Symmetry of function defined by integral

Question

Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as

$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$

One can use, for example, the Residue Theorem to show that

$$ f(\alpha, \beta) = \frac{\pi \sin{\left (\pi \alpha \beta\right )}}{ \sin{\left (\pi \alpha\right )} \, \sin{\left (\pi \beta\right )}} $$

Clearly, from this latter expression, $f(\alpha, \beta) = f(\beta, \alpha)$. My question is, can one see this symmetry directly from the integral expression?

Answer 1

Very interesting question! But, alas, not an answer. Only few representations for the integral obtained. One of them evaluated to the form claimed in the question.

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First, transform the integral into a form, symmetric under $\alpha \mapsto -\alpha$: $$ \int_0^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x = \int_0^1 \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x + \int_1^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x $$ Make a change of variables $x \to x^{-1}$ in the last integral to obtain: $$ f(\alpha,\beta) = \int_0^1 \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1} $$ Now, making a change of variables $x = \exp(-t)$ we have: $$ f(\alpha,\beta) = \int_0^\infty \frac{\cosh(\alpha t)}{\cosh(t) + \cos(\beta \pi)} \mathrm{d} t \tag{2} $$ Using $$ \int_0^\infty \exp\left(-u \left( \cosh t + \cos \pi \beta \right) \right) \mathrm{d}u = \frac{1}{\cosh(t) + \cos(\beta \pi)} $$ and the integral representation of the modified Bessel function of the second kind: $$ \int_0^\infty \cosh(\alpha t) \exp\left( - u \cosh t \right) \mathrm{d}t = K_\alpha(u) $$ we arrive at a compact representation: $$ f(\alpha,\beta) = \int_0^\infty K_\alpha(u) \mathrm{e}^{-u \cos\left(\pi \beta\right)} \mathrm{d} u \tag{3} $$ expanding the exponential into series and using $\int_0^\infty u^n K_\alpha(u) \mathrm{d} u = 2^{n-1} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right)$ we get: $$ f(\alpha,\beta) = \sum_{n=0}^\infty \frac{2^{n-1}}{n!} \left(-\cos \pi \beta\right)^{n} \Gamma\left(\frac{n}{2} + \frac{1+\alpha}{2} \right)\Gamma\left(\frac{n}{2} + \frac{1-\alpha}{2} \right) \tag{4} $$ summing over even and over odd integers: $$ f(\alpha, \beta) = \frac{\pi}{2} \frac{ \cos\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \cos \left( \frac{\pi \alpha}{2} \right)} - \frac{\pi}{2} \frac{ \sin\left( \alpha \arcsin \cos(\pi \beta) \right) }{ | \sin(\pi \beta) | \sin \left( \frac{\pi \alpha}{2} \right)} = \pi \frac{\sin \left( \alpha \left( \frac{\pi}{2} - \arcsin \cos(\pi \beta) \right) \right)}{ | \sin \pi \beta | \sin(\pi \alpha)} $$ Now $\frac{\pi}{2} - \arcsin \cos(\pi \beta) = \arccos \cos(\pi \beta) = \pi | \beta |$ for $-1<\beta<1$. Thus, restoring parity, we recover the OP's expression: $$ f(\alpha, \beta) = \pi \frac{ \sin(\pi \alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} = \frac{\operatorname{sinc}(\pi \alpha \beta)}{\operatorname{sinc}(\pi \alpha) \operatorname{sinc}(\pi \beta)} \tag{5} $$

Answer 2

This question is fantastic! I have not found an integral displaying the symmetry but I wanted to show that it points to a symmetry of the Lerch zeta function, if this is known then perhaps that could explain the symmetry, if it is not known it is very interesting I think.

