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).
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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.