My question is related to this problem asked by me . Unable to find the poles of a function similar to gamma function.
Question >I am unable to think if $F(t, u) = R_n(t) ( π/\sin(πt))^3 e^{ut} $ , Then how can author write $$\frac{1}{2\pi i} \int_{R_T} F(t, u) dt = \sum_{k=n+1}^{[T]} \text{Res}_{t=k} (F(t, u)),$$ where $$\text{Res}_{t=k} (F(t, u)) =\frac{π^2 +u^2} {2} R_{n(k)} (-e^u)^k + uR'_{n}(k)(-e^u)^k + \frac{1}{2} R''_{n}(k) (-e^u)^k.$$
I am unable to think if I assume poles to be integer $\geq n+1$, then how is residue the above mentioned. Can someone please help.
Can someone please help how to think about it. I shall be really thankful.
Let's do a quick residue computation at $k$ for the function in the OP.
Let $F(t)=a(t)b(t)c(t)$ where $b(t)=\frac{(\pi(t-k))^3}{\sin^3{\pi t}}$ and $a(t)= R_n(t), c(t)=e^{tu}$
Since the pole has order $3$, the residue is $\frac{1}{2}F''(k)$ and we claim that $b(k)=(-1)^k, b'(k)=0, b''(k)=(-1)^k\pi ^2$, which immediately gives (the terms with $b'(k)$ are $0$!):
$F''(k)=a''(k)b(k)c(k)+2a'(k)b(k)c'(k)+a(k)(b''(k)c(k)+b(k)c''(k))=$
$R_n''(k)(-e^u)^k+2uR_n'(k)(-e^u)^k+(\pi^2+u^2)R_n(k)(-e^u)^k$,
which is exactly the required answer.
Now making the change of variable $u=t-k$ and using the periodicity of the sine, we get that the computation for $b$ at $k$ is just $(-1)^k$ the computation when $t=0$, so it is enough to consider $b(t)=\frac{(\pi t)^3}{\sin^3{\pi t}}$ and show the claimed results. Obviously $b(0)=1$, while for the rest we use that $\sin{\pi t}=\pi t-\frac{(\pi t)^3}{6}+O(t^5)$ so $\sin^3{\pi t}=\pi^3 t^3 -\frac{\pi^5}{2}t^5+O(t^6)$, hence $b(t)=\frac{1}{1-\frac{\pi^2}{2}t^2+O(t^3)}=1+\frac{\pi^2}{2}t^2+O(t^3)$ hence $b'(0)=0, b''(0)=\pi^2$ and we are done!