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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?

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    $\begingroup$ I'm tempted to say "no". Seems like the simples way to see this symmetry is to evaluate the integral... $\endgroup$
    – Fabian
    Jan 1, 2013 at 22:33
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    $\begingroup$ Invariance under $\alpha \to -\alpha$ is not obvious either. $\endgroup$
    – Maesumi
    Jan 1, 2013 at 23:16
  • $\begingroup$ @user7530: thanks. This question has vexed me for some time. You'd think that symmetry like this has a simple explanation. But I wonder what I am missing. $\endgroup$
    – Ron Gordon
    Jan 2, 2013 at 0:57
  • $\begingroup$ $f(\alpha, \beta) = \int_{-\infty}^{\infty} du \: \frac{e^{u\alpha}}{e^{-u}+e^u+e^{i \pi \beta}+e^{-i \pi \beta}}$ after $u=\ln x$ and expanding cosine $\endgroup$
    – ebsddd
    Jan 3, 2013 at 22:56
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    $\begingroup$ The question is too old to migrate (there's a 60 day limit on migration). You can re-ask it on MO (with a link here, so people can see whether they have something new). $\endgroup$ Jun 21, 2016 at 13:28

6 Answers 6

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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.


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: $$ f(\alpha, \beta) = \pi \frac{ \sin(\pi \alpha \beta)}{\sin(\pi \alpha) \sin(\pi \beta)} = \frac{\operatorname{sinc}(\pi \alpha \beta)}{\operatorname{sinc}(\pi \alpha) \operatorname{sinc}(\pi \beta)} \tag{5} $$

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  • $\begingroup$ I am very impressed with what you have derived, nonetheless! I love the integral form in Equation (2), and I am out of practice enough to need to review my MacDonald functions to see how you derived that infinite sum, and then evaluated that sum. Nonetheless, as you said, not quite there yet, although I hope that maybe the expression in Equation (2) (user7530's observation taken a few steps further) may help. $\endgroup$
    – Ron Gordon
    Jan 4, 2013 at 20:19
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    $\begingroup$ I am giving you the bounty, although nobody really answered the question. You did provide new insights into this integral for which I am grateful. $\endgroup$
    – Ron Gordon
    Jan 10, 2013 at 22:01
  • $\begingroup$ @rlgordonma You could have awarded the bounty without accepting the answer. Acceptance marks the question with green background that tells other users "nothing to do here, the matter is settled". $\endgroup$
    – user53153
    Jan 11, 2013 at 1:23
  • $\begingroup$ Ah. Thanks for pointing that out. I will fix that if possible. You should have the bounty nonetheless. $\endgroup$
    – Ron Gordon
    Jan 11, 2013 at 1:34
  • $\begingroup$ @Sasha Exquisite! $\endgroup$ Nov 1, 2013 at 17:32
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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. ${\partial_{\alpha}}{\partial_{\beta}}$.

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    $\begingroup$ Good work, Graham. I'll be interested very much to hear what you discover. $\endgroup$
    – Ron Gordon
    Nov 1, 2013 at 18:45
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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.

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  • $\begingroup$ This seems like progress, but it is still far from clear what it buys us. By using this $\alpha \leftrightarrow - \alpha$ symmetry, we can get a cosine-like structure in the numerator (by adding contributions from the equal $\alpha$/$-\alpha$ pieces). But I am still stuck. $\endgroup$
    – Ron Gordon
    Jan 2, 2013 at 16:51
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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

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

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, $$ \fermi\pars{\mu,0} = \fermi\pars{0,\mu} =\lim_{\nu \to 0^{-}}\fermi\pars{\mu,\nu} =\lim_{\mu \to 0^{-}}\fermi\pars{\mu,\nu} $$

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  • $\begingroup$ However I find some issues with signs. if $\beta \in (0, 1)$ and $y \in (0, \infty)$ then $\dfrac{y^{\beta}}{y + 1} > 0$ so that $\int_{0}^{\infty}(y^{\beta})/(y + 1)\,dy > 0$, but you have put that as $-\pi/\sin(\pi\beta)$. I believe you should get $f(\alpha, \beta) = \dfrac{\sin(\pi\alpha\beta)}{\sin(\pi\beta)}\int_{0}^{\infty}\dfrac{x^{\alpha}}{x + 1}\,dx$ because here also we don't need a $-$ sign if we restrict $\alpha \in (0, 1), \beta \in (0, 1)$. $\endgroup$
    – Paramanand Singh
    Feb 16, 2014 at 10:55
  • $\begingroup$ The right formula is $\int_{0}^{\infty}(y^{-\beta})/(y + 1)\, dy = \pi/\sin(\pi\beta)$ where $0 < \beta < 1$. Perhaps there is some issue in the last few lines. $\endgroup$
    – Paramanand Singh
    Feb 16, 2014 at 11:30
  • $\begingroup$ @ParamanandSingh I'll check and comment later. Some little issues will be clear. Right now this page is working very bad. Thanks. $\endgroup$ Feb 16, 2014 at 22:25
  • $\begingroup$ @ParamanandSingh I just checked every thing and rewrote the answer to deal with the worries you commented about it. Thanks. $\endgroup$ Feb 17, 2014 at 16:47
  • $\begingroup$ Felix, I thought about accepting this answer, but it seems a little off. Not that it's wrong - it is not - but that you had to evaluate the integral and cast it into an alternative form. The problem I have with this is that the symmetry is indeed palpable once you evaluate the integral - but the challenge was to avoid doing that in the first place. Maybe if you can show how to derive a similar form from the integral directly, I may accept. $\endgroup$
    – Ron Gordon
    Nov 27, 2014 at 7:50
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I hope this answers your question.

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

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

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

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


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

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 $y=1$.


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

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

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    $\begingroup$ Great stuff. Very happy to see this demonstrated nearly 7 yrs later! $\endgroup$ Dec 31, 2019 at 0:20
  • $\begingroup$ @GrahamHesketh Thanks for the compliment :) $\endgroup$
    – Szeto
    Dec 31, 2019 at 0:44

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