Ramanujan相关问题研究 - Eufisky - The lost book
Euler Sum的若干研究
$\sum{\frac{n}{{{e^{2\pi n}} - 1}}}$型的级数求解

Ramanujan相关问题研究

Eufisky posted @ 2014年7月23日 19:12 in 数学分析 with tags Ramanujan , 1825 阅读

问题一:

\[\int_0^\infty  {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {r^2}{x^2}} \right)\left( {1 + {r^4}{x^2}} \right)\left( {1 + {r^6}{x^2}} \right) \cdots }}}  = \frac{\pi }{{2\left( {1 + r + {r^3} + {r^6} + {r^{10}} +  \cdots } \right)}}.\]

Proof.If we set
$$ f(x)=\prod_{n=0}^{+\infty}(1+r^{2n}x^2) $$
we have:
$$\int_{0}^{+\infty}\frac{dx}{f(x)}=\pi i\sum_{m=0}^{+\infty}\operatorname{Res}\left(f(z),z=\frac{i}{r^m}\right)=\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}\tag{1}$$
but since 
$$ \prod_{n=0}^{+\infty}(1-x^n z)^{-1}=\sum_{n=0}^{+\infty}\frac{z^n}{(1-x)\cdot\ldots\cdot(1-x^n)}$$
is one of the Euler's partitions identities, and:
$$\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\sum_{m=0}^{+\infty}\frac{(1/r)^m}{(1-(1/r^2))\cdot\ldots\cdot(1-(1/r^2)^m)}$$
we have:
$$\int_{0}^{+\infty}\frac{dz}{f(z)}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\prod_{m=0}^{+\infty}\left(1-\frac{1}{r^{2m+1}}\right)^{-1}\tag{2}$$
and the claim follows from the Jacobi triple product identity:
$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$
 

参阅:http://math.stackexchange.com/questions/876106/how-find-this-integral-int-0-infty-fracdx1x21r2x21r4x21

问题二:Ramanujan Log-Trigonometric Integrals

\begin{align*}R_n^ -  &= \frac{2}{\pi }\int_0^{\frac{\pi }{2}} {{{\left( {{\theta ^2} + {{\ln }^2}\cos \theta } \right)}^{ - 2\left( { - n - 1} \right)}}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} +  \cdots +\frac{1}{2}\sqrt {\frac{{{{\ln }^2}\cos \theta }}{{{\theta ^2} + {{\ln }^2}\cos \theta }}} } } d\theta }  = {\left( {\ln 2} \right)^{ - {2^{ - n}}}}\\R_n^ +  &= \frac{2}{\pi }\int_0^{\frac{\pi }{2}} {{{\left( {{\theta ^2} + {{\ln }^2}\cos \theta } \right)}^{2\left( { - n - 1} \right)}}\sqrt {\frac{1}{2} + \frac{1}{2}\sqrt {\frac{1}{2} +  \cdots +\frac{1}{2}\sqrt {\frac{{{{\ln }^2}\cos \theta }}{{{\theta ^2} + {{\ln }^2}\cos \theta }}} } } d\theta }  = {\left( {\ln 2} \right)^{{2^{ - n}}}}.\end{align*}
问题二:Ramanujan's eccentric Integral formula
\[\int_0^\infty  {\frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 1} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \times \frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 2} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{{\left( {a + 1} \right)}^2}}}}} \times  \cdots dx}  = \frac{{\sqrt \pi  }}{2} \times \frac{{\Gamma \left( {a + \frac{1}{2}} \right)\Gamma \left( {b + 1} \right)\Gamma \left( {b - a + \frac{1}{2}} \right)}}{{\Gamma \left( a \right)\Gamma \left( {b + \frac{1}{2}} \right)\Gamma \left( {b - a + 1} \right)}}.\]
Proof.Andrey Rekalo:This is one of those precious cases when Ramanujan himself provided (a sketch of) a proof. The identity was published in his paper "Some definite integrals"  (Mess. Math. 44 (1915), pp. 10-18) together with several related formulae. 
 
It might be instructive to look first at the simpler identity (i.e. the limiting case when $b\to\infty$; the identity mentioned in the original question can be obtained by a similar approach):
$$\int\limits_{0}^{\infty} \prod_{k=0}^{\infty}\frac{1}{  1 + x^{2}/(a+k)^{2}}dx = \frac{\sqrt{\pi}}{2} \frac{ \Gamma(a+\frac{1}{2})}{\Gamma(a)},\quad a>0.\qquad\qquad\qquad(1)$$
Ramanujan  derives (1) by using a partial fraction decomposition of the product  $\prod_{k=0}^{n}\frac{1}{  1 + x^{2}/(a+k)^{2}}$, integrating term-wise, and passing to the limit $n\to\infty$. He also indicates that alternatively (1) is implied by the factorization 
$$\prod_{k=0}^{\infty}\left[1+\frac{x^2}{(a+k)^2}\right] = \frac{ [\Gamma(a)]^2}{\Gamma(a+ix)\Gamma(a-ix)},$$
which follows readily from  Euler's product formula for the gamma function. Thus (1) is equivalent to the formula 
$$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$
 
------------------------------------------------------------------
 
There is a nice paper "Wallis-Ramanujan-Schur-Feynman" by Amdeberhan et al (American Mathematical Monthly 117 (2010), pp. 618-632) that discusses interesting combinatorial aspects of formula (1) and its generalizations. 
 
