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This is the formula

$$ \zeta(s) = \frac{1}{s-1}+\frac{1}{2} + 2\int_0^{\infty} \frac{\sin{(s \arctan{t})}}{(1+t^2)^{\frac{s}{2}}\left(e^{2\pi t}- 1 \right)}\, dt $$

Can this expression be simplified for any s or we can only use numerical integration like trapezoidal rule to solve this?

pshmath0
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Rohan Asif
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  • For your information, I have tried to evaluate the integral in the above equation but so far no substitution or reduction methods work. Can it be solved by the trick used to evaluate gamma function of 1/2 i.e. root of pi? – Rohan Asif Oct 18 '19 at 09:49
  • What do you mean by "simplified"? Is it a hope to find a simple-yet-unknown representation of $\zeta(s)$? – metamorphy Oct 18 '19 at 12:41
  • @metamorphy i mean to solve the integral and find an expression in terms of s – Rohan Asif Oct 18 '19 at 12:51
  • @metamorphy suppose that s=3 we can substitute in above expression and solve the the integral for s=3 and it would give us a finite answer. Maybe its not possible to get a generalized expression for any s – Rohan Asif Oct 18 '19 at 12:56
  • You want the MSE community to "solve" $\zeta(3)$. Well, wait a bit. – metamorphy Oct 18 '19 at 12:58
  • It was just an example to solve for a particular value of s. All I want is to find some trick to be able to evaluate the integral for any value of s – Rohan Asif Oct 18 '19 at 13:01
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    Since $\zeta(2n)$ and $\zeta(-n)$ have closed form solutions, why not start trying to solve it using $s=2n$ and $s=-n$ for $n\in\mathbb{N}$ ? See here: https://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values – pshmath0 Oct 18 '19 at 13:02
  • As i said, I read about the proof of gamma of 1/2 and how the variables are changed to polar ones to solve and get the value of root of pi, Similarly is there any such trick to be used for this integral – Rohan Asif Oct 18 '19 at 13:02
  • @Pixel Can you send me a link where I can read about the closed forms of these expressions? – Rohan Asif Oct 18 '19 at 13:03
  • @Pixel thanks I guess I have to first learn about bernoulli numbers now!... maths is hard – Rohan Asif Oct 18 '19 at 13:07
  • @RohanAsif The Bernoulli numbers are a sequence of rational numbers generated by the exponential generating function $$f(x)=\sum_{n=0}^\infty\frac{B_n}{n!}x^n=\frac{x}{e^x-1}.$$ Maclaurin series expansion of the RHS, e.g., will enable you to obtain the Bernoulli numbers, but you don't need to as they are already known. PS maths is fun :-D – pshmath0 Oct 18 '19 at 13:07
  • @Pixel thanks I'll read up on it and get to this post again – Rohan Asif Oct 18 '19 at 13:09

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