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DreamWeaver
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Initially, the purpose of this tutorial will be to explore and evaluate various lower order derivatives of the Hurwitz Zeta function. In each case, the Hurwitz Zeta function will be differentiated with respect to its first parameter. A little later on - although this will take some time! - these derivatives will be integrated with respect to the second parameter. In yet further, later cases, the integrand will contain an additional function, such as a power of the variable. For example, on the understanding that
\(\displaystyle \zeta' (s,x) = \frac{d}{ds} \zeta(s,x)\)
\(\displaystyle \zeta'' (s,x) = \frac{d^2}{ds^2} \zeta(s,x)\)
etc.
We will evaluate definite, parametric integrals of the form
\(\displaystyle \int_0^z x^n\, \zeta' (s,x)\, dx\)
\(\displaystyle \int_0^z x^n\, \zeta'' (s,x)\, dx\)
and so on. --------------------------
Definitions and notation:
--------------------------The Hurwitz Zeta function:
\(\displaystyle \zeta(s,x) = \sum_{k=0}^{\infty} \frac{1}{(k+x)^s}\)
CAUTION: some authors transpose the order of the two parameters in the Hurwitz Zeta function. And, even worse, some do so without explicitly defining it by an unambiguous series like that given above, so it's worth bearing in mind if searching on-line for additional reading material.
For \(\displaystyle \mathscr{Re}(s) < 0\,\) and \(\displaystyle 0 \le x \le 1\, \) the Hurwitz Zeta function has the following Fourier Series representation:
\(\displaystyle \zeta(s,x) = \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] \)The vast majority of what follows will - directly or indirectly - be consequences of this Fourier Series.
The Polygamma functions:
\(\displaystyle \psi_0(x) = \frac{d}{dx}\log \Gamma(x) = -\gamma +\, \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+x-1} \right)\)
\(\displaystyle \psi_{m\ge 1}(x) = \frac{d^{m+1}}{dx^{m+1}} \log \Gamma (x) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+x)^{m+1}}\)
Where \(\displaystyle \gamma\, \)is the Euler-Mascheroni constant. Note that, comparing that
last expression for higher order Polygamma functions with the series definition for the Hurwitz Zeta function, we also have
\(\displaystyle \psi_{m\ge 1}(x) = (-1)^{m+1}m!\, \zeta(m+1,x)\)
The main property of the Polygamma functions used below will be the incrementation formula for the Digamma function
\(\displaystyle \psi_0(1+x) = \psi_0(x) + \frac{1}{x}\)
and the higher order analogues obtained from direct differentiation of this result, eg.
\(\displaystyle \psi_1(1+x) = \psi_1(x) - \frac{1}{x^2}\)
\(\displaystyle \psi_2(1+x) = \psi_2(x) + \frac{2}{x^3}\)
\(\displaystyle \psi_3(1+x) = \psi_3(x) - \frac{6}{x^4}\)
etc. These results are easily extended. Let \(\displaystyle m \in \mathbb{Z} \ge 0\,\) and \(\displaystyle n \in \mathbb{Z} \ge 1\,\) then
\(\displaystyle \psi_{m}(x+n) = \psi_m(x) + (-1)^mm! \, \sum_{j=0}^{n-1} \frac{1}{(x+j)^{m+1}}
\)These results can be simplified further, by using the Harmonic Numbers. The Harmonic numbers are simply finite, truncated versions of the (regular) Riemann Zeta function:
\(\displaystyle \zeta (s) = \zeta(s,1) = \sum_{k=1}^{\infty}\frac{1}{k^s}\)
Notation for the Harmonic numbers varies somewhat, from one author to the next, so I'll use what seems to be the most prevalent form, namely
\(\displaystyle H_n^{(m)} = \sum_{j=0}^{n-1}\frac{1}{(j+1)^m} =
\)
\(\displaystyle 1+ \frac{1}{2^m} + \frac{1}{3^m} + \frac{1}{4^m} +\, \cdots \, + \frac{1}{n^m}\)In addition, for reasons that will become clear shortly, the term
\(\displaystyle \log 2\pi + \gamma - H_{n}^{(1)} \)
will appear in our results with (almost) alarming frequency. That being the case, once the appearance of such a term is apparent - in any given calculation - I won't hesitate to save space by using the short-hand notation
\(\displaystyle \mathscr{H}_n = \log 2\pi + \gamma - H_{n}^{(1)}
\)
If we set \(\displaystyle x=1\,\) in that last Polygamma formula, then - in terms of the Harmonic numbers - we get:
\(\displaystyle \psi_{m}(n+1) = \psi_m(1) + (-1)^mm! \, \sum_{j=0}^{n-1} \frac{1}{(j+1)^{m+1}} = \)
\(\displaystyle \psi_m(1) + (-1)^mm! \, H_n^{(m+1)}
\)The Generalized Clausen Functions:
Apologies, but notation is one of those things that tends to be - or in my case often is - quite fluid until you settle on a preferred form. Consequently, I have used slight variations of the following notations - on here - before. I hope this won't cause any confusion.
