Is there an easy way to calculate this infinite summation

Question

Is there an easy way to calculate this summation of integral:

$$\sum_{n=0}^\infty \int_{r=0}^1 \frac {(r-\frac{1}{2})\cos(c\cdot\ln(r+n))} {(r+n)^{1-b}} dr $$

The most obvious approach is to calculate the integral which I did by using https://www.integral-calculator.com/ and got a terribly long answer for a single term.

$-\dfrac{\left(n+1\right)^b\left(\left(4b+2\right)cn-c^3+\left(-b^2+2b+1\right)c\right)\sin\left(c\ln\left(n+1\right)\right)+\left(n+1\right)^b\left(\left(-2c^2+2b^2+2b\right)n+\left(-b-2\right)c^2-b^3+b\right)\cos\left(c\ln\left(n+1\right)\right)+n^b\cdot\left(\left(-4b-2\right)cn-c^3+\left(-b^2-2b-1\right)c\right)\sin\left(c\ln\left(n\right)\right)+n^b\cdot\left(\left(2c^2-2b^2-2b\right)n-bc^2-b^3-2b^2-b\right)\cos\left(c\ln\left(n\right)\right)}{2\left(c^4+\left(2b^2+2b+1\right)c^2+b^4+2b^3+b^2\right)}$

But this seems an absurd way to even try. There has to be a better and simpler way to solve this. So can someone please help with calculating this series? I don't have much clue how to approach it?

We are assuming that the constants $b, c$ are such that the series is convergent if that helps.

If simplifying this in not possible how do we proceed further. How do we find the summation of this infinite series? Is it possible to convert it into an integral that can be explicitly calculated? If so can someone please demonstrate it how ?

Answer 1

Using Mathematica one gets

  ((r - 1/2) Cos[c Log[r + n]])/(r + n)^(1 - b) //  TrigToExp // FullSimplify

$$\frac{1}{4} (2 r-1) \left(1+(n+r)^{2 i c}\right) (n+r)^{b-i c-1}$$

int= Integrate [1/4 (n+r)^(-1+b-I c) (-1+2 r) (1+(n+r)^(2 I c)),r]//Together//FullSimplify

$$\text{int}=\frac{1}{4} \ (n+r)^{b-i c} \left(-\frac{(n+r)^{2 i c} (-2 b r+b+i c (1-2 r)+2 n+1)}{(b+i c) (b+i c+1)}-\frac{-2 b r+b+i c (2 r-1)+2 n+1}{(b-i c) (b-i c+1)}\right)$$

Taking the limits r=1 - r =0

$$\frac{1}{4} \left(\frac{(b+i c+2 n+1) n^{b+i c}}{(b+i c) (b+i c+1)}+\frac{(b-i c+2 n+1) n^{b-i c}}{(b-i c) (b-i c+1)}+\frac{(n+1)^{b+i c} (b+i c-2 n-1)}{(b+i c) (b+i c+1)}+\frac{(n+1)^{b-i c} (b-i c-2 n-1)}{(b-i c) (b-i c+1)}\right)$$

Ignoring the constants

$$\sum _{n=0}^{\infty } (n+1)^{b-i c} =\text{Zeta}[-b + I c]$$

and

$$\sum _{n=0}^{\infty } n (n+1)^{b-i c} = \sum _{n=0}^{\infty } ((n+1)-1) (n+1)^{b-i c}$$

etc.