The product $\sin(\pi s/2)\Gamma(1 - s)$ How to show that $\sin(\pi s/2)\Gamma(1 - s)$ is analytic when $s$ is an even positive integer?
 A: Hint: The poles of $\Gamma(s)$ are $s=0,-1,-2,-3,\ldots,$ and they are all simple.
Edit:
Gamma has simple poles at the nonpositive integers: 
To see this, first observe that the integral $\Gamma(s)=\int_0^\infty e^{-t}t^{s-1}dt$ is analytic for $\operatorname{Re}(s)>0.$ Integrating by parts yields the functional equation $\Gamma(s+1)=s\Gamma(s).$ This allows us to analytically continue gamma to $\operatorname{Re}(s)>-1,s\neq 0,$ by the formula $\Gamma(s)=\dfrac{\Gamma(s+1)}{s}.$ Since $\Gamma(s+1)$ is analytic in this region, we see that $\Gamma(s)$ must have a simple pole at $s=0.$ 
We can continue this method, using $\Gamma(s+2)=(s+1)\Gamma(s+1)$ to analytically continue gamma to $\operatorname{Re}(s)>-2,s\neq 0,-1.$ Namely, we have $\Gamma(s+2)=(s+1)s\Gamma(s),$ which gives $\Gamma(s)=\dfrac{\Gamma(s+2)}{s(s+1)}.$ Again, the numerator is analytic, and we see that gamma must have simple poles at $s=0,-1.$
We do this ad infinitum, showing that $s=0,-1,-2,\ldots$ are simple poles of gamma, and that gamma is analytic elsewhere.
$\sin(\pi s/2)\Gamma(1-s)$ is analytic near even positive integers:
By the previous, we know that the poles of $\Gamma(1-s)$ occur at $1-s=0,-1,-2,\ldots,$ or in other words at $s=1,2,3,\ldots.$ Of course $\sin(\pi s/2)$ has zeros at $2n$ for any $n\in\Bbb N$ since $\sin(\pi(2n)/2)=\sin(n\pi)=0.$ Thus, we can write a Taylor expansion for $\sin(\pi s/2)$ near $s=2n,$ which will have the form $$\sin(\pi s/2)=a_m(s-2n)^{m}+a_{m+1}(s-2n)^{m+1}+\cdots = a_m(s-2n)^m+f(s)$$
where $f(s)$ is the tail of the expansion, and $m\ge 1,a_m\neq 0$.
We can also write a Laurent expansion for $\Gamma(1-s)$ near $s=2n$ of the form $$\Gamma(1-s)=a_{-1}(s-2n)^{-1}+a_0+a_1(s-2n)+\cdots=a_{-1}(s-2n)^{-1}+g(s)$$
where $g(s)$ is the holomorphic part of the expansion, and $a_{-1}\neq 0$. This follows because we know that the pole $s=2n$ is simple. Hence, we easily see that $\sin(\pi s/2)\Gamma(1-s)$ has a Laurent expansion of the form $$\big(a_m(s-2n)^m+f(s)\big)\big(a_{-1}(s-2n)^{-1}+g(s)\big)=a_ma_{-1}(s-2n)^{m-1}+  O((s-2n)^m)$$ where I'm just using the "big oh" to mean higher order terms.
Recall, that $m\ge 1$ since $\sin(\pi s/2)$ vanishes at $s=2n.$ Hence, this Laurent series is actually a Taylor series and it follows that the product $\sin(\pi s/2)\Gamma(1-s)$ is analytic in a neighbourhood of $s=2n,$ as initially desired.
A: Use the identity,
$$ \sin\left(\frac{\pi s}{2} \right) = \frac{\pi}{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}\,. $$