David Hansen:Here is a proof of Ramanujan's identity (thanks to Todd Trimble for encouraging me to post this!).  As Andrey Rekalo notes, we have the identity $$\prod_{k=0}^{\infty}(1+\frac{x^2}{(k+a)^2})=\frac{\Gamma(a)^2}{|\Gamma(a+ix)|^2}$$.  In particular, the integrand in Ramanujan's integral is $\frac{\Gamma(b+1)^2 |\Gamma(a+ix)|^2}{\Gamma(a)^2 |\Gamma(b+1+ix)|^2}$.  Hence, after a little algebra (and also changing $b$ to $b-1$; I personally think Ramanujan made the wrong aesthetic choice here), we need to prove the integral evaluation $$I=\int_{-\infty}^{\infty} \frac{|\Gamma(a+ix)|^2}{|\Gamma(b+ix)|^2}dx=\sqrt{\pi}\frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b-a-1/2)}{\Gamma(b-1/2)\Gamma(b)\Gamma(b-a)}$$.
 
Now, if $f(x)$ has Mellin transform $F(s)$, then one form of Parseval's theorem for Mellin transforms is the identity $$\int_{0}^{\infty}f(x)x^{-1}dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}|F(it)|^2 dt$$ (under suitable conditions of course).  Applying this with the Mellin pair $$f(x)=\Gamma(b-a)^{-1}x^{a}(1-x)^{b-a-1} \; \mathrm{if} \; 0\leq x \leq 1$$ (and $f=0$ otherwise), $$F(s)=\frac{\Gamma(s+a)}{\Gamma(s+b)}$$ gives
 
\begin{align*}I&=2\pi \Gamma(b-a)^{-2} \int_{0}^{\infty}x^{2a-1}(1-x)^{2b-2a-2}dx\\&=2\pi \Gamma(b-a)^{-2} \frac{\Gamma(2a) \Gamma(2b-2a-1)}{\Gamma(2b-1)}.\end{align*}
 
Next, apply the formula $$\Gamma(2z)=2^{2z-1}\pi^{-1/2}\Gamma(z)\Gamma(z+1/2)$$ to each of the $\Gamma$-functions in the quotient here, getting 
 
$$I=\sqrt{\pi} \frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b-a-1/2)\Gamma(b-a)}{\Gamma(b-a)^2 \Gamma(b-1/2) \Gamma(b)}$$, and cancelling a $\Gamma(b-a)$ concludes the proof.
 
Exercise: Give a proof, along similar lines, of the formula
 
$$\int_{-\infty}^{\infty} |\Gamma(a+ix)\Gamma(b+ix)|^2 dx=\sqrt{\pi}\frac{\Gamma(a)\Gamma(a+1/2)\Gamma(b)\Gamma(b+1/2)\Gamma(a+b)}{\Gamma(a+b+1/2)}$$, 
 
and determine for what range of $a,b$ it holds.
 
 
问题二:Ramanujan's eccentric Integral formula
If $\alpha$ and $\beta$ are positive numbers such that $\alpha\dots \beta=\pi^2$ then,
\[\alpha  \cdot \sum\limits_{n = 1}^\infty  {\frac{n}{{{e^{2n\alpha }} - 1}}}  + \beta \sum\limits_{n = 1}^\infty  {\frac{n}{{{e^{2n\beta }} - 1}}}  = \frac{{\alpha  + \beta }}{{24}} - \frac{1}{4}.\]
Proof.I believe this formula is true, provided the $\alpha$ in the second sum is changed to a $\beta$, as suggested by Todd Trimble's comment. Let
$$ P(x) = \prod_{n=1}^\infty \frac{1}{1-x^n} $$
be the generating function for the number of partitions of a non-negative integer $n$. Dedekind proved that $P$ satisfies the transformation formula
$$ \log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr)+ \frac{1}{2} \log t $$
for $t > 0$.
Differentiating this formula with respect to $t$ gives
$$ -\sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n t}-1} - \frac{1}{t^2} \sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n/t} -1} = \frac{\pi}{12} \Bigl( -\frac{1}{t^2} - 1\Bigr) + \frac{1}{2t} $$
Now multiply through by $-t/2$ and substitute $\alpha = \pi t$, $\beta = \pi /t$ to get
$$ \sum_{n=1}^\infty \frac{\alpha n}{e^{2n\alpha}-1} + \sum_{n=1}^\infty \frac{\beta n}{e^{2n\beta}-1} = \frac{1}{24}(\beta+\alpha) - \frac{1}{4}$$
which is Ramanujan's formula. 
 
The transformation formula for $P$ is related to the theory of modular forms, of which
the Eisenstein series mentioned in Derek Jennings' answer to your question on math.stackexchange are important examples. Briefly, if we define
$$ \eta(\tau) = \frac{e^{2\pi i \tau/24}}{P(e^{2\pi i \tau})} = e^{2\pi i \tau/24} \prod_{n=1}^\infty (1-e^{2\pi i n \tau}), $$
then $\eta(\tau)^{24}$ is a modular form of weight $12$. As such, $\eta$ satisfies the identity
$$ \eta(-1/\tau) = \sqrt{-i \tau}\; \eta(\tau). $$
The transformation formula for $P$ follows by setting $\tau = it$ and taking logs.
 
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