Let
\(\displaystyle \mathcal{C}_m(n, \theta) = \lim_{s \to m^{+}} (-1)^n \frac{d^n}{ds^n}\, \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^s}\)
\(\displaystyle \mathcal{S}_m(n, \theta) = \lim_{s \to m^{+}} (-1)^n \frac{d^n}{ds^n}\, \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^s}\)Explicitly, when \(\displaystyle m \in \mathbb{Z} \ge 1\,\) and \(\displaystyle n \in \mathbb{Z} \ge 0\,\) \(\displaystyle \mathcal{C}_m(n, \theta) = \sum_{k=1}^{\infty} \frac{(\log k)^n}{k^m} \cos k\theta\)
\(\displaystyle \mathcal{S}_m(n, \theta) = \sum_{k=1}^{\infty} \frac{(\log k)^n}{k^m} \sin k\theta\)Much like the Modified Harmonic Numbers \(\displaystyle \mathscr{H}_n\, \)above, certain variations of these Generalized Clausen functions will occur with such frequency as to deserve their own short-hand. With the understanding that \(\displaystyle 0 \le z \le 1 \in \mathbb{R}\, \)in the general sense, but the cases we will deal with are 'small' vulgar fractions of the form \(\displaystyle z=p/q \in \mathbb{Q}\,\), then I will use the following short-hands:
\(\displaystyle \mathcal{C}_j(z)= \mathcal{C}_j(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^j} \cos 2\pi kz\)
\(\displaystyle \mathcal{S}_j(z)= \mathcal{S}_j(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^j} \sin 2\pi kz\)NOTE: I will, occasionally, also write
\(\displaystyle \mathcal{C}_{1-s}(z)= \mathcal{C}_{1-s}(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \cos 2\pi kz\)
\(\displaystyle \mathcal{S}_{1-s}(z)= \mathcal{S}_{1-s}(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \sin 2\pi kz\)
Where \(\displaystyle s\, \) is not necessarily an integer, and the understanding is that \(\displaystyle \mathscr{Re}(s) <0\, \).-----------------
Proposition 1.0:
-----------------
The first derivative of the Hurwitz Zeta function - with respect to its first parameter - is given by: \(\displaystyle \zeta'(s,x) = \)\(\displaystyle \frac{\pi\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] + \)
\(\displaystyle \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \sin \frac{\pi s}{2}\, \mathcal{C}_{1-s}(x) + \cos \frac{\pi s}{2}\, \mathcal{S}_{1-s}(x) \right] + \)
\(\displaystyle \Bigg[\log 2\pi - \psi_0(1-s) \Bigg]\, \zeta(s,x)\)Proof:
This follows directly by differentiating the Fourier Series expansion of the Hurwitz Zeta function. Personally, I'd break it down into the product of two functions, and then differentiate their product. The two functions are given in large brackets here:\(\displaystyle \zeta'(s,x) = \frac{d}{ds}\, \Bigg[ \mathscr{F}(s) \Bigg]\, \Bigg[ \mathscr{G}(s) \Bigg]=\)\(\displaystyle \frac{d}{ds}\, \left[ \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \right]\, \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right]\)The derivative of the 'first' function is
\(\displaystyle \frac{d}{ds}\, \mathscr{F}(s) = \frac{d}{ds}\, \left[ \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \right] = \mathscr{F}'(s) = \)
