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$\phi=\dfrac{1+\sqrt5}{2}$, the golden ratio.

$\zeta(s)=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s}$ is the Zeta function valid for Re(s)>1

Where $ n \ge 1$

$$\int_0^\infty\sum_{n=1}^{\infty}\frac{1}{(n^s+x^\phi)^{\phi}}dx=\zeta(s)$$

My dear friend claims that this is true and I told him that here a maths site that can help answer these sort of questions. Is my friend joking with me? It is his integral above correct? Can anyone please answer it?

Narasimham
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  • Did he only claim this to be true, or did he actually calculate the left side ? – Peter May 03 '16 at 09:12
  • He just make a claim and told me to verify it, left me puzzle for a while –  May 03 '16 at 09:15
  • I checked some values with PARI/GP numerically, and the claim COULD be true. However, I have no idea how to verify it. Because of the good fitness for relatively large $s$ (Say $5$ or $6$), it can be ruled out that your friend just produced a random formula as a joke. It has to be taken serious. – Peter May 03 '16 at 09:18
  • @Peter (+1). So he wasn't joking with me after all. How can zeta function link with golden ratio? –  May 03 '16 at 09:21
  • No idea, but there should be suitable substitutions for the integral ... – Peter May 03 '16 at 09:22
  • Another possibility. He knows another representation for $\zeta(s)$, that has something to do with $\phi$. Look at wikipedia or mathworld, maybe you find something interesting. – Peter May 03 '16 at 09:27
  • I have see $\phi$ with $\pi$ And other but not with zeta. maybe there is another representation with $\phi$ –  May 03 '16 at 09:30
  • @pisquare, See my answer here, there is the golden ratio involved. – Neves May 03 '16 at 09:39
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    You want to prove that $$I(a)=\int_0^\infty \frac1{(a+x^\phi)^\phi}dx=\frac1{a}.$$ The substitution $\large x \mapsto x ,a^{\frac1{\phi}}$ shows that you only need to calculate $I(1)$, which was done in more than one way here: http://math.stackexchange.com/questions/1328423/how-to-compute-int-0-infty-frac11x-varphi-varphi-dx – Noam Shalev - nospoon May 03 '16 at 09:43
  • Bravo @nospoon, it just then take the sum and we arrive at the zeta function –  May 03 '16 at 09:50

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