# user6818

 2d awarded Popular Question Dec 9 asked What are the AdS/CFT papers which study the stringy effects in the bulk? Dec 8 comment Does Mathematica know if an integral is convergent or not?Just because $f(x)$ decreases with increasing $x$ is no guarantee that $\int _0 ^\infty f(x) dx$ exists. I don't think such a theorem exists. But here if the $N-integrate$ converges it seems to be integrable for at least the values of $q$ that you are testing but its likely that the two things are unrelated. I have never used the manipulate command - let me see if I can learn about it from documentation. [...or can you write a brief explanation of what those {q,0,5,1,Appearance -> Labvelled} mean? :D...] Dec 8 comment Does Mathematica know if an integral is convergent or not?Okay..I changed that - still there is no answer - Mathematica just displays the stuff back - does it mean that its at least convergent? Dec 8 comment Does Mathematica know if an integral is convergent or not?Can you give an example? I have uniformly used [] for all function arguments Log, Exp, Tanh ,Integrate, Sqrt, Assuming . (...somehow Mathematica is not giving any errors to me!...) Dec 8 comment Does Mathematica know if an integral is convergent or not?I copy pasted my text back into the file and there are no errors - I am on Mathematica 9 BTW Dec 8 comment Does Mathematica know if an integral is convergent or not?Can you see my reformatted pasting of the input? Its not really [pi] its actually "escape p i escape" and so on and so forth Dec 8 revised Does Mathematica know if an integral is convergent or not?added 7 characters in body Dec 8 comment Does Mathematica know if an integral is convergent or not?I am doing here "copy as Input Text" - and Mathematica doesn't give any error messages on these - Integrate [ Tanh[ [Pi] Sqrt[[Lambda]]]* Log[ 1 - Exp[-2 [Pi] 3 Sqrt [ [Lambda] + 1/4 ] ]] , {[Lambda], 0, Infinity }] Dec 8 comment Does Mathematica know if an integral is convergent or not?@Nasser Why do you think so? There is no error message. And this displays correctly on WolframAlpha - (the second integral with q=3) wolframalpha.com/input/?i=integral_0^%E2%88%9E+tanh%28%CF%80+sqrt%28%­CE%BB%29%29+log%281-exp%28-2+%CF%80+3+sqrt%28%CE%BB%2B1%2F4%29%29%29+d%CE%BB&lk=1&a=ClashPrefs_*Math- Dec 8 asked Does Mathematica know if an integral is convergent or not? Dec 8 comment Some identities with the Riemann-Hurwitz zeta function@MustafaSaid May be you can elaborate... Dec 8 comment Some identities with the Riemann-Hurwitz zeta function@user17762 Yes! Sorry for the typo! Dec 8 revised Some identities with the Riemann-Hurwitz zeta functiondeleted 10 characters in body Dec 7 asked Some identities with the Riemann-Hurwitz zeta function Dec 5 awarded Supporter Dec 5 accepted zeta-function regularized integrals Dec 5 comment zeta-function regularized integrals- I guess a similar proof you have given for $\xi(3)$ will also give this above. Dec 5 comment zeta-function regularized integralsThanks for the efforts. Now I remember that there is this general statement that for $Re(q)>0$ one can write, $\xi(q) = \frac{\pi^q}{2^{1-2q}(2^q - 2) \Gamma(q)} \int _0 ^\infty dx \frac{ e^{-2\pi\sqrt{x}} x^{\frac{q}{2} -1 } }{1 + e^{-2\pi\sqrt{x}} }$ - now for $q=3$ this seems to match the first integral equality I wrote down - and now if I power-series expand the denominator of this integrand and integrate term-by-term I get, $\xi(q) = \frac{2^q}{2-2^q}\sum_{s=1}^{\infty} \frac{ (-1)^s }{s^q }$ - which matches your expression for $\xi(3)$ Dec 5 comment zeta-function regularized integralsI am a bit confused about what you are saying - (1) Why is $\xi(3) = - \frac{4}{3} \sum_{k=1}^\infty (-1)^k \frac{1}{k^3}$ ? (...that doesn't naively seem to be the usual definition of the Riemann zeta function...) (2) Are you proving the first integration equality also somehow?