\(\displaystyle 2\, \frac{ (2\pi)^{1-s} [-\Gamma'(1-s)] - \Gamma(1-s)\, [-(2\pi)^{1-s}\log 2\pi] }{ [(2\pi)^{1-s}]^2 } = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{ \Gamma'(1-s) }{ \Gamma(1-s)} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{ \Gamma'(x) }{ \Gamma(x)} \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{d}{dx} \log \Gamma(x) \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \psi_0(x) \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \Bigg[ \log 2\pi - \psi_0(1-s) \Bigg] \)The derivative of the 'second' function is given by:\(\displaystyle \frac{d}{ds}\, \mathscr{G}(s) = \frac{d}{ds}\, \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] = \mathscr{G}'(s) =\)\(\displaystyle \frac{\pi}{2} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}} \right] +\)\(\displaystyle \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \cos 2\pi kx + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \sin 2\pi kx \right] = \)\(\displaystyle \frac{\pi}{2} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}} \right] +\)
\(\displaystyle \left[ \sin \frac{\pi s}{2}\, \mathcal{C}_{1-s}(x) + \cos \frac{\pi s}{2}\, \mathcal{S}_{1-s}(x) \right]\)Hence, by the rule for differentiation of a product of functions,\(\displaystyle \zeta'(s,x) = \frac{d}{ds}\, \mathscr{F}(s)\, \mathscr{G}(s) = \mathscr{F}'(s)\, \mathscr{G}(s) + \mathscr{F}(s)\, \mathscr{G}'(s)\)Evaluation of the above by brute force yields Proposition 1. \(\displaystyle \, \Box \)
http://mathhelpboards.com/commentary-threads-53/commentary-quot-derivatives-integrals-hurwitz-zeta-function-quot-10777.html
\(\displaystyle \zeta' (s,x) = \frac{d}{ds} \zeta(s,x)\)
\(\displaystyle \zeta'' (s,x) = \frac{d^2}{ds^2} \zeta(s,x)\)
etc.
We will evaluate definite, parametric integrals of the form
\(\displaystyle \int_0^z x^n\, \zeta' (s,x)\, dx\)
\(\displaystyle \int_0^z x^n\, \zeta'' (s,x)\, dx\)
and so on. --------------------------
Definitions and notation:
--------------------------The Hurwitz Zeta function:
\(\displaystyle \zeta(s,x) = \sum_{k=0}^{\infty} \frac{1}{(k+x)^s}\)
CAUTION: some authors transpose the order of the two parameters in the Hurwitz Zeta function. And, even worse, some do so without explicitly defining it by an unambiguous series like that given above, so it's worth bearing in mind if searching on-line for additional reading material.
For \(\displaystyle \mathscr{Re}(s) < 0\,\) and \(\displaystyle 0 \le x \le 1\, \) the Hurwitz Zeta function has the following Fourier Series representation:
\(\displaystyle \zeta(s,x) = \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] \)The vast majority of what follows will - directly or indirectly - be consequences of this Fourier Series.
The Polygamma functions:
\(\displaystyle \psi_0(x) = \frac{d}{dx}\log \Gamma(x) = -\gamma +\, \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+x-1} \right)\)
\(\displaystyle \psi_{m\ge 1}(x) = \frac{d^{m+1}}{dx^{m+1}} \log \Gamma (x) = (-1)^{m+1}m!\, \sum_{k=0}^{\infty} \frac{1}{(k+x)^{m+1}}\)
Where \(\displaystyle \gamma\, \)is the Euler-Mascheroni constant. Note that, comparing that
last expression for higher order Polygamma functions with the series definition for the Hurwitz Zeta function, we also have
\(\displaystyle \psi_{m\ge 1}(x) = (-1)^{m+1}m!\, \zeta(m+1,x)\)
The main property of the Polygamma functions used below will be the incrementation formula for the Digamma function
\(\displaystyle \psi_0(1+x) = \psi_0(x) + \frac{1}{x}\)
and the higher order analogues obtained from direct differentiation of this result, eg.
