MHB A generalized Clausen Function, and associated loggamma integrals

[DreamWeaver]

I've recently been working on a number of integrals related to the loggamma function, so I thought I'd share my results here. I'll have to post as and when I have time, and there will be a fair bit of preliminary work before we get to the final results, but - loosely speaking - the main aim here is to find a closed form for parametric integrals of the form:$$\Gamma i_m(q)=\int_0^qx^m\log\Gamma(x)\,dx$$Where $$0 < q \le 1$$. The special case where $$q=1$$ appears in a number of contemporary papers, but - thus far - I've not encountered the general case for $$q$$.

In the process of evaluating the above integral, we are led to consider a two generalized Clausen Functions, that in turn, can be expressed as finite sums of derivatives of the Hurwitz Zeta function $$\zeta(z,a)$$, where:

$$\zeta(z,a)=\sum_{k=0}^{\infty}\frac{1}{(k+a)^z}$$Definition:

For the purposes of this tutorial, we define the (possibly new? - although I doubt it) generalized Poly-Clausen Functions:$$C_m(n;\theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos k\theta}{k^m}=\lim_{\,z\to m}\frac{d^n}{dz^n}\sum_{k=1}^{\infty} \frac{\cos k\theta}{k^z}$$$$S_m(n;\theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin k\theta}{k^m}=\lim_{\,z\to m}\frac{d^n}{dz^n}\sum_{k=1}^{\infty} \frac{\sin k\theta}{k^z}$$Following the usual convention for CL and SL-type Clausen Functions, where applicable, we will use the following notation, which is entirely consistent with regular Clausen Functions:$$\text{Cl}_{2m}(n;\theta)=\sum_{k=1}^{\infty} \frac{(\log k)^n \sin k \theta}{k^{2m}}$$

$$\text{Cl}_{2m+1}(n;\theta)=\sum_{k=1}^{\infty} \frac {(\log k)^n \cos k \theta}{k^{2m+1}}$$

$$\text{Sl}_{2m}(n;\theta)=\sum_{k=1}^{\infty} \frac{(\log k)^n \cos k \theta}{k^{2m}}$$

$$\text{Sl}_{2m+1}(n;\theta)=\sum_{k=1}^{\infty} \frac {(\log k)^n \sin k \theta}{k^{2m+1}}$$Evaluations of the Generalized Clausen Functions:With regards to evaluating our parametric loggamma integral, $$\Gamma i_m(q)$$, we will have a particular need for the Generalized Clausen functions at $$q=1/p$$, and $$p \in \mathbb{Z}^{+}$$. In other words, we will need to evaluate the functions:$$C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}$$

$$S_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin \left( \frac{2\pi k}{p}\right)}{k^m}$$Actually, we will only really need the case where $$n=1$$, but there's no harm in evaluating the general case for $$n$$.

As a starting point, we want to split those sums on the RHS into a finite composition of infinite sums. Indeed, we want each sum to split into exactly p-parts, with denominators corresponding to $$(kp+1)$$, $$(kp+2)$$, etc, up to $$(kp +p)$$.

For example, splitting the cosine-type generalized Poly-Clausen Function in such a way, we obtain:

$$C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{j=1}^p \left( \sum_{k=0}^{\infty}\frac{\log^n(kp+j) \cos \left( \frac{2\pi (kp+j)}{p}\right)}{(kp+j)^m} \right)=$$

$$\frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \left( \sum_{k=0}^{\infty} \frac{\log^n \left[p(k+j/p) \right]}{(k+j/p)^m} \right)$$Next, we expand the logarithmic term via the binomial theorem, like so:

$$\log^n[p(k+j/p)]=\sum_{l=0}^n \binom{n}{l}(\log p)^{n-l}\log^l(k+j/p)$$Which gives us the partial evaluation:

$$\frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{l=0}^n \binom{n}{l}(\log p)^{n-l} \left( \sum_{k=0}^{\infty} \frac{\log^l (k+j/p) }{(k+j/p)^m} \right)$$That final, rightmost (doubly 'nested') sum - within the large brackets - can be expressed in terms of derivatives of the Hurwitz Zeta function.

