I am interested in the integral over the log-zeta function$$I=\int_0^1 \ln\left(\zeta(s)\right)ds$$  My attempts:  Sub $u=1-s$ to get $$I=\int_0^1 \ln\left(\zeta(1-u)\right)du=\int_0^1 \ln(\zeta(u))-u\ln2-(u-1)\ln\pi-\ln\left(\sin(\frac{\pi u}{2})\right)-\ln\left(\Gamma(1-u)\right) du$$ This doesn't seem to lead anywhere unfortunately. I also tried to set up  a parameter $a$ as  $$I(a)=\int_0^1 \ln\left(\zeta(s+a)\right)ds$$ $$I'(a)=\int_0^1 \frac{\zeta'(s+a)}{\zeta(s+a)}ds=\ln\left(\zeta(s+a)\right)|_{s=0}^{s=1}=\ln\left(\zeta(a+1)\right)-\ln\left(\zeta(a)\right)$$ From this point on I am not sure of what to do as integrating with respect to $a$ to recover $I(a)$ does not simplify the integral. If anybody has any idea of how to approach finding a closed form of this integral then please let me know. Thank you in advance.
121678
With the help of  I know good approximations of the real part of the integral $$\int_0^1 \log \left(\zeta(x)\right)dx,$$ where $\zeta(x)$ denotes the Riemann zeta function (evaluated over real numbers).  See it as int log(zeta(x))dx, from x=0 to 1 or 15 digits of int log(zeta(x))dx, from x=0 to 1     Question. I'm curious to know if this constant $$\Re\int_0^1 \log \left(\zeta(x)\right)dx$$ is in the literature. Can you provide me the reference? Then answer this question as a reference request adding what is the article showing/defining the value of this real constant, and I am going to search and read such statement. Many thanks.   My problem here is that I can not find the numeric value of this constant in  The On-Line Encyclopedia of Integer Sequences or in Internet.
105432
Is there a way I can derive the value of the integral  $ \int_0^1 \ln(x)\ln(1-x)dx$  using the fact that  $\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  ? (the actual value of the integral is $2-\frac{\pi^2}{6}$)  Thanks in advance
77909
I want to show that we can continue Riemann's zeta function to Re$(s)&gt;0$, $s\neq 1$ by the following formula \begin{align} (1-2^{1-s})\zeta(s)&amp;=\left(1-2\frac{1}{2^s}\right)\left(\frac1{1^s}+\frac1{2^s}+\ldots \right) \\ &amp;=\frac1{1^s}+\frac1{2^s}+\ldots -2\left(\frac1{2^s}+\frac1{4^s}+\ldots \right)\\ &amp;=\frac1{1^s}-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\ldots  \\ &amp;=\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}. \end{align} In order to do that, I need to show that the series converges for Re$(s)&gt;0$, except $s=\frac{2k\pi i}{\ln 2}+1$, $k\in \mathbb{Z}$, which are removable singularities. Any ideas on how to do that?
68586
Does anybody know how to prove this series?  $$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$  I arrived at this through Mathematica.  I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.
92005
I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate:  $$\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $$
73390
A few days ago, I posted   Prove that \begin{equation} \int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt] -\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3) \end{equation}  and the OP receives some good answers even I then could answer it.    My next question is finding the closed-forms for     \begin{align} \int_0^{\pi/4}\ln^2(\sin x)\,dx\tag1\\[20pt] \int_0^{\pi/4}\ln^2(\cos x)\,dx\tag2\\[20pt] \int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt\tag3 \end{align}   I have a strong feeling that the closed-forms exist because we have nice closed-forms for \begin{equation} \int_0^{\pi/4}\ln(\sin x)\ dx=-\frac12\left(C+\frac\pi2\ln2\right)\\ \text{and}\\ \int_0^{\pi/4}\ln(\cos x)\ dx=\frac12\left(C-\frac\pi2\ln2\right). \end{equation} The complete proofs can be found .  As shown by Mr. Lucian in his answer below, the three integrals are closely related, so finding the closed-form one of them will also find the other closed-forms. How to find the closed-forms of the integrals? Could anyone here please help me to find the closed-form, only one of them, preferably with elementary ways (high school methods)? If possible, please avoiding contour integration and double summation. Any help would be greatly appreciated. Thank you.