\(\displaystyle \psi_1(1+x) = \psi_1(x) - \frac{1}{x^2}\)
\(\displaystyle \psi_2(1+x) = \psi_2(x) + \frac{2}{x^3}\)
\(\displaystyle \psi_3(1+x) = \psi_3(x) - \frac{6}{x^4}\)
etc. These results are easily extended. Let \(\displaystyle m \in \mathbb{Z} \ge 0\,\) and \(\displaystyle n \in \mathbb{Z} \ge 1\,\) then
\(\displaystyle \psi_{m}(x+n) = \psi_m(x) + (-1)^mm! \, \sum_{j=0}^{n-1} \frac{1}{(x+j)^{m+1}}
\)These results can be simplified further, by using the Harmonic Numbers. The Harmonic numbers are simply finite, truncated versions of the (regular) Riemann Zeta function:
\(\displaystyle \zeta (s) = \zeta(s,1) = \sum_{k=1}^{\infty}\frac{1}{k^s}\)
Notation for the Harmonic numbers varies somewhat, from one author to the next, so I'll use what seems to be the most prevalent form, namely
\(\displaystyle H_n^{(m)} = \sum_{j=0}^{n-1}\frac{1}{(j+1)^m} =
\)
\(\displaystyle 1+ \frac{1}{2^m} + \frac{1}{3^m} + \frac{1}{4^m} +\, \cdots \, + \frac{1}{n^m}\)In addition, for reasons that will become clear shortly, the term
\(\displaystyle \log 2\pi + \gamma - H_{n}^{(1)} \)
will appear in our results with (almost) alarming frequency. That being the case, once the appearance of such a term is apparent - in any given calculation - I won't hesitate to save space by using the short-hand notation
\(\displaystyle \mathscr{H}_n = \log 2\pi + \gamma - H_{n}^{(1)}
\)
If we set \(\displaystyle x=1\,\) in that last Polygamma formula, then - in terms of the Harmonic numbers - we get:
\(\displaystyle \psi_{m}(n+1) = \psi_m(1) + (-1)^mm! \, \sum_{j=0}^{n-1} \frac{1}{(j+1)^{m+1}} = \)
\(\displaystyle \psi_m(1) + (-1)^mm! \, H_n^{(m+1)}
\)The Generalized Clausen Functions:
Apologies, but notation is one of those things that tends to be - or in my case often is - quite fluid until you settle on a preferred form. Consequently, I have used slight variations of the following notations - on here - before. I hope this won't cause any confusion.
Let
\(\displaystyle \mathcal{C}_m(n, \theta) = \lim_{s \to m^{+}} (-1)^n \frac{d^n}{ds^n}\, \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^s}\)
\(\displaystyle \mathcal{S}_m(n, \theta) = \lim_{s \to m^{+}} (-1)^n \frac{d^n}{ds^n}\, \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^s}\)Explicitly, when \(\displaystyle m \in \mathbb{Z} \ge 1\,\) and \(\displaystyle n \in \mathbb{Z} \ge 0\,\) \(\displaystyle \mathcal{C}_m(n, \theta) = \sum_{k=1}^{\infty} \frac{(\log k)^n}{k^m} \cos k\theta\)
\(\displaystyle \mathcal{S}_m(n, \theta) = \sum_{k=1}^{\infty} \frac{(\log k)^n}{k^m} \sin k\theta\)Much like the Modified Harmonic Numbers \(\displaystyle \mathscr{H}_n\, \)above, certain variations of these Generalized Clausen functions will occur with such frequency as to deserve their own short-hand. With the understanding that \(\displaystyle 0 \le z \le 1 \in \mathbb{R}\, \)in the general sense, but the cases we will deal with are 'small' vulgar fractions of the form \(\displaystyle z=p/q \in \mathbb{Q}\,\), then I will use the following short-hands:
\(\displaystyle \mathcal{C}_j(z)= \mathcal{C}_j(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^j} \cos 2\pi kz\)
\(\displaystyle \mathcal{S}_j(z)= \mathcal{S}_j(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^j} \sin 2\pi kz\)NOTE: I will, occasionally, also write
\(\displaystyle \mathcal{C}_{1-s}(z)= \mathcal{C}_{1-s}(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \cos 2\pi kz\)