$$\frac{d^l}{dz^l}\zeta(z,a)=\frac{d^l}{dz^l}\sum_{k=0}^{\infty}\frac{1}{(k+a)^z}=(-1)^l\sum_{k=0}^{\infty} \frac{\log^l (k+a) }{(k+a)^z}$$So, with the understanding that - henceforth - ALL derivatives of the Hurwitz Zeta function included here are derivatives with respect to the first parameter, we may write:$$C_m \left(n;\frac{2\pi}{p}\right)=$$$$\frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{l=0}^n \binom{n}{l}(\log p)^{n-l} (-1)^l\zeta^{(l)}\left(m, \frac{j}{p} \right)$$

Which - with a re-labelling of the summation index - is the result we required:

$$C_m \left(n;\frac{2\pi}{p}\right)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}=$$

$$\frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)$$An equivalent calculation gives us the result for the other Poly-Clausen Function:Results:$$C_m \left(n;\frac{2\pi}{p}\right)=$$$$(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\cos \left( \frac{2\pi k}{p}\right)}{k^m}=$$$$\frac{(-1)^n}{p^m} \sum_{j=1}^p \cos \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)$$

$$S_m \left(n;\frac{2\pi}{p}\right)=$$$$(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n\sin \left( \frac{2\pi k}{p}\right)}{k^m}=$$$$\frac{(-1)^n}{p^m} \sum_{j=1}^{p-1} \sin \left( \frac{2\pi j}{p} \right) \sum_{k=0}^n \binom{n}{k}(\log p)^{n-k} (-1)^k\zeta^{(k)}\left(m, \frac{j}{p} \right)$$

**Notice that, when $$j=p$$ in that last result, we get a vanishing term corresponding to $$\sin 2\pi$$, hence the reduction in the upper limit of the summation index, with $$p$$ being replaced by $$p-1$$.

Comments and/or questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-generalized-clausen-function-associated-loggamma-7506.html

 

[DreamWeaver]

For certain small, rational arguments of the second parameter, derivatives of the Hurwitz Zeta function can variously be expressed in terms of (slightly simpler) related functions. These relations will be explored more thoroughly later on, but for now, I simply give their definitions, and a few basic functional relations.
(1) The Riemann Zeta function:

$$\zeta(z)=\sum_{k=0}^{\infty}\frac{1}{(k+1)^z} \equiv \sum_{k=1}^{\infty}\frac{1}{k^z}$$(2) The Dirichlet Beta function:

$$\beta(z)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^z}$$(3) The Eta Function:

$$\eta(z)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^z}$$(4) The Legendre Chi function:

$$\chi(z)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^z}$$(5) Functional relations:

$$\eta(z)=\left(1-\frac{2}{2^z}\right)\zeta(z)$$

$$\chi(z)=\left(1- \frac{1}{2^z}\right)\zeta(z)$$$$\chi(z)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^z}= \frac{1}{2^z} \sum_{k=0}^{\infty}\frac{1}{\left( k+\frac{1}{2} \right)^z}= \frac{1}{2^z} \zeta\left(z, \frac{1}{2} \right) $$$$\eta(z) + \zeta(z) = 2 \chi (z)$$

 

[DreamWeaver]

We almost have all of the requisite results needed to evaluate the integral function $$\Gamma i_m(q)$$, but a tiny bit more groundwork is required first...

For reasons that will soon become apparent, we will need closed form evaluations of the following trigonometric integrals ($$m \in \mathbb{Z}^{+}$$):$$\int x^m\sin x\,dx$$

$$\int x^m\cos x\,dx$$These can be found in many texts on elementary indefinite integrals. Indeed, the proof requires nothing more than repeated integration by parts, and a modicum of induction, so the proofs are omitted. We simply state the results:$$(1) \quad \int x^{2n}\sin x\,dx=$$

$$(2n)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j+1}\frac{x^{2n-2j}}{(2n-2j)!} \cos x + \sum_{j=0}^{n-1}(-1)^j\frac{x^{2n-2j-1}}{(2n-2j-1)!} \sin x \Bigg\}$$
$$(2) \quad \int x^{2n+1}\sin x\,dx=$$

$$(2n+1)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j+1}\frac{x^{2n-2j+1}}{(2n-2j+1)!} \cos x + \sum_{j=0}^{n}(-1)^j\frac{x^{2n-2j}}{(2n-2j)!} \sin x \Bigg\}$$
$$(3) \quad \int x^{2n}\cos x\,dx=$$

