Proving $\int\limits_{0}^{\infty}\frac{1-\text{e}^{-x}\cos(ax)}{x^{r+1}}\operatorname d\!x = \frac{\Gamma(1-r)}{r}(1-a^2)^{r/2} \cos(r \arctan(a))$ does anyone have an idea or a guess how to prove the following equation:  
$$\int\limits_{0}^{\infty}\frac{1-\text{e}^{-x}\cos(ax)}{x^{r+1}}\operatorname d\!x = \dfrac{\Gamma(1-r)}{r}(1-a^2)^{\frac{r}{2}} \cos(r \arctan(a))$$
for a fixed real-valued $a$ and  $r \in (0,1)$.
I cannot come up with a way to solve the integral. Thanks!
 A: I think your expression is off.  Begin by writing
$$\int_0^{\infty} dx \frac{1-e^{-x} \cos{(a x)}}{x^{r+1}} = \Re{\left [\int_0^{\infty} dx \frac{1-e^{-(1-i a) x}}{x^{r+1}} \right ]}$$
Then recognize that
$$\frac{1-e^{-(1-i a) x}}{x} = (1-i a) \int_0^1 du \, e^{-(1-i a) x u}$$
Assume that we may change the order of integration; this is justified because the integrals absolutely converge.  Then the integral is
$$(1-i a) \int_0^1 du \, \int_0^{\infty} dx \, x^{-r} e^{-(1-i a) x u}$$
The inner integral may be evaluated by subbing $y=(1-i a) x u$; note that we end up with an expression for the gamma function:
$$(1-i a) (1-i a)^{r-1} \Gamma(1-r) \int_0^1 du \, u^{r-1} = (1-i a)^{r} \frac{\Gamma(1-r)}{r}$$
Write 
$$1-i a = \sqrt{1+a^2} e^{-i \arctan{a}}$$
and the integral may be written as the real part of the above expression:
$$\frac{\Gamma(1-r)}{r} (1+a^2)^{r/2} \cos{(r \arctan{a})}$$
Note that the pythagorean piece has a plus and not a minus.
EDIT
I will add some more detail to the gamma function integral, which is
$$\int_0^{\infty} dx \, x^{-r} \, e^{-(1-i a) x u}$$
We want to prove that the result is equal to $(1-i a)^{r-1} u^{r-1} \Gamma(1-r)$ as asserted above.  One way to do this is, as pointed out, to deform the integration contour in the complex plane and apply Cauchy's theorem.  Recall that the substitution $y=(1-i a) x u$ placed the integration contour along a ray of angle $\arctan{a}$ in the complex plane.  So we complete a closed circuit by forming a wedge of radius $R$ and angle $\arctan{a}$ with the positive real axis.
We then consider the integral
$$\oint_C dz \, z^{-r} \, e^{-(1-i a) u z}$$
where $C$ is the above-mentioned wedge.  The contour integral is then equal to
$$ \int_0^{R} dx \, x^{-r} e^{-(1-i a) x u} + (1-i a)^{r-1} u^{r-1} \int_R^0 dy \, y^{-r} \, e^{-y} \\ + i R^{1-r} \int_0^{\arctan{a}} d\phi \, e^{i (1-r) \phi} \, e^{-(1-i a) u R e^{i \phi}}$$
We simply need to show that the third integral vanishes as $R \to \infty$ because the contour integral is zero by Cauchy's theorem.  (Rigorously, we should introduce a small arc around the origin, but the integral around that small arc must vanish because $r \in (0,1)$.)
The magnitude of that third integral is bounded by the quantity
$$R^{1-r} \int_0^{\arctan{a}} d\phi \, e^{-(\cos{\phi}+a \sin{\phi}) u R} = R^{1-r} \int_0^{\arctan{a}} d\phi \, e^{-u R\sqrt{a^2+1} \cos{(\phi-\arctan{a})}}$$
The exponent is never zeros within the integration region, so the integral is merely bounded by the ML lemma, or
$$ R^{1-r} e^{-u R} \arctan{a}$$
which clearly vanishes as $R \to \infty$.  Thus, the substitution made in the derivation is justified.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\int_{0}^{\infty}{1 - \expo{-x}\cos\pars{ax} \over x^{r + 1}}\,\dd x:\
{\Large ?}.\qquad a\ \in {\mathbb R}\,,\quad r\ \in \pars{0,1}}$.

$$
\int_{0}^{\infty}{1 - \expo{-x}\cos\pars{ax} \over x^{r + 1}}\,\dd x
=\Re\int_{0}^{\infty}{1 - \expo{-\mu x} \over x^{r + 1}}\,\dd x\,,
\qquad\mu\equiv 1 - \verts{a}\ic\tag{1}
$$