\(\displaystyle \mathcal{S}_{1-s}(z)= \mathcal{S}_{1-s}(1, 2\pi z) = \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \sin 2\pi kz\)
Where \(\displaystyle s\, \) is not necessarily an integer, and the understanding is that \(\displaystyle \mathscr{Re}(s) <0\, \).-----------------
Proposition 1.0:
-----------------
The first derivative of the Hurwitz Zeta function - with respect to its first parameter - is given by: \(\displaystyle \zeta'(s,x) = \)\(\displaystyle \frac{\pi\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] + \)
\(\displaystyle \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \left[ \sin \frac{\pi s}{2}\, \mathcal{C}_{1-s}(x) + \cos \frac{\pi s}{2}\, \mathcal{S}_{1-s}(x) \right] + \)
\(\displaystyle \Bigg[\log 2\pi - \psi_0(1-s) \Bigg]\, \zeta(s,x)\)Proof:
This follows directly by differentiating the Fourier Series expansion of the Hurwitz Zeta function. Personally, I'd break it down into the product of two functions, and then differentiate their product. The two functions are given in large brackets here:\(\displaystyle \zeta'(s,x) = \frac{d}{ds}\, \Bigg[ \mathscr{F}(s) \Bigg]\, \Bigg[ \mathscr{G}(s) \Bigg]=\)\(\displaystyle \frac{d}{ds}\, \left[ \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \right]\, \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right]\)The derivative of the 'first' function is
\(\displaystyle \frac{d}{ds}\, \mathscr{F}(s) = \frac{d}{ds}\, \left[ \frac{2\, \Gamma(1-s)}{(2\pi)^{1-s}} \right] = \mathscr{F}'(s) = \)
\(\displaystyle 2\, \frac{ (2\pi)^{1-s} [-\Gamma'(1-s)] - \Gamma(1-s)\, [-(2\pi)^{1-s}\log 2\pi] }{ [(2\pi)^{1-s}]^2 } = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{ \Gamma'(1-s) }{ \Gamma(1-s)} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{ \Gamma'(x) }{ \Gamma(x)} \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \frac{d}{dx} \log \Gamma(x) \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \left[ \log 2\pi - \psi_0(x) \Bigg|_{x=1-s} \right] = \)
\(\displaystyle \frac{2 \Gamma(1-s)}{(2\pi)^{1-s}}\, \Bigg[ \log 2\pi - \psi_0(1-s) \Bigg] \)The derivative of the 'second' function is given by:\(\displaystyle \frac{d}{ds}\, \mathscr{G}(s) = \frac{d}{ds}\, \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}}\right] = \mathscr{G}'(s) =\)\(\displaystyle \frac{\pi}{2} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}} \right] +\)\(\displaystyle \left[ \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \cos 2\pi kx + \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\log k}{k^{1-s}} \sin 2\pi kx \right] = \)\(\displaystyle \frac{\pi}{2} \left[ \cos \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\cos 2\pi kx}{k^{1-s}} - \sin \frac{\pi s}{2}\, \sum_{k=1}^{\infty} \frac{\sin 2\pi kx}{k^{1-s}} \right] +\)
\(\displaystyle \left[ \sin \frac{\pi s}{2}\, \mathcal{C}_{1-s}(x) + \cos \frac{\pi s}{2}\, \mathcal{S}_{1-s}(x) \right]\)Hence, by the rule for differentiation of a product of functions,\(\displaystyle \zeta'(s,x) = \frac{d}{ds}\, \mathscr{F}(s)\, \mathscr{G}(s) = \mathscr{F}'(s)\, \mathscr{G}(s) + \mathscr{F}(s)\, \mathscr{G}'(s)\)Evaluation of the above by brute force yields Proposition 1. \(\displaystyle \, \Box \)
http://mathhelpboards.com/commentary-threads-53/commentary-quot-derivatives-integrals-hurwitz-zeta-function-quot-10777.html
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) - rather long-winded, they are more cumbersome than they are fiendishly complicated. In short, evaluating that Zeta Integral above won't be half as hard as it might first appear. Proof: