Ramanujan相关问题研究

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:

$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$

 

\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

\[\int_0^\infty  {\frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 1} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \times \frac{{1 + \frac{{{x^2}}}{{{{\left( {b + 2} \right)}^2}}}}}{{1 + \frac{{{x^2}}}{{{{\left( {a + 1} \right)}^2}}}}} \times  \cdots dx}  = \frac{{\sqrt \pi  }}{2} \times \frac{{\Gamma \left( {a + \frac{1}{2}} \right)\Gamma \left( {b + 1} \right)\Gamma \left( {b - a + \frac{1}{2}} \right)}}{{\Gamma \left( a \right)\Gamma \left( {b + \frac{1}{2}} \right)\Gamma \left( {b - a + 1} \right)}}.\]

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.

 

链接：http://mathoverflow.net/questions/66299/ramanujans-incorrect-formula