=2\) we find the well-known result \(\displaystyle\int_0^{\infty}\frac{dx}{x^2+1}=\arctan(x)\bigg|_0^{\infty}=\tfrac{1}{2}\pi\). For \(\alpha>1\) we obtain using the substitution \(x^{\alpha}=t\):

\[\int_0^{\infty}\frac{dx}{x^{\alpha}+1}=\frac{1}{\alpha}\int_0^{\infty}\frac{t^{\frac{1}{\alpha}-1}}{t+1}\,dt=\frac{1}{\alpha}B(\tfrac{1}{\alpha},1-\tfrac{1}{\alpha}) =\frac{1}{\alpha}\cdot\frac{\Gamma(\frac{1}{\alpha})\Gamma(1-\frac{1}{\alpha})}{\Gamma(1)}=\frac{\pi}{\alpha\sin\left(\frac{\pi}{\alpha}\right)}.\]

For instance, this implies: \(\displaystyle\int_0^{\infty}\frac{dx}{x^4+1}=\frac{\pi}{2\sqrt{2}}\), \(\displaystyle\int_0^{\infty}\frac{dx}{x^6+1}=\tfrac{1}{3}\pi\) and \(\displaystyle\int_0^{\infty}\frac{dx}{x\sqrt{x}+1}=\frac{\pi}{\frac{3}{2}\cdot\frac{1}{2}\sqrt{3}}=\frac{4\pi}{3\sqrt{3}}\).

Consider the integral \(\displaystyle\int_0^{\infty}\frac{x^{\beta}\ln(x)}{x^{\alpha}+1}\,dx\). This integral is convergent for \(\alpha>\beta+1>0\). Note that \(x^{\beta}\ln(x)=\dfrac{d}{d\beta}x^{\beta}\). Using the substitution \(x^{\alpha}=t\) we obtain as above:

\[\int_0^{\infty}\frac{x^{\beta}}{x^{\alpha}+1}\,dx=\frac{1}{\alpha}\int_0^{\infty}\frac{t^{\frac{\beta+1}{\alpha}-1}}{t+1}\,dt =\frac{1}{\alpha}B(\tfrac{\beta+1}{\alpha},1-\tfrac{\beta+1}{\alpha})=\frac{\pi}{\alpha\sin\left(\frac{(\beta+1)\pi}{\alpha}\right)}.\]

Differentiating with respect to \(\beta\) this leads to:

\[\int_0^{\infty}\frac{x^{\beta}\ln(x)}{x^{\alpha}+1}\,dx=-\frac{\pi\cos\left(\frac{(\beta+1)\pi}{\alpha}\right)}{\alpha\sin^2\left(\frac{(\beta+1)\pi}{\alpha}\right)}\cdot\frac{\pi}{\alpha} =-\frac{\pi^2}{\alpha^2\sin\left(\frac{(\beta+1)\pi}{\alpha}\right)\tan\left(\frac{(\beta+1)\pi}{\alpha}\right)}\]

for \(\alpha>\beta+1>0\). For instance, this implies that \(\displaystyle\int_0^{\infty}\frac{\sqrt{x}\ln(x)}{x^2+1}\,dx =-\frac{\pi^2}{4\sin\left(\frac{3}{4}\pi\right)\tan\left(\frac{3}{4}\pi\right)}=\frac{\pi^2}{2\sqrt{2}}\) and \(\displaystyle\int_0^{\infty}\frac{x\ln(x)}{x^3+1}\,dx=-\frac{\pi^2}{9\sin\left(\frac{2}{3}\pi\right)\tan\left(\frac{2}{3}\pi\right)}=\frac{2}{27}\pi^2\). The special case \(\alpha=2\beta+2>0\) leads to \(\sin\left(\frac{(\beta+1)\pi}{\alpha}\right)=\sin(\frac{1}{2}\pi)=1\) and \(\tan\left(\frac{(\beta+1)\pi}{\alpha}\right)=\tan(\frac{1}{2}\pi)=\infty\). This implies that \(\displaystyle\int_0^{\infty}\frac{x^{\beta}\ln(x)}{x^{2\beta+2}+1}\,dx=0\) for \(\beta>-1\). The case \(\beta=0\) we already encountered in: another remarkable integral. Generally, the substitution \(t=1/x\) leads to: \(\displaystyle\int_1^{\infty}\frac{t^{\beta}\ln(t)}{t^{2\beta+2}+1}\,dt=\int_0^1\frac{x^{-\beta}\ln(1/x)}{x^{-2\beta-2}+1}\left(-\frac{1}{x^2}\right)\,dx =-\int_0^1\frac{x^{\beta}\ln(x)}{x^{2\beta+2}+1}\,dx\) for \(\beta>-1\). Earlier, we have seen, using \(\displaystyle\int_0^1x^n\ln(x)\,dx=\frac{1}{n+1}x^{n+1}\ln(x)\bigg|_0^1-\frac{1}{n+1}\int_0^1x^n\,dx=-\frac{1}{(n+1)^2}\) for \(n=0,1,2,\ldots\), that \(\displaystyle\int_0^1\frac{\ln(x)}{x^2+1}\,dx=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=-G\), where \(G\) denotes Catalan's constant. Generally, we now have:

\begin{align*} \int_0^1\frac{x^{\beta}\ln(x)}{x^{2\beta+2}+1}\,dx&=\sum_{n=0}^{\infty}(-1)^n\int_0^1x^{(2\beta+2)n+\beta}\ln(x)\,dx =-\sum_{n=0}^{\infty}\frac{(-1)^n}{((2\beta+2)n+\beta+1)^2}\\[2.5mm] &=-\frac{1}{(\beta+1)^2}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=-\frac{1}{(\beta+1)^2}G,\quad\beta>-1. \end{align*}

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Last modified on April 17, 2024