Return to Question

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{-1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Quine, Heydai and Song 1993 stating $P_{1,b}$ as mine but

$$ P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma \left( 1+\frac{b}{a}\right )}. $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{-1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Quine, Heydari and Song 1993 stating $P_{1,b}$ as mine but

$$ P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma \left( 1+\frac{b}{a}\right )}. \tag{18} $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Zeta Regularized Product by Quine, Heydai & Song in AMS 338, 1 jul 93 stating $P_{1,b}$ as mine but

$$

P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{-1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Quine, Heydai and Song 1993 stating $P_{1,b}$ as mine but

$$ P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma \left( 1+\frac{b}{a}\right )}. $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Zeta Regularized Product stating $P_{1,b}$ as mine but

$$

P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?

I was trying to compute the product

$$ P_{a,b} = \prod_{n=1}^\infty(an + b), $$

after I computed

$$ P_{1,b} = \prod_{n=1}^\infty(n + b) = \frac{\sqrt{2\pi}}{\Gamma(b+1)}, $$

and the well-known

$$ \prod_{n=1}^\infty a = \exp\left\{\log(a)\sum_{n=1}^\infty n^0 \right\} = \exp\left\{\log(a)\zeta(0) \right\} = a^{-1/2}. $$

So I have

$$ P_{a,b} = \prod_{n=1}^\infty a \prod_{n=1}^\infty\left(n + \frac{b}{a} \right) = a^{1/2}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

However I found this article Zeta Regularized Product by Quine, Heydai & Song in AMS 338, 1 jul 93 stating $P_{1,b}$ as mine but

$$

P_{a,b} = a^{-1/2 - b/a}\frac{\sqrt{2\pi}}{\Gamma\left(1+\frac{b}{a}\right)}. $$

Of course this formula is not compatible with product of infinite products, but it seems to work rather than mine when computing some partition function by path integrals as

$$ \int\mathcal{D}[\phi,\phi^\dagger]\exp\left\{-\int_0^\beta\mathrm{d}t\phi^\dagger(t)(\partial_t + w)\phi(t) \right\}, $$

with $\phi,\phi^\dagger$ bosonic fields. Notice that in this case

$$ \phi(t) = \sum_{n=-\infty}^\infty\phi_n e^{\frac{2\pi i}{\beta}n t} $$

so that to evaluation of path integral boils up to some gaussian one.

Can anyone help me?