115828
$$ \displaystyle \int_{0}^{\infty}{\frac{{\log}^{2}(1-{e}^{-x}){x}^{5}}{{e}^{x}-1} dx} $$ What I have done - $ \displaystyle I(k) =  \int_{0}^{\infty}{\frac{{x}^{5}}{{e}^{x}{(1-{e}^{-x})}^{k}}}$ And then, $ I''(1) $ will be our integral. Now, we need to compute $ I(k) $ $ \displaystyle I(k) = \int_{0}^{\infty}{\sum_{m=0}^{\infty}{\binom{k+m-1}{m} {e}^{-(m+1)x}{x}^{5}}} $ $ \displaystyle I(k) = \sum_{m=0}^{\infty}{\binom{k+m-1}{m} \frac{\Gamma(6)}{{(m+1)}^{6}}} $ Now, I computed $ \displaystyle I''(1) = 120 \left(\sum_{m=0}^{\infty}{\frac{{({H}_{m}^{(1)})}^{2} - {H}_{m}^{(2)}}{{(m+1)}^{6}}}\right)$ Or, in style, $ \displaystyle 120\left({s}_{h}(2,6) - {\sigma}_{h}(2,6)\right) $ I want to ask if the above computation is correct or not. EDIT I am updating the answer now. We can go by solving Euler sums from formula  as was suggested by   and get the form as given by Marco(watch out his answer). Then, his form can be expanded by hand as - $ 14\zeta(8) - 5\zeta(6)\zeta(2) - 6\zeta(5)\zeta(3) - 3{\zeta(4)}^{2} + \zeta(4){\zeta(2)}^{2} + {\zeta(3)}^{2}\zeta(2) $ which on simplification gives a really nice form - $$ 20{\pi}^{2}{\zeta(3)}^{2} - 720\zeta(3)\zeta(5) + 61{\pi}^{2}\zeta(6) $$
6458
/A problem from the 2012 MIT Integration Bee is $$ \int_0^1\frac{x^7-1}{\log(x)}\mathrm dx $$ The answer is $\log(8)$. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand?
38053
Let $n \in \mathbb{N}$. We know that:  $$\int_0^1 x^n \log(1-x) \, {\rm d}x = - \frac{\mathcal{H}_{n+1}}{n+1}$$  Now, let $m , n \in \mathbb{N}$. What can we say about the integral  $$\int_0^1 x^n \log^m (1-x) \, {\rm d}x$$  For starters we know that $\displaystyle \log^m (1-x)=m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k$ where $s(k, m)$ are the .   Thus  \begin{align*} \int_{0}^{1} x^n \log^m (1-x) \, {\rm d}x &amp;=m! \int_{0}^{1}x^n \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k  \\   &amp;= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \int_{0}^{1}x^{n+m} \, {\rm d}x\\   &amp;= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \frac{1}{m+n+1} \end{align*}  Can we simplify? I know that Striling numbers are related to the Harmonic number but I don't remember all identities.
49414
I'm looking for a reference for (or an elementary proof of) $$ \lim_{s \rightarrow 1} \left( \zeta(s) - \frac{1}{s-1} \right) = \gamma$$ Thanks for your help.
12916
How do I compute      $$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$   What I am doing is take   $$f(z)=\frac{(\log z)^2}{1+z^2}$$  and calculating   $\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$ which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$  Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$  Any help will be appreciated. thank you in advanced.
47607
I'm trying to find $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx$ where $\alpha&gt;0$ is real.  My approach was to take an integral along the real line from $1/R$ to $R$, around the circle counterclockwise to $-R$, along the real line to $-1/R$, and then around the circle clockwise to $1/R$.  I have encountered 2 problems with this:   This path encloses one pole, at $z=\alpha i$.  I found the residue at $z=\alpha i$ to be $\frac{\ln(\alpha)+i\pi/2}{2\alpha i}$.  However, this gives me that $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}dx=\frac{\pi(\ln(\alpha)+i\pi/2)}{2\alpha}$.  Since I have a real function integrated over the real line, there cannot be an imaginary part.  Where did I go wrong?  (Also, doing a few examples, the correct answer seems to be $\frac{\pi\ln(\alpha)}{2\alpha}$, the same as I have but without the imaginary part.) At first chose my path so instead of going all the way around the upper semicircle, it only went 3/4 of the way around, as I wanted to avoid anything that might go wrong with the discontinuity of $\log(x)$ at the negative real axis.  When I do this, though, I get a different answer than before (the denominator of the fraction is $\alpha(1-e^{3\pi i/4})$ instead of $2\alpha$.  What am I doing wrong that gives me different answers?
85817
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
86374
How to prove the following   $$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$  I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.  Any other ideas ?
116381
Prove that:  $\displaystyle\int_{\pi/4}^{\pi/2} \ln (\ln(\tan x))dx =\frac{\pi}{2}\ln \left( \frac{\sqrt{2\pi} \Gamma \left(\dfrac{3}{4} \right)}{\Gamma \left(\dfrac{1}{4} \right)}\right)$  I know that the Vardi's Integral can be evaluated in terms of derivatives of Hurwitz Zeta Function. I would like to see a method which uses differentiation under the integral sign.
102100
Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution )  1)$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} $$  2)$$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} $$  3) $$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$  and lastly,  4) $$\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\!$$  I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used.  $\gamma$ is the Euler-Mascheroni constant and $\zeta(s)$ is the Riemannian Zeta function.  Thanks in advance.
91157
Question:   Compute $$\int _0 ^{2\pi} \left(\sum_{n=1}^\infty \frac {\cos(nx)}{2^n}\right)^2 dx$$   Thoughts: Tried interchanging the integral and the sum, but then the integral turned out to be zero...