How find this sum closed form $I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx$

Question

Find this closed form? $$I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx,m\ge 1,a>0$$ use $$\cos{(2kx)}e^{-ax}=e^{-ax}\cdot\dfrac{e^{2kix}+e^{-2kix}}{2}=\dfrac{1}{2}(e^{(2ki-a)x}+e^{-(a+2ki)x})$$ so $$I=\dfrac{1}{2}\sum_{k=1}^{n}\int_{0}^{\infty}(e^{(2ki-a)x}+e^{-(a+2ki)x})x^{m-1}dx$$ note $$\int_{0}^{\infty}e^{(2ki-a)x}x^{m-1}dx=\dfrac{\Gamma{(m)}}{(a-2ki)^m}$$ so we must find this closed form $$I=\dfrac{1}{2}\Gamma{(m)}\sum_{k=1}^{n}\left(\dfrac{1}{(a-2ki)^m}+\dfrac{1}{(a+2ki)^m}\right)$$

This problem is from:AMM problem:http://www.mat.uniroma2.it/~tauraso/AMM/amm.html

Answer 1

For $m=1$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^1}+\frac{1}{(a+2 i k)^1}\right)=\frac{1}{2} i \left(\psi ^{(0)}\left(1+\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(0)}\left(1+\frac{i a}{2}\right)+\psi ^{(0)}\left(1-\frac{i a}{2}\right)\right)$$ For $m=2$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^2}+\frac{1}{(a+2 i k)^2}\right)$$

$$=\frac{1}{4} \left(\psi ^{(1)}\left(1+\frac{i a}{2}+n\right)+\psi ^{(1)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(1)}\left(1+\frac{i a}{2}\right)-\psi ^{(1)}\left(1-\frac{i a}{2}\right)\right)$$

For $m=3$ $$\sum _{k=1}^n \left(\frac{1}{(a-2 i k)^3}+\frac{1}{(a+2 i k)^3}\right)$$

$$=-\frac{1}{16} i \left(\psi ^{(2)}\left(1+\frac{i a}{2}+n\right)-\psi ^{(2)}\left(1-\frac{i a}{2}+n\right)-\psi ^{(2)}\left(1+\frac{i a}{2}\right)+\psi ^{(2)}\left(1-\frac{i a}{2}\right)\right)$$

Do you figure out the pattern? You should get all these by differentiating with respect to $a$ in the case $m=1$ and carefully treating the signs.