$$(2n)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j}\frac{x^{2n-2j}}{(2n-2j)!} \sin x + \sum_{j=0}^{n-1}(-1)^j\frac{x^{2n-2j-1}}{(2n-2j-1)!} \cos x \Bigg\}$$
$$(4) \quad \int x^{2n+1}\cos x\,dx=$$

$$(2n+1)! \, \Bigg\{ \sum_{j=0}^n(-1)^{j}\frac{x^{2n-2j+1}}{(2n-2j+1)!} \sin x + \sum_{j=0}^{n}(-1)^j\frac{x^{2n-2j}}{(2n-2j)!} \cos x \Bigg\}$$Use of the floor function allows us to condense (1) and (2), respectively (3) and (4) into the following:
$$(5) \int x^m\sin x \, dx=$$$$m! \, \Bigg\{ \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{x^{m-2j}}{(m-2j)!} \cos x +
\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \sin x \Bigg\}$$
$$(6) \int x^m\cos x \, dx=$$$$m! \, \Bigg\{ \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{x^{m-2j}}{(m-2j)!} \sin x +
\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \cos x \Bigg\}$$
-----------------------------In terms of the above trigonometric integrals, the particular examples we will need later on are as follows:$$\int_0^{q}x^m\sin 2\pi kx \, dx$$$$\int_0^{q}x^m\cos 2\pi kx \, dx$$
The substitution $$y=2\pi k x$$ then yields:$$\int_0^{q}x^m\sin 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\sin x\, dx$$$$\int_0^{q}x^m\cos 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\cos x\, dx$$A little caution is required here, when evaluating the cosine series in (5) and (6) at $$x=0$$, the reason being that these cosine series vanish except for the term where the exponent of $$x$$ is equal to $$0$$. In both cases, we treat the term $$x^n\cos(0)$$ as a limiting value, and define:$$x^n\cos x \, \Bigg|_{x=0}=\cos (0) \, \lim_{\epsilon \to 0^{+}}\epsilon ^n =1$$Referring back to integrals (1) and (2), we note that the exponent of $$x$$ is only $$0$$ in the final term of the cosine series when $$m=2n$$ is even. Conversely, looking back at integrals (3) and (4), we note that the cosine series vanishes except for the final term in (4), when $$m=2n+1$$ is odd. For this reason, we have need of two generalized integer functions, where one is zero for odd numbers, 1 for even numbers, and the second is the reverse. Such integer functions are easily found. Define$$\mathcal{F}(m) = \frac{[1+(-1)^m]}{2}=
\begin{cases}
0, & \text{if }m\text{ is odd} \\
1, & \text{if }m\text{ is even}
\end{cases}$$$$\mathcal{G}(m) = \frac{[1+(-1)^{m+1}]}{2}=
\begin{cases}
1, & \text{if }m\text{ is odd} \\
0, & \text{if }m\text{ is even}
\end{cases}$$In light of these generalized integer functions, we can now find a neat closed form for our two key trigonometric integrals.

$$\int_0^{q}x^m\sin 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\sin x\, dx=$$$$ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{x^{m-2j}}{(m-2j)!} \cos x \, \Bigg|_0^{2\pi k q} +$$

$$\frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \sin x \, \Bigg|_0^{2\pi k q} =$$$$ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \cos 2\pi kq \, +$$$$\frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \sin 2\pi kq \, -\mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \frac{m!}{(2\pi k)^{m+1}}$$Equivalently, our cosine integral evaluates to:$$\int_0^{q}x^m\cos 2\pi kx \, dx=\frac{1}{(2\pi k)^{m+1}}\int_0^{2\pi kq}x^m\cos x\, dx=$$$$ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{x^{m-2j}}{(m-2j)!} \sin x \, \Bigg|_0^{2\pi k q} +$$

$$\frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{x^{m-2j-1}}{(m-2j-1)!} \cos x \, \Bigg|_0^{2\pi k q} =$$
$$ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \sin 2\pi kq +$$

$$\frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \cos 2\pi kq -\mathcal{G}(m) (-1)^{ \lfloor \frac{m-1}{2} \rfloor } \frac{m!}{(2\pi k)^{m+1}} $$More to follow shortly... (Heidy)

 

[MarkFL]

This topic is for comments and/or questions pertaining to the tutorial topic:

http://mathhelpboards.com/math-notes-49/generalized-clausen-function-associated-loggamma-integrals-7461.html

 

[DreamWeaver]

Mark... I wonder if your good self or another admin will take pity on me and make a small but significant correction to my first post of the Generalized Clausen Function thread...?

