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I am looking for help with doing the following integral : $\frac{1}{2\pi i}\int_{1}^{\infty}\ln\left(\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}} \right )\frac{dx}{x\left(\ln x+z\right)}\;\;\;\;z\in \mathbb{C}$ i tried to transform it into a complex integral along a 'keyhole contour', with a branch cut along the +ive real line $\left[1,\infty\right)$. but then $\;\ln x\;$ would be transformed into $\;\ln x+2\pi i\;$ when doing the integral along the segment parallel to and below $\left(\infty,1\right]\;$ which doesn't add up nicely to the portion along the segment parallel to and above $\left[1,\infty\right)$ . any insights are appreciated.

EDIT:

the above integral is equivalent to: $\int_{1}^{\infty}\sum_{n=1}^{\infty}\frac{\sin(2\pi n x)}{n\pi}\left(\frac{1}{x\left(\ln x+z\right)}\right)dx$ And $\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx$ Also, we can prove that it's equivalent to the limit: $e^{-z}\text{Ei}(z)+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left(n+\frac{1}{2} \right )\ln\left(\frac{\ln(n+1)+z}{\ln(n)+z} \right )-e^{-z}\text{Ei}(z+\ln (N+1)) $ or: $e^{-z}\text{Ei}(z)-\frac{3}{2}\ln(z)-\left(\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\ln\left(\ln(n+1)+z \right )+e^{-z}\text{Ei}\left(\ln (N+1) + z\right)-\frac{2N+1}{2}\ln(\ln(N+1)+z)\right)$ EDIT: using the definition of the zeta function - eq. 3 - : $\zeta(s)=\frac{s}{s-1}-s\int_{1}^{\infty}(x-\left \lfloor x \right \rfloor )x^{-s-1}dx\;\;\;\;\;\Re(s)>0$ We have: $\frac{\zeta(s)}{s}+\frac{1}{2s}-\frac{1}{s-1}=\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor\right)x^{-s-1}dx$ And: $\int_{1}^{\infty}\left(\frac{1}{2}-x+\left \lfloor x \right \rfloor \right )\left(\frac{1}{x\left(\ln x+z\right)}\right)dx=\int_{0}^{\infty}\left(\frac{\zeta(s)}{s}+\frac{1}{2s}-\frac{1}{s-1}\right)e^{-zs}ds$

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    another way to think of it, is to take the Taylor expansion of the log: $$\frac{1}{2\pi i }\left(\ln(1-e^{-2\pi i x})-\ln(1-e^{2\pi i x})\right)=\frac{1}{2\pi i}\sum_{k=1}^{\infty}\frac{e^{2\pi i kx}-e^{-2\pi i kx}}{k}=\sum_{k=1}^{\infty}\frac{\sin(2\pi kx)}{k}$$. which in turn is the Fourier expansion of $\frac{1}{2}-\left \{ x \right \}$2012-12-13

2 Answers 2

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Let's consider your integral : $\tag{1}I(z)=\int_1^\infty \frac{\frac 12-x+\lfloor x \rfloor}{x\left(\ln x+z\right)}\,dx$

Your derivation at the end is right (for $\Re(s)>0$) so let's reproduce it here : $\tag{2}\zeta(s)=\frac s{s-1}-s\int_1^\infty(x-\lfloor x \rfloor)\,x^{-s-1}dx$ so that : $\tag{3}\frac{\zeta(s)}s+\frac 1{2s}-\frac 1{s-1}=\int_1^\infty\left(\frac 12-x+\lfloor x \rfloor\right)\,x^{-s-1}\,dx$ At this point you used (for $x>1\;$ and $\;\Re(z)>0$) the integral : $-\frac{(x e^z)^{-s}}{x\left(\ln(x e^z)\right)}=\int \frac{(x e^z)^{-s}}x\,ds$ between the limits $\,s=0$ and $+\infty\ $ to get : $\tag{4}I(z)=\int_1^\infty\frac{\frac 12-x+\lfloor x \rfloor}{x\left(\ln x+z\right)}\,dx=\int_0^\infty\left(\frac{\zeta(s)}s+\frac 1{2s}-\frac 1{s-1}\right)\,e^{-zs}ds$ (you should and still can provide this as an answer...)

