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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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Well $\ \dfrac {1-e^{-2i\pi x}}{1-e^{2i\pi x}}=-e^{-2i\pi x}=e^{-2i\pi (x+1/2)}\ $ so that the initial $\ \ln\ $ term should return $-2i\pi (x+1/2 +k)$ (with $k$ chosen in $\mathbb{Z}$). The asymptotic form of the integrand will be $-\frac x{x\ln(x)}\sim -\frac 1{\ln(x)}$ so that I fear that the integral will be divergent (from the $+\infty$ limit). –  Raymond Manzoni Dec 12 '12 at 15:25
    
Let's add that the indefinite integral exists. In the case $k=-1$ for example we get $\frac 12\ln(\ln(x)+z)-e^{-z}\operatorname{Ei}(\ln(x)+z)+C$. Obtained by Alpha. You may rewrite $\operatorname{Ei}(\ln(x)+z)=\operatorname{li}(x\,e^z)$. –  Raymond Manzoni Dec 12 '12 at 15:47
    
@RaymondManzoni .. i think it's more accurate to say : $$\frac{1-e^{-2\pi i x}}{1-e^{2\pi i x}}=-e^{-2\pi i x}=e^{-2\pi i(x-1/2)}= e^{-2\pi i \left(\left \{ x \right \}-1/2\right)}$$. i'm afraid i don't understand your second comment !! –  Mohammad Al Jamal Dec 12 '12 at 16:18
    
Your equality is true but when you'll take the logarithm you'll introduce a discontinuity at every integer value. This means that you are not considering the continuous version of the logarithm but the principal branch (that's done by calculators and computer software as in : log(exp(2*Pi*I*0.500))= 0 + 3.1416*I, log(exp(2*Pi*I*0.501))= 2.3283 E-10 - 3.1353*I nearly. Note the $-2\pi i$ jump for $x=0.5, 1.5,\cdots$). You have the right to chose your variant I merely pointed out a more usual one. In my other comment I computed $-\int \frac{x-1/2}{x(\ln(x)+z)}dx$ and not $\int_1^\infty$. –  Raymond Manzoni Dec 12 '12 at 17:12
    
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 $$\frac{1}{2}-\left \{ x \right \}$$ –  Mohammad Al Jamal Dec 13 '12 at 14:00
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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

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