From $(1)$ in @Sasha: $$\begin{align} f(\alpha,\beta) = \int_0^1 \frac{x^\alpha + x^{-\alpha}}{1+2 x \cos(\pi \beta) + x^2} \mathrm{d} x \tag{1}\\ \end{align} $$ The integrand is comparable to the generating function of the Chebyshev polynomials of the second kind and in fact: $$ \begin{align} {\frac {{x}^{\alpha}+{x}^{-\alpha}}{1+2\,x\cos \left( \pi \,\beta \right) +{x}^{2}}}&=\sum _{n=0}^{\infty }U_n \left( - \cos \left( \pi \,\beta \right) \right) \left( {x}^{n+\alpha}+{x}^{n -\alpha} \right) \tag{2}\\ \end{align} $$ and the Chebyshev polynomial of the second kind satsifies: $$U_n \left( -\cos \left( \pi \,\beta \right)\right)={\frac { \left( -1 \right) ^{n}\sin \left( \left( 1+n \right) \pi \, \beta \right) }{\sin \left( \pi \,\beta \right) }} \tag{3} $$ so after using $(2,3)$ in $(1)$ and switching integration and summation order we obtain a Fourier series: $$ \begin{align} f(\alpha,\beta)&=\frac{1}{\sin \left( \pi \,\beta \right)} \sum _{n=0}^{\infty }\left( -1 \right) ^{n}\sin \left( \left( 1+n \right) \pi \, \beta \right) \left( \dfrac{1}{1+n+\alpha }+ \dfrac{1}{1+n-\alpha } \right)\tag{4}\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \sum _{n=-\infty\,(n\ne0)}^{\infty } \dfrac{\left( -1 \right) ^{n}\sin \left( n \pi \, \beta \right)}{n+\alpha } \\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[ \sum _{n=1}^{\infty }\left( -1 \right) ^{n}e^{i n\pi \beta } \left( \dfrac{1}{n+\alpha }+ \dfrac{1}{n-\alpha } \right)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[\Phi(-e^{i \pi \beta },1,\alpha)+\Phi(-e^{i \pi \beta },1,-\alpha)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[\Phi_{+}(-e^{i \pi \beta },1,\alpha)\right]\quad:\quad(n=-\infty..+\infty,n\ne 0)\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[L\left(\frac{\beta+1}{2},\alpha,1\right)+L\left(\frac{\beta+1}{2},-\alpha,1\right)\right]\\ &=-\frac{1}{\sin \left( \pi \,\beta \right)} \mathfrak{I}\left[L_{+}\left(\frac{\beta+1}{2},\alpha,1\right)\right]\\ &=-\frac{1}{2\sin \left( \pi \,\beta \right)}\left[L_{+}\left(\frac{\beta+1}{2},\alpha,1\right)-L_{+}\left(\frac{\beta+1}{2},-\alpha,1\right)\right] \end{align}$$ where $L$ is the Lerch zeta function and $\Phi$ the Lerch transcendant. I can't believe this Fourier series is invariant under $\alpha \leftrightarrow \beta$. I have not as of yet found that symmetry for the Lerch zeta function on line. This series together with the previous demonstration that: $$f(\alpha, \beta) = \pi \frac{ \sin(\pi\alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} \tag{5}$$ show immediately that the following special cases hold which are interesting in their own right: $$\sum _{n=-\infty }^{\infty }{\frac { \left( -1 \right) ^{n}\sin \left( \pi \,xn \right) }{x-n}}={\frac {\pi\sin \left( \pi{x}^{ 2} \right) }{\sin \left( \pi x \right) }}\tag{6}$$ $$\sum _{n=-\infty }^{\infty }{\frac { \left( -1 \right) ^{n}\cos \left( \pi \,xn \right) }{1- \left( n+x \right) ^{2}}}=\pi\sin \left( \pi {x}^{2} \right)\tag{7}$$

