more questions: Integral $\int_{0}^{A}\frac{\exp(-2\pi iwx)}{x-c}dx$

This is a continue of my previous question Integral $\int_{0}^{A}\frac{\exp(-2\pi iwx)}{x-i}dx$.

Here $A>0$, $w$-real, $c$-complex number.

I have realized, that I need a precise answer for any $w$, not only an approximation for big $w$

By definition incomplete Gamma function is:$$\Gamma(0,x)=\int_{x}^{\infty}t^{-1}e^{-t}dt$$ Why does the simple change of variable doesn't work? $$\int_{0}^{A}\frac{e^{-2\pi iw(x-c+c)}}{x-c}dx=e^{-2\pi iwc}\int_{-2\pi iwc}^{2\pi iw(A-c)}\frac{e^{-t}}{t}dt=e^{-2\pi iwc}[\Gamma(0,-2\pi iwc)-\Gamma(0,2\pi iw(A-c))]$$

(change of the variable $2\pi iw(x-c)=t$).

Instead, the correct result, which mathematica gives, is : $$\begin{multline} e^{-2\pi iwc}[-\Gamma(0,2i(A-c)\pi w)+\Gamma(0,-2\pi icw)\\ +\log(A-c)- log(-c)-\log(i(A-c)w)+\log(-icw)] \end{multline}$$ The part with logarithm is just a sum of corresponding arguments. For some $c$ this sum is zero, but not always.

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It looks like a problem with the branch cut of the gamma function. –  Fabian May 18 '11 at 9:57

The incomplete gamma function $\Gamma(0,z)$ has branch points at $z=0$. The (standard) implementation of Mathematica implements the function with a branch cut along the negative real line. Thereby, the expression $$e^{-2\pi iwc}\int_{-2\pi iwc}^{2\pi iw(A-c)}\frac{e^{-t}}{t}dt=e^{-2\pi iwc}[\Gamma(0,-2\pi iwc)-\Gamma(0,2\pi iw(A-c))]$$ is valid as long as the integral does not cross the negative real line. Otherwise, you have to correct the expression by the jump across the branch cut. The jump across the branch cut is given by $$\lim_{\epsilon \to 0^+}[ \Gamma(0,-x+i\epsilon)- \Gamma(0,-x-i \epsilon)] = \oint dz \frac{e^{-z}}{z} = -2\pi i, \text{ for } x>0.$$ This is equivalent to the logarithms given by Mathematica. However, you can write it simpler by noting that $$\gamma(t) = -2\pi i w c t + 2\pi i w (A-c) (1-t)$$ parametrizes the integration path with $t\in[0,1]$. The path crosses the real line when $\text{Im} \gamma(t) =0$ which is equivalent to $$A w (1-t) - w\,\text{Re} c =0$$ with the solution $t=1- \,\text{Re}c/A$. In order that this point is in the integration region, $0<\text{Re}\,c < A$. Furthermore, $\text{Re} \gamma(t) = w \,\text{Im}\,c < 0$ in order that the integration crosses the branch cut.
Putting everything together, we obtain the result $$\begin{multline} \int_{0}^{A}\frac{e^{-2\pi iw(x-c+c)}}{x-c}dx = e^{-2\pi iwc}[\Gamma(0,-2\pi iwc)-\Gamma(0,2\pi iw(A-c))] \\- \begin{cases}2\pi i e^{-2\pi i w c} & w \,\text{Im}\,c < 0 \text{ and } 0<\text{Re}\,c < A, \\ 0 & \text{else}. \end{cases} \end{multline}$$
Should it be $0<=Re[c]<=A$? –  Katja May 18 '11 at 14:50
@Katja: The boundary of the region in which the $2\pi i$ term appears has to be treated separately. E.g., for $\text{Re}\,c =0$ or $A$ it depends how the function $\Gamma(0,0)$ is defined (this depends on the implementation incomplete gamma function but the value is anyway diverging). For $w\,\text{Im}c =0$, it depends how $\Gamma(0,z)$ is defined on the negative real axis, i.e., whether it assumes the value continous from the upper or the lower half plane. –  Fabian May 18 '11 at 15:10