This integral is convergent (for $\ \Re(z)>0\ $ I think) and $(4)$ may be integrated by parts to get further for $\displaystyle f(s):=\frac{\zeta(s)}s+\frac 1{2s}-\frac 1{s-1}$ (this applies to general Laplace transforms as well) : $I(z)=\int_0^\infty f(s)\,e^{-zs}ds=\left[\frac{f(s)e^{-zs}}{-z}\right]_0^\infty +\int_0^\infty f'(s)\frac{e^{-zs}}z\,ds$ $\tag{5}I(z)=\frac{\lim_{s\to 0}f(s)}z +\frac 1z \int_0^\infty f'(s)\,e^{-zs}\,ds$ We may continue this game to get an asymptotic expansion : $I(z)\sim\frac{f(0^+)}z+\frac{f'(0^+)}{z^2}+\frac{f''(0^+)}{z^3}+\cdots$ The limit at $f(0^+)$ as well as the derivatives $f^{(n)}(0^+)$ were evaluated in your 'regularization of a divergent integral' thread where we got $\ K_n:=f^{(n)}(0^+)=\frac{n!+\zeta^{(n)}(0)}n$. The asymptotic expansion will simply be :

$\tag{6}I(z)\sim\sum_{n>0} \frac{K_n}{z^n}\sim \sum_{n>0} \frac{n!+\zeta^{(n)}(0)}{n\,z^n}$ This has of course the drawback of the divergence problems we emphasized here (with a table of values of $\,n\,K_n$) and implies that the sum should be interrupted for $n$ near $20$ (or $30$).

Let's add that the $K_n$ divergence problem appears 'regularized' here since $I(1)$ for example has a well defined and finite value.

We may too split the initial integral in integrals $\int_{n-\frac 12}^{n+\frac 12}\cdots dx$ and rewrite $I(z)$ as an integral $\int_0^{\frac 12}$ over a series (see the script at the bottom) to get some accurate numerical evaluations (all the methods return the same numerical results !) :

\begin{array} {cl} z&I(z)\\ \hline 10^{-15} &16.3096530675478979183202495497193802845495251563\\ 10^{-10} &10.5531903374319358641649617819163355585942889453\\ 10^{-4} &3.64642265723654724788211497515174434549406723370\\ 10^{-3} &2.50171392362613726727194292910979673196605938961\\ 10^{-2} & 1.39543999432476638672924784183965200139077802817\\ 10^{-1} &0.486868432497953417060515317199037503343873648836\\ 1/8 &0.422105200530486066954436439320203842530811825240\\ 1/4 &0.256624396604535894012861595020444840056178229095\\ 1/2 &0.115112905757527639323106492399718643046468630496\\ 1 &0.0769186991412752978246055672366920464084421892313\\ 2 &0.0395820464350666838193632888947748021440219942919\\ 3 &0.0266177747964158089008118870276107101328689178125\\ 4 &0.0200449423033013057537716317592213987247899100373\\ 10 &0.00807286582120089521388366887280155614313771306934\\ 100 &0.000810295282006106142951971110178216745088801945013\\ 1000 &0.00008105828699770816796829446558\\ \end{array}

The digits at the end correspond to the first digits of $\ 1-\log(2\pi)/2=0.081061466795\cdots\ $ as they should.

Hoping this helped or entertained a little,

// pagi/gp script for numerical evaluation of the initial integral f(z,x)=sumpos(n=2,1/((n+x)*(log(n+x)+z))-1/((n-x)*(log(n-x)+z))) I(z)=intnum(x=0,1/2,(1/2-x)*f(z,x))+intnum(x=1,3/2,(1/2-x+floor(x))/(x*(log(x)+z))) 
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If we just look at the integral, we get that integral (ln((1-e^(-2 i (pi x)))/(1-e^(2 i (pi x)))))/(x (ln(x)+z))dx = (2ipix+ln(-e^(-2ipix)))ln(ln(x)+z)-2ipie^(-z)Ei(z+ln(x))+constant where Ei is the exponential integral. Because of your upper limit, the integral will not converge regardless of z so long as it is an element of C