\begin{align}
&\int_{0}^{\infty}{1 - \expo{-\mu x} \over x^{r + 1}}\,\dd x
=\int_{x = 0}^{x \to \infty}\pars{1 - \expo{-\mu x}}\,\dd\pars{-\,{1 \over rx^{r}}}
\\[3mm]&=\overbrace{\left.-\,{1 - \expo{-\mu x} \over rx^{r}}
\right\vert_{x = 0}^{x \to \infty}}^{\ds{=\ 0}}\
-\
\int_{0}^{\infty}\pars{-\,{1 \over rx^{r}}}\bracks{-\expo{-\mu x}\pars{-\mu}}\,\dd x
={\mu \over r}\int_{0}^{\infty}x^{-r}\expo{-\mu x}\,\dd x
\\[3mm]&={\mu^{r} \over r}\int_{0}^{\pars{1 - \verts{a}\ic}\infty}
x^{-r}\expo{-x}\,\dd x
\\[3mm]&={\mu^{r} \over r}\lim_{R \to \infty}\bracks{\int_{0}^{R}x^{-r}\expo{-x}\,\dd x
+ \int_{0}^{-R\verts{a}\ic}\pars{R + y\ic}^{r}\expo{-\pars{R + y\ic}}\ic\,\dd y}
\end{align}

The second integral, in the right hand side, vanishes out when $R \to \infty$.
  Then,
  \begin{align}
\int_{0}^{\infty}{1 - \expo{-\mu x} \over x^{r + 1}}\,\dd x
&={\mu^{r} \over r}\int_{0}^{\infty}x^{\pars{1 - r} - 1}\expo{-x}\,\dd x
={1 \over r}\,\mu^{r}\,\Gamma\pars{1 - r}
={1 \over r}\,\Gamma\pars{1 - r}\pars{1 - \verts{a}\ic}^{r}
\\[3mm]&={1 \over r}\,\Gamma\pars{1 - r}\pars{1 + a^{2}}^{r/2}
\expo{-\ic r\arctan\pars{-\verts{a}}}
\end{align}

By replacing this result in $\pars{1}$, we'll find
$$
\color{#00f}{\large\int_{0}^{\infty}{1 - \expo{-x}\cos\pars{ax} \over x^{r + 1}}\,\dd x
=
{\Gamma\pars{1 - r} \over r}\,\pars{1 + a^{2}}^{r/2}
\cos\pars{r\arctan\pars{a}}}
$$
A: $$
\begin{align}
\int_0^\infty\frac{1-e^{-x}\cos(ax)}{x^{r+1}}\,\mathrm{d}x
&=\mathrm{Re}\left(\int_0^\infty\frac{1-e^{-x(1+ia)}}{x^{r+1}}\,\mathrm{d}x\right)\tag{1}\\
&=\mathrm{Re}\left((1+ia)^r\int_{\gamma(a)}\frac{1-e^{-z}}{z^{r+1}}\,\mathrm{d}z\right)\tag{2}\\
&=\mathrm{Re}\left((1+ia)^r\int_{\gamma(0)}\frac{1-e^{-z}}{z^{r+1}}\,\mathrm{d}z\right)\tag{3}\\
&=\mathrm{Re}\left((1+ia)^r\int_0^\infty\frac{1-e^{-x}}{x^{r+1}}\,\mathrm{d}x\right)\tag{4}\\
&=\mathrm{Re}\left(-\frac{(1+ia)^r}{r}\int_0^\infty(1-e^{-x})\,\mathrm{d}x^{-r}\right)\tag{5}\\
&=\mathrm{Re}\left(\frac{(1+ia)^r}{r}\int_0^\infty x^{-r}e^{-x}\,\mathrm{d}x\right)\tag{6}\\
&=\mathrm{Re}\left(\frac{(1+ia)^r}{r}\Gamma(1-r)\right)\tag{7}\\
&=-\Gamma(-r)\sqrt{1+a^2}^{\raise{2pt}{\large\,r}}\cos(r\tan^{-1}(a))\tag{8}
\end{align}
$$
Justification:
$(1)$: $e^{iax}=\cos(ax)+i\sin(ax)$
$(2)$: substitute $z=x(1+ia)$ and $\gamma(a)=(0,\infty)(1+ia)$
$(3)$: since there are no singularities inside the wedge and
$\hphantom{(3):}$the integrand decays to order $-r-1$ at $\infty$ and blows up to order $-r$ near $0$
$(4)$: substitute $x=z$
$(5)$: prepare to integrate by parts
$(6)$: integrate by parts
$(7)$: definition of $\Gamma$
$(8)$: take the real part and $\Gamma(1-r)=-r\Gamma(-r)$
Let me explain a bit more in detail what is going on in step $(3)$. We are using contour integration to justify the change of path of integration from $\gamma(a)=(0,\infty)(1+ia)$ in $(2)$ to $\gamma(0)=(0,\infty)$ in $(3)$. The difference of the integral along these two paths is the limit the integral around the wedge-shaped contour
$$
\overbrace{[\epsilon,R]}^{\gamma(0)}\cup\overbrace{R(1+ia[0,1])}^{\text{vanishes}}\cup\overbrace{(1+ia)[R,\epsilon]}^{\gamma(a)\text{ reversed}}\cup\overbrace{\epsilon(1+ia[1,0])}^{\text{vanishes}}
$$
as $\epsilon\to0$ and $R\to\infty$ since the integrals along the paths $R(1+ia[0,1])$ and $\epsilon(1+ia[1,0])$ vanish. Since there are no singularities of the integrand inside the contour, the integral around the contour is $0$.
