Integral $\int\limits_{0}^{1} \prod\limits_{k=0}^{\infty}(1-q^{2k})(1+q^{2k+1})(1+q^{2k+3})dq$

Asked Feb 12 '19 at 15:09

Active Feb 12 '19 at 19:51

Viewed 135 times

$$ \int\limits_0^1 \prod\limits_{k=0}^{\infty}\left(1-q^{2k}\right)\left(1+q^{2k+1}\right)\left(1+q^{2k+3}\right)dq = \frac{\pi}{\sqrt{a}} \left(\frac{e^{\pi a^{5/2}/ b}-c} {e^{\pi a^{5/2}/ b}+e^{\pi a^{3/2}/ b}+c}\right) $$

I want to find the value of $a,b,c$

Sorry I have nothing much to show in effort. Thanks for any hint.

edited Feb 12 '19 at 19:51

rtybase

16,907

asked Feb 12 '19 at 15:09

Shweta Aggrawal

5,501
2
15
48

are all exponential terms on the RHS the same by intent or is it a typo? – gt6989b Feb 12 '19 at 15:25
@gt6989b You are right. I have edited the post. – Shweta Aggrawal Feb 12 '19 at 15:29
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Possible useful: https://math.stackexchange.com/q/3040367/515527 and https://math.stackexchange.com/q/1883140/515527 – Zacky Feb 12 '19 at 15:30
1

Even more useful: http://mathworld.wolfram.com/JacobiTripleProduct.html – Zacky Feb 12 '19 at 15:43
@Zacky Thanks.. – Shweta Aggrawal Feb 12 '19 at 15:44
I think I saw it somewhere on brilliant.org I guess – Rohan Shinde Feb 12 '19 at 17:07
Indeed @Digamma, this is an open problem, with a quick search on brilliant I found it. See here: https://brilliant.org/problems/a-q-triple-product-integral-2/?ref_id=1561212 – Zacky Feb 12 '19 at 17:11
1

A fundamental problem: the product on the LHS is equal to $0$ hence for $k=0$ the we got that $(1-q^0)(1+q^1)(1+q^3)=(1-1)(1+q)(1+q^3)=0$. – mrtaurho Feb 12 '19 at 17:46

Integral $\int\limits_{0}^{1} \prod\limits_{k=0}^{\infty}(1-q^{2k})(1+q^{2k+1})(1+q^{2k+3})dq$

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