# Help with the Euler-type integral $\int_{0}^{m}\frac{1-e^{2\pi i x}}{x-j}\frac{x^{s-1}}{(1+x)^{z}}dx$

Consider the integral : $$I=\int_{0}^{m}\frac{1-e^{2\pi i x}}{x-j}\frac{x^{s-1}}{(1+x)^{z}}dx\;\;\;\;s,z\in\mathbb{C}\;\;\;\;j,m \in \mathbb{N}\;\;0\leq j\leq m$$ i have tried using the Mellin integral representation of $$\frac{1}{(1+x)^{z}}$$, which is given in terms of the beta function, but that didn't get me anywhere. I have tried using the generalized binomial theorem, and split the integral at $$1$$, but that didn't get me anywhere either. Any help is highly appreciated.

EDIT

Using the Taylor expansion : $$\frac{1-e^{2\pi i x}}{x-j}=-\sum_{n=0}^{\infty}\frac{(2\pi i)^{n+1}}{(n+1)!}(x-j)^{n}$$ And substituting $$x=my$$, we have : $$I=-m^{s}\sum_{n=0}^{\infty}\frac{(2\pi i)^{n+1}}{(n+1)!}\int_{0}^{1}(my-j)^{n}\frac{y^{s-1}}{(1+my)^{z}}dy$$ Using the integral representation of the Appell's hypergeometric function : $$F_{1}(a,b_{1},b_{2},c,x,y)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b_{1}}(1-yt)^{-b_{2}}dt$$ we have : $$I=\frac{m^{s}}{sj}\sum_{n=0}^{\infty}\frac{(-2\pi i j)^{n+1}}{(n+1)!}F_{1}\left(s,z,-n,s+1,-m,\frac{m}{j}\right)$$ Now, is there any way to simplify this, say, using the exponential generating function of $$F_{1}(\cdot)$$ ? i have looked for one, but could find any !

$$\color{brown}{\textbf{Transformations and splitting.}}$$

Since $$j\in\mathbb N,$$ then $$e^{2\pi i j} = 1,$$ \begin{align} &I=\int\limits_0^m \dfrac{1-e^{2\pi i x}}{x-j}\,\dfrac{x^{s-1}}{(1+x)^z}\,\mathrm dx = -\int\limits_{0}^{m} \dfrac{e^{2\pi i(x-j)}-1}{x-j}\,\dfrac{x^{s-1}}{(1+x)^z}\,\mathrm dx \\ & = -2\pi i\int\limits_{0}^{m}\left(\int\limits_0^1e^{2\pi i t(x-j)}\mathrm dt\right)\dfrac{x^{s-1}}{(1+x)^z}\,\mathrm dx,\\ &I = -2\pi i\int\limits_{0}^{1}\int\limits_0^me^{-2\pi i t(j-x)}\dfrac{x^{s-1}}{(1+x)^z}\,\mathrm dx\,\mathrm dt,\tag1\\ &I = -2\pi i\sum\limits_{n=0}^{m-1}\int\limits_{0}^{1}\int\limits_n^{n+1} e^{-2\pi i t(j-x)}\dfrac{x^{s-1}}{(1+x)^z}\,\mathrm dx\,\mathrm dt,\\ &I = -2\pi i\sum\limits_{n=0}^{m-1}\int\limits_{0}^{1}\int\limits_0^1 e^{-2\pi i t(j-n-x)}\dfrac{(x+n)^{s-1}}{(x+n+1)^z}\,\mathrm dx\,\mathrm dt\tag2. \end{align}

$$\color{brown}{\textbf{Exponent expansion.}}$$

Using Maclaurin series for the exponent, the further transformation of the inner integral can be achieved:

\begin{align} & \int\limits_0^1 e^{-2\pi i t(j-n-x)}\dfrac{(x+n)^{s-1}}{(x+n+1)^z}\,\mathrm dx = e^{-2\pi i t(j-n)} \sum_{k=0}^\infty \dfrac{(2\pi i t)^k}{k!} \int\limits_0^1 \dfrac{x^k (x+n)^{s-1}}{(x+n+1)^z}\,\mathrm dx. \end{align}

Therefore, \begin{align} &\color{brown}{\mathbf{I = \sum\limits_{n=0}^{m-1} \sum_{k=0}^\infty P_{nk} Q_{nk},\tag3}}\\ &\text{where}\quad P_{nk}=-\dfrac{(2\pi i)^{k+1}}{k!}\,\int\limits_{0}^{1} t^ke^{-2\pi i\, (j-n) t}\,\mathrm dt,\quad Q_{nk} = \int\limits_0^1 \dfrac{x^k (x+n)^{s-1}}{(x+n+1)^z}\,\mathrm dx.\tag4\\ \end{align}

$$\color{brown}{\textbf{First factors.}}$$

Taking in account presentations of the Exponential integral and Gamma function in the forms of $$\operatorname E_{k}(y)=\int\limits_1^\infty t^{-k}e^{-yt}\,\mathrm dt,\quad \Gamma(k+1) = y^{k+1}\int\limits_0^\infty t^k e^{-yt}\,\mathrm dt,\tag5$$ one can get \begin{align} &\int\limits_0^1 t^{k}e^{-yt}\,\mathrm dt = \int\limits_0^\infty t^{k}e^{-yt}\,\mathrm dt - \int\limits_1^\infty t^{k}e^{-yt}\,\mathrm dt = k!y^{-(k+1)} - \operatorname E_{-k}(y), \\ &P_{nk}= -\dfrac{(2\pi i)^{k+1}}{k!}\left(k!(2\pi i\, (j-n))^{-(k+1)} - \operatorname E_{-k}(2\pi i\, (j-n))\right), \\ \end{align}

$$\color{brown}{\mathbf{P_{nk}= \dfrac{(2\pi i)^{k+1}}{k!}\operatorname E_{-k}(2\pi i\,(j-n)) - (j-n)^{-(k+1)}.\tag6}}$$

$$\color{brown}{\textbf{Second factors.}}$$

Since $$x^k = (x+n-n)^k = \sum_{d=0}^k \binom kd (-n)^{k-d}(x+n)^d,$$

then $$Q_{nk} = \sum_{d=0}^k \binom kd (-n)^{k-d}\int\limits_0^1 \dfrac{(x+n)^{s+d-1}}{(x+n+1)^z}\,\mathrm dx.\tag7$$

Taking in account a presentation of the incomplete Beta function in the form of $$\operatorname B_{y}(a,b) = \int\limits_0^y u^{a-1}(1-u)^{b-1}\,\mathrm du,\tag8$$

one can get $$\int\limits_0^1 \dfrac{(x+n)^{s+d-1}}{(x+n+1)^z}\,\mathrm dx =(-1)^{s+d-1}\int\limits_{-n-1}^{-n} u^{-z}(1-u)^{s+d-1}\,\mathrm du,$$

$$\color{brown}{\mathbf{Q_{nk} = \sum_{d=0}^k \binom kd (-1)^{k+s-1}n^{k-d} \left(\operatorname B_{-n}(1-z,s+d)-\operatorname B_{-n-1}(1-z,s+d)\right).\tag9}}$$

$$\color{brown}{\textbf{Result.}}$$

Formulas $$(3),(6),(9)$$ define the value of the given integral.

Taking in account that the exponent expansion does not provide calculation stability in the wide intervals, considered approach looks the most accurate of possible.