Please and thank you! :DI made a typo, and wrote:$$C_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \cos k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^m}$$

and

$$S_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \sin k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^m}$$These should have been:$$ C_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \cos k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^z} $$and$$ S_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \sin k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^z} $$respectively...

Ooops!

 

[MarkFL]

Happy to oblige! :D

 

[DreamWeaver]

Thanks Mark! You're a gent... :D

 

[polygamma]

I'm curious exactly how you express it in terms of the Hurwitz zeta function.

I evaluated it for $m=0$ because I wanted to derive a closed form expression for the Barnes G function for $\text{Re}(z) > 0$.

See here.But what I did can be generalized.Specifically, $\displaystyle \int_{0}^{z} x^{m} \log \Gamma(x) \ dx = \frac{z^{m+1}}{(m+1)^{2}} - \frac{z^{m+1}\log z}{m+1} - \frac{\gamma z^{m+2}}{m+2} + \sum_{k=2}^{\infty} \frac{(-1)^{k} \zeta(k)}{k(k+m+1)} z^{k+m+1}$.For $m=1$ the infinite sum appears to be expressible in terms of $\zeta'(-1,z)$ and $\zeta'(-2,z)$.

 

[DreamWeaver]

Our goal is now in sight...! Or, to put it another way, the end is nigh (if you're a pessimist, that is lol (Hug) )

Just a tiny bit more housekeeping, and then we derive a first evaluation for $$\Gamma i_m(q)$$, in the special case where $$q=1/p$$ and $$p \in \mathbb{Z}^{+}$$. The general case for $$\Gamma i_m (q/p)$$ will follow a little later.

We just need a few more results first...

Firstly - and I really should have added these nearer the start - we define the (regular) Clausen functions as follows:$$\text{Cl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m}}$$

$$\text{Cl}_{2m+1}(\theta)=\sum_{k=1}^{\infty}\frac{ \cos k\theta}{k^{2m+1}}$$

$$\text{Sl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \cos k\theta}{k^{2m}}$$

$$\text{Sl}_{2m+1}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m+1}}$$Although these definitions will be of considerable use later on, of particular interest are the CL-type Clausen Functions $$\text{Cl}_2(\theta)$$ and $$\text{Cl}_1(\theta)$$. The former has the integral representation:

$$\text{Cl}_2(\theta)=-\int_0^{\theta}\log \Bigg| 2\sin \frac{x}{2} \Bigg| \, dx$$Differentiation with respect to $$\theta$$ gives:

$$\frac{d}{d \theta} \text{Cl}_2(\theta)=-\log \Bigg| 2\sin \frac{x}{2} \Bigg| \equiv \sum_{k=1}^{\infty}\frac{\cos k\theta}{k}$$The final piece in the puzzle is Kummer's Fourier expansion for the loggamma function:$$\log \left( \frac{ \Gamma(x)}{ \sqrt{2\pi} } \right)=$$

$$\frac{1}{2}(\gamma +\log 2\pi)(1-2x)-\frac{1}{2}\log(2\sin \pi x) + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx$$Kummer's expansion is valid for $$0 < x < 1$$, and at $$x=1$$ we take the limiting value. It is no understatement to say that everything we develop below is dependent upon Kummer's result, so it would be remiss of me to omit a proof. However, I would quite like to move things along a little, so I'll add a proof a bit later on...

Re-write Kummer's expansion as:

$$\log \Gamma(x) = \frac{1}{2}(\gamma+ 2 \log 2\pi)-(\gamma + \log 2\pi)x $$

$$-\frac{1}{2}\log(2\sin \pi x) + \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx
$$

