Proving $\int\limits_{0}^{\infty}\frac{1-\text{e}^{-x}\cos(ax)}{x^{r+1}}\operatorname d\!x = \frac{\Gamma(1-r)}{r}(1-a^2)^{r/2} \cos(r \arctan(a))$

Question

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!

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{1 - \expo{-x}\cos\pars{ax} \over x^{r + 1}}\,\dd x:\ {\Large ?}.\qquad a\ \in {\mathbb R}\,,\quad r\ \in \pars{0,1}}$.

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

\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 $$ \color{#00f}{\large\int_{0}^{\infty}{1 - \expo{-x}\cos\pars{ax} \over x^{r + 1}}\,\dd x = {\Gamma\pars{1 - r} \over r}\,\pars{1 + a^{2}}^{r/2} \cos\pars{r\arctan\pars{a}}} $$

Answer 3

$$ \begin{align} \int_0^\infty\frac{1-e^{-x}\cos(ax)}{x^{r+1}}\,\mathrm{d}x &=\mathrm{Re}\left(\int_0^\infty\frac{1-e^{-x(1+ia)}}{x^{r+1}}\,\mathrm{d}x\right)\tag{1}\\ &=\mathrm{Re}\left((1+ia)^r\int_{\gamma(a)}\frac{1-e^{-z}}{z^{r+1}}\,\mathrm{d}z\right)\tag{2}\\ &=\mathrm{Re}\left((1+ia)^r\int_{\gamma(0)}\frac{1-e^{-z}}{z^{r+1}}\,\mathrm{d}z\right)\tag{3}\\ &=\mathrm{Re}\left((1+ia)^r\int_0^\infty\frac{1-e^{-x}}{x^{r+1}}\,\mathrm{d}x\right)\tag{4}\\ &=\mathrm{Re}\left(-\frac{(1+ia)^r}{r}\int_0^\infty(1-e^{-x})\,\mathrm{d}x^{-r}\right)\tag{5}\\ &=\mathrm{Re}\left(\frac{(1+ia)^r}{r}\int_0^\infty x^{-r}e^{-x}\,\mathrm{d}x\right)\tag{6}\\ &=\mathrm{Re}\left(\frac{(1+ia)^r}{r}\Gamma(1-r)\right)\tag{7}\\ &=-\Gamma(-r)\sqrt{1+a^2}^{\raise{2pt}{\large\,r}}\cos(r\tan^{-1}(a))\tag{8} \end{align} $$ 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 $0$.