Now, from $(4)$ and $(5)$ and the variable change $\alpha=x,\, \beta=2y-1$, with $-1<x<1,\,0\le y<1$, we have: $$L_{+}(y,x,1)=-L_{+}(y,-x,1)-2\pi i\dfrac{\sin(2\pi x(y-\frac{1}{2})}{\sin(x)} \tag{8}$$ where we recognise the trigonometric term as the Dirichlet Kernel (with $x\rightarrow2x$ and for $y$ generalised to non-integer). If we then use differentiation with respect to $x$ as a raising operator we obtain the reflection formula in the $x$ variable: $$L_{+}(y,x,k)=\left( -1 \right) ^{k-1}L_{+}(y,-x,k) -2\pi i \dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{9}$$ where the order becomes $k$ from simple differentiation of the function definition, and it also follows from reversing summation order in the function definition that: $$L_{+}(y,x,k)=\sum_{n=-\infty (n\ne0)}^{\infty}\dfrac{e^{2\pi in y}}{(n+x)^k}=(-1)^k L_{+}(-y,-x,k)\tag{10}$$ and so $(9)$ can also be viewed as a reflection formula in $y$: $$L_{+}(y,x,k)=L_{+}(-y,x,k) -2\pi i \dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{11}$$ or as an explicit formula for the imaginary part: $$\mathfrak{I}\left(L_{+}(y,x,k)\right)= \pi\dfrac{ \left( -1 \right) ^{k-1}}{(k-1)!}{\frac {\partial ^{k-1}}{\partial {x}^{k-1}}} {\frac {\sin \left( 2\pi x \left( y-\frac{1}{2} \right) \right) } {\sin \left( \pi x \right) }} \tag{11}$$ As a final application, evaluating $(9)$ and $(10)$ at $x=0$ and recognising that $(10)$ is then proportional to the Fourier series of the Bernoulli polynomials $B(m,y)$, we obtain the Taylor series for the Dirichlet kernel in the $x$ variable: $$2\sum _{m=0}^{\infty }\left( -1 \right) ^{m}{\frac { B \left( 2m+1,y \right) }{ \left( 2m+1 \right) !}}{x}^{2m}={ \frac {\sin \left( x \left( y-\frac{1}{2} \right) \right) }{\sin \left( \frac{x}{2} \right) }} \tag{12}$$

Still no closer to displaying the symmetry as an integral but I just wanted to show some interesting consequences of the symmetry and the relation itself. Also, if we wanted to preserve the symmetry and generalise equation $(1)$ then we could do so by applying any symmetric differential operator as a raising operator e.g. ${\partial_{\alpha}}{\partial_{\beta}}$.

Answer 3

Not an answer, but a reply to Maesumi's comment --

Invariance under $\alpha \to -\alpha$ is not so difficult to see: let $x=y^{-1}$, then \begin{align*}\int_0^{\infty} \frac{x^{\alpha}}{1+2x\cos(\pi \beta) + x^2} dx &= \int_{\infty}^0 \frac{y^{-\alpha} }{1+2y^{-1}\cos(\pi\beta)+y^{-2}}(-y^{-2})dy\\ &=\int_0^{\infty} \frac{y^{-\alpha}}{1+2y\cos(\pi \beta) + y^2} dy. \end{align*}

No idea about $\alpha \leftrightarrow \beta$ though.

Answer 4

I hope this answers your question.

$$\begin{align} \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2} &= \int_0^{\infty} x^{\alpha}\,\mathscr M\circ\mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\} dx\\ &= \int_0^{\infty} \int^\infty_0 x^{\alpha}y^{\beta-1}\mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\}(y)\,\, dydx\\ &= \int_0^{\infty} \int^\infty_0 x^{\alpha}y^{\beta}\cdot\frac1y \mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\}(y)\,\, dydx\\ \end{align} $$

Ideally, if $\displaystyle{\frac1y \mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\}(y)}$ is symmetric in $x$ and $y$, then we got what the OP wants.

Surprisingly, $$\frac1y \mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\}(y)=\frac2{\pi}\frac{\sin\left(\frac{\ln x\ln y}{\pi}\right)}{(x^2-1)(y^2-1)}\qquad(\star)$$ which is obviously symmetric.

The rest of the answer is devoted to the proof of $(\star)$.

