# Tag Archives: Gamma

## Integration related to gamma function using rectangle contour

[Ex 9] Watson’s complex integration $$\int^{\infty}_{-\infty}e^{-x^2}\,\cos(2ax)dx=e^{-a^2}\sqrt{\pi}$$ $$proof$$ Integrate the following function $$f(z) = e^{-z^2}$$ Use the following contour Note that the function is entire, hence $$\int^{L}_{-L} e^{-t^2}\,dt+\int^{L+ai}_{L} e^{-t^2}\,dt+\int^{-L}_{-L+ai} e^{-t^2}\,dt+\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=0$$ For the forth integral use the substitution $x= t-ai$ $$\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=\int^{-L}_{L}e^{-(x+ai)^2}\,dx=-e^{a^2}\int^{L}_{-L}e^{-x^2}\,e^{-2iax}dx$$ Take … Continue reading

## Bromwich contour integration of the gamma function

$$\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\Gamma(a+t)\Gamma(b-t) s^{-t}\,dt= \frac{\Gamma(a+b)}{(1+s)^{a+b}}s^a$$ $$\textit{proof}$$ Consider the following function $$f(z) = \Gamma(z+a)\Gamma(b-z) s^{-z}$$ Suppose that $a,b \in \mathbb{R}$ and $a < b$. Note that the Gamma function has a pole of order 1 at each non-positive integer where … Continue reading

## The most ugly looking integral

Prove the following $$I= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$   where   $$I = \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$ $$\textit{proof}$$ First note that since there is a log in the denominator that gives as an idea to use differentiation under … Continue reading

Posted in Digamma function, Gamma function | Tagged , , , , , | Leave a comment

## Gamma Euler Reflection formula proof

$$\Gamma \left({z}\right) \Gamma \left({1 – z}\right) = \frac \pi {\sin \left({\pi z}\right)}\,\,\, \, , \forall z \notin \mathbb{Z}$$ $$\textit{proof}$$ We can use the sine product formula $$\frac{\pi}{\sin(\pi z)} = \frac{1}{z}\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1}$$ Now we start by noting that $$\Gamma(z)\Gamma(1-z) = -z\Gamma(z)\Gamma(-z)$$ … Continue reading

Posted in Gamma function | Tagged , , | 1 Comment

## Gamma duplication formula proof

$$\Gamma \left(\frac{1}{2}+n\right) = {(2n)! \over 4^n n!} \sqrt{\pi}$$ $$\textit{proof}$$ For the proof we use induction by assuming $n\geq 0$. If $n=0$ we have our basic identity $$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$ Now we need … Continue reading

## Proof of beta function using convolution

Prove the following $$\beta(x, y)=\int^{1}_{0}t^{x-1}\, (1-t)^{y-1}\,dt= \frac{\Gamma(x)\Gamma{(y)}}{\Gamma{(x+y)}}$$ $$\textit{proof}$$ Let us choose some functions $f$ and $g$ $$f(t) = t^{x} \,\, , \, g(t) = t^y$$ Hence we get $$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds$$ So by the convolution rule we have the … Continue reading

$$\Gamma(z)\,= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}$$ $\textit{where } \gamma \textit{ is the Euler constant}$ $$\textit{proof}$$ Take logarithm to the Euler representation $$\log z\Gamma(z) =\lim_{n\to \infty}z\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right)-\sum_{k=1}^n\log\left(1+\frac{z}{k}\right)$$ Note the alternating sum $$\sum_{k=1}^n \left(\log\left(1+k\right)-\log(k)\right) = \log(n+1)$$ Hence we have … Continue reading
$$\Gamma(z) = \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = \frac{1}{z}\prod^\infty_{k=1}\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}$$ $$\textit{proof}$$ Note that $$\Gamma(z+n+1) = \Gamma(z+1) \prod^{n}_{k=1}(k+z)$$ Which indicates that $$\prod^{n}_{k=1}(k+z) = \frac{\Gamma(z+n+1)}{z\Gamma(z)}$$ Also note that $$\prod^{n}_{k=1}k = n!$$ Hence we have \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = … Continue reading