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$$I:=\frac1y \mathscr M^{-1}\left\{\frac{1}{1+2 x \cos{(\pi \beta)} + x^2}\right\}(y) =\frac1{2\pi iy}\int^{i\infty}_{-i\infty}\underbrace{\frac{y^{-\beta}}{1+2 x \cos{(\pi \beta)} + x^2}}_{:= f(\beta)}d\beta$$

Let's look at the poles and residues of the integrand, so that this integral can be evaluated by residue theorem later.

The poles are all simple, and are located at $$p_n^{\pm}=\pm\left(\frac{\ln x}{\pi i}+1\right)+2n:=\pm c\pm 1+2n\qquad n\in\mathbb Z$$

Moreover, the residues are $$\operatorname{Res}(f(\beta),\beta=p_n^{\pm})=\pm\frac{y^{\mp c\mp 1-2n}}{\pi i(1-x^2)}$$

For $y>1$, $f(\beta)$ decays exponentially on the right half plane. By residue theorem, $$\begin{align} I &=\frac1{2\pi i y}\cdot -2\pi i\sum\text{residue on the right half plane} \\ &=-\frac1y\left(\sum^\infty_{n=0}\operatorname{Res}(f(\beta),p_n^{+})+\sum^\infty_{n=1}\operatorname{Res}(f(\beta),p_n^{-})\right) \\ &=-\frac1{\pi i(1-x^2)y}\left[\frac{y^{-c-1}}{1-y^{-2}}+\frac{y^{c+1}}{1-y^2}\right] \\ &=-\frac1{\pi i(1-x^2)}\left[\frac{y^{-c-2}}{1-y^{-2}}+\frac{y^{c}}{1-y^2}\right] \\ &=-\frac1{\pi i(1-x^2)}\left[\frac{y^{-c}}{y^{2}-1}+\frac{y^{c}}{1-y^2}\right] \\ &=-\frac1{\pi i(1-x^2)(1-y^2)}\left(e^{\ln y\ln x/(\pi i)}-e^{-\ln y\ln x/(\pi i)}\right)\\ &=\frac2{\pi(1-x^2)(1-y^2)}\sin\left(\frac{\ln x\ln y}{\pi}\right)\\ \end{align} $$

For $0<y<1$, $f(\beta)$ decays exponentially on the left half plane. By residue theorem, $$I=\frac1{2\pi i y}\cdot 2\pi i\sum\text{residue on the left half plane}$$

It is not a surprise that the sum is also equal to $\displaystyle{\frac2{\pi(1-x^2)(1-y^2)}\sin\left(\frac{\ln x\ln y}{\pi}\right)}$. This can be seen as an instance of analytic continuation, due to the removable singularity at $y=1$.

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In conclusion, $$\frac1{2\pi i}\int^{i\infty}_{-i\infty}\frac{y^{-\beta-1}}{1+2 x \cos{(\pi \beta)} + x^2}d\beta=\frac2{\pi}\frac{\sin\left(\frac{\ln x\ln y}{\pi}\right)}{(x^2-1)(y^2-1)}$$

$$ \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2} = \frac2{\pi}\int_0^{\infty} \int^\infty_0 x^{\alpha}y^{\beta}\frac{\sin\left(\frac{\ln x\ln y}{\pi}\right)}{(x^2-1)(y^2-1)}dxdy$$

Answer 5

This is not the answer of your problem Mr. Ron Gordon but it only the other way to prove the integral. You may also find the other ways, really beautiful methods, to prove the integral here, my OP. If you don't mind, I would like to present an alternative approach that makes use of the fact that $$\int^\infty_0\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin{p\pi}}$$ Simply factorise the denominator and decompose the integrand into partial fractions. \begin{align} \int^\infty_0\frac{x^\alpha}{x^2+2(\cos{\pi\beta})x+1}dx &=\int^\infty_0\frac{x^\alpha}{(x+e^{i\pi\beta})(x+e^{-i\pi\beta})}dx\\ &=\frac{1}{-e^{i\pi\beta}+e^{-i\pi\beta}}\int^\infty_0\frac{x^\alpha}{e^{i\pi\beta}+x}dx+\frac{1}{-e^{-i\pi\beta}+e^{i\pi\beta}}\int^\infty_0\frac{x^\alpha}{e^{-i\pi\beta}+x}dx\\ &=\frac{1}{-2i\sin{\pi\beta}}\int^\infty_0\frac{(e^{i\pi\beta}u)^\alpha}{1+u}du+\frac{1}{2i\sin{\pi\beta}}\int^\infty_0\frac{(e^{-i\pi\beta}u)^\alpha}{1+u}du\\ &=\frac{e^{i\alpha\pi\beta}}{-2i\sin{\pi\beta}}\frac{\pi}{\sin(\alpha\pi+\pi)}+\frac{e^{-i\alpha\pi\beta}}{2i\sin{\pi\beta}}\frac{\pi}{\sin(\alpha\pi+\pi)}\\ &=\frac{\pi}{\sin \alpha\pi\sin{\pi\beta}}\left(\frac{e^{i\alpha \pi\beta}-e^{-i\alpha \pi\beta}}{2i}\right)\\ &=\frac{\pi\sin{\alpha \pi\beta}}{\sin{\alpha\pi}\sin{\pi\beta}} \end{align}

Credit answer: Mr. Superabound

Answer 6

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

Let's consider the integral $\ds{\fermi\pars{\alpha,\beta,\Lambda}\equiv\int_{0}^{\Lambda} {x^{\alpha} \over 1 + 2\cos\pars{\pi\beta}x + x^{2}}\,\dd x}$, with $\Lambda > 0$, such that $\ds{\fermi\pars{\alpha,\beta} = \lim_{\Lambda \to \infty}\fermi\pars{\alpha,\beta\Lambda}}$.

Note that \begin{align} &1 + 2x\cos\pars{\pi\beta} + x^{2} =\bracks{x + \cos\pars{\pi\beta}}^{2} + \sin^{2}\pars{\beta} \\[3mm]&=\bracks{x + \cos\pars{\pi\beta} + \ic\sin\pars{\pi\beta}} \bracks{x + \cos\pars{\pi\beta} - \ic\sin\pars{\pi\beta}} =\pars{x + \expo{\ic\pi\verts{\beta}}}\pars{x + \expo{-\ic\pi\verts{\beta}}} \end{align}

\begin{align} \fermi\pars{\alpha,\beta,\Lambda}&=\int_{0}^{\Lambda}x^{\alpha} \pars{{1 \over x + \expo{-\ic\pi\verts{\beta}}} - {1 \over x + \expo{\ic\pi\verts{\beta}}}} \,{1 \over \expo{\ic\pi\verts{\beta}} - \expo{-\ic\pi\verts{\beta}}}\,\dd x \\[3mm]&={1 \over 2\ic\sin\pars{\pi\verts{\beta}}}\bracks{% \int_{0}^{\Lambda}{x^{\alpha} \over x + \expo{-\ic\pi\verts{\beta}}}\,\dd x - \int_{0}^{\Lambda}{x^{\alpha} \over x + \expo{\ic\pi\verts{\beta}}}\,\dd x} \\[3mm]&= {1 \over 2\ic\sin\pars{\pi\verts{\beta}}}\bracks{% \expo{-\ic\pi\alpha\verts{\beta}} \int_{0}^{\Lambda\expo{\ic\pi\verts{\beta}}}{x^{\alpha} \over x + 1}\,\dd x - \expo{\ic\pi\alpha\verts{\beta}} \int_{0}^{\Lambda\expo{-\ic\pi\alpha\verts{\beta}}}{x^{\alpha} \over x + 1}\,\dd x} \\[3mm]&={1 \over 2\ic\sin\pars{\pi\beta}}\times \\[3mm]&\left\lbrace% \expo{-\ic\pi\alpha\verts{\beta}}\bracks{% \int_{0}^{\Lambda}{x^{\alpha} \over x + 1}\,\dd x + \int_{0}^{\Lambda\sin\pars{\pi\verts{\beta}}} {\bracks{\Lambda\cos\pars{\pi\beta} + \ic y}^{\alpha} \over \Lambda\cos\pars{\pi\beta} + \ic y + 1}\,\ic\,\dd y}\right. \\[3mm]&\phantom{\braces{}}- \\[3mm]&\phantom{\braces{}}\left.% \expo{\ic\pi\alpha\verts{\beta}}\bracks{% \int_{0}^{\Lambda}{x^{\alpha} \over x + 1}\,\dd x + \int_{0}^{-\Lambda\sin\pars{\pi\verts{\beta}}} {\bracks{\Lambda\cos\pars{\pi\beta} + \ic y}^{\alpha} \over \Lambda\cos\pars{\pi\beta} + \ic y + 1}\,\ic\,\dd y}\,\right\rbrace \end{align}

\begin{align} \fermi\pars{\alpha,\beta,\Lambda}&= -\,{\sin\pars{\pi\alpha\beta} \over \sin\pars{\pi\beta}} \int_{0}^{\Lambda}{x^{\alpha} \over x + 1}\,\dd x \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,\,\pars{1} \\[3mm]&\phantom{=}+ {1 \over \sin\pars{\pi\verts{\beta}}}\Re\bracks{\expo{-\ic\pi\alpha\verts{\beta}}% \int_{0}^{\Lambda\sin\pars{\pi\verts{\beta}}} {\bracks{\Lambda\cos\pars{\pi\beta} + \ic y}^{\alpha} \over \Lambda\cos\pars{\pi\beta} + \ic y + 1}\,\dd y}\qquad\qquad\pars{2} \end{align}

$$ \mbox{When}\ \Lambda \to \infty\,,\quad\left\lbrace% \begin{array}{rl} \bullet & \mbox{The integral in}\ \pars{1}\ \mbox{converges when}\ \alpha < 0.\ \mbox{Since}\ \alpha\in\pars{-1,1}\,, \\&\mbox{the result is valid when}\ \alpha\in\pars{-1,0} \\[3mm] \bullet & \mbox{The integral in}\ \pars{2}\ \mbox{vanishes out in this limit.} \end{array}\right. $$ Then, $$ \lim_{\Lambda \to \infty}\fermi\pars{\alpha,\beta,\Lambda} =-\,{\sin\pars{\pi\alpha\beta} \over \sin\pars{\pi\beta}}\int_{0}^{\infty} {x^{\alpha} \over x + 1}\,\dd x\,,\qquad -1 < a < 0 $$

$$ \mbox{Also,}\quad {1 \over \sin\pars{\pi\beta}}= -\,{1 \over \pi}\int_{0}^{\infty}{y^{\beta} \over y + 1}\,\dd y\,,\qquad -1 < \beta < 0\tag{3} $$ $\tt @user7530$ has already proved the original integral symmetry under $\alpha \to -\alpha$ and it is clearly symmetric under $\beta \to -\beta$. The cases $\alpha = 0$ or $\beta = 0$ are easily handled $\pars{~\mbox{see}\ \pars{3}~}$ with the limit $$ \lim_{\mu \to 0^{-}}\bracks{\mu\int_{0}^{\infty}{x^{\mu} \over x + 1}}\,\dd x = -1\tag{4} $$

$$ \fermi\pars{\alpha,\beta} = {\sin\pars{\pi\verts{\alpha\beta}} \over \pi}\int_{0}^{\infty}\int_{0}^{\infty} {x^{-\verts{\alpha}}y^{-\verts{\beta}} \over xy + x + y + 1}\,\dd x\,\dd y \quad\mbox{if}\quad\alpha, \beta \in \pars{-1,0}\cup\pars{0,1} $$ Otherwise, $$ \fermi\pars{\mu,0} = \fermi\pars{0,\mu} =\lim_{\nu \to 0^{-}}\fermi\pars{\mu,\nu} =\lim_{\mu \to 0^{-}}\fermi\pars{\mu,\nu} $$