In light of the previous definitions of the (regular) Clausen functions, we note that the logsine term in Kummer's expansion may be written as:

$$-\frac{1}{2}\log(2\sin \pi x) = -\frac{1}{2}\log \left( 2\sin \frac{2\pi x}{2}\right)=\frac{1}{2}\text{Cl}_1(2\pi x)=\frac{1}{2} \sum_{k=1}^{\infty}\frac{\cos 2\pi k x}{k}$$So Kummer's expansion becomes:$$\log \Gamma(x) = \frac{1}{2}(\gamma+ 2 \log 2\pi)-(\gamma + \log 2\pi)x +\frac{1}{2} \sum_{k=1}^{\infty}\frac{\cos 2\pi k x}{k}
+ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \sin 2\pi kx$$Next, we multiply both sides by $$x^m$$, and then integrate over the interval $$[0,q]$$:$$\Gamma i_m(q)=\int_0^qx^m\log\Gamma(x) \, dx=$$$$\frac{1}{2}(\gamma+ 2 \log 2\pi)\int_0^q x^m \, dx -(\gamma + \log 2\pi) \int_0^q x^{m+1}

\, dx$$

$$+\frac{1}{2} \sum_{k=1}^{\infty}\frac{1}{k} \int_0^q x^m \cos 2\pi k x \, dx
+ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^q x^m \sin 2\pi kx
$$The first two integrals on the RHS are trivial, and give the partial result:

$$\frac{q^{m+1}}{2(m+1)}(\gamma+ 2 \log 2\pi) - \frac{q^{m+2}}{(m+2)}(\gamma + \log 2\pi)$$

There is an obvious temptation to split these terms, and write them in the form

$$A \gamma + B \log 2\pi$$

Where $$A$$ and $$B$$ are constants in terms of $$m$$ and $$q$$. Take my word for it: it's not worth it... The results are far from elegant, or efficient, so I'll just stick with the form above.

The third integral on the RHS is:

$$\sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)$$

We already have a neat closed form for this integral, and may write

$$\sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)=$$

$$\sum_{k=1}^{\infty} \frac{1}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{(2\pi kq)^{m-2j}}{(m-2j)!} \sin 2\pi kq \Bigg\}+$$

$$\sum_{k=1}^{\infty} \frac{1}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{(2\pi kq)^{m-2j-1}}{(m-2j-1)!} \cos 2\pi kq \Bigg\}+$$

$$(-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \sum_{k=1}^{\infty} \frac{1}{k^{m+2}}=$$
$$ m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \sum_{k=1}^{\infty} \frac{\sin 2\pi kq}{k^{2j+2} } +$$

$$ m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \sum_{k=1}^{\infty} \frac{\cos 2\pi kq}{k^{2j+3} } +$$

$$(-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)$$In light of the series definitions for the regular Clausen functions, this becomes:$$\sum_{k=1}^{\infty} \frac{1}{k} \left( \int_0^q x^m \cos 2\pi kx \right)=$$

$$ m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \text{Cl}_{2j+2} (2\pi q) +$$

$$ m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \text{Cl}_{2j+3} (2\pi q) +$$

$$(-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)$$Three down, one to go... Be right back! (Heidy)

 

[DreamWeaver]

Combining all of the previous results, we have the following partial (almost complete) evaluation:Theorem 1.0:$$\Gamma i_m (q)=$$

$$\frac{q^{m+1}}{2(m+1)}(\gamma+ 2 \log 2\pi) - \frac{q^{m+2}}{(m+2)}(\gamma + \log 2\pi)+$$

$$\frac{1}{2} m! \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{q^{m-2j}}{(2\pi)^{2j+1} (m-2j)!} \text{Cl}_{2j+2} (2\pi q) +$$

$$\frac{1}{2} m! \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{q^{m-2j-1}}{(2\pi)^{2j+2} (m-2j-1)!} \text{Cl}_{2j+3} (2\pi q) +$$

$$\frac{1}{2} (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \frac{m!}{(2\pi)^{m+1}} \zeta (m+2)+$$

$$\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^q x^m \sin 2\pi kx$$As mentioned before, we will evaluate the general q-case shortly, for $$q \in \mathbb{Q}^{+}$$, where $$0 < q \le 1$$. For now, however, let's set $$q=1/p$$, $$p \in \mathbb{Z}^{+}$$ in Theorem 1.0 to obtain:

$$\Gamma i_m(1/p)=\int_0^{1/p}x^m\log\Gamma(x)\,dx=$$$$\frac{(\gamma+ 2 \log 2\pi)}{2p^{m+1}(m+1)} - \frac{(\gamma + \log 2\pi)}{p^{m+2}(m+2)}+
\frac{m!}{2} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^j \frac{ \text{Cl}_{2j+2} \left( \frac{2\pi}{p} \right) }{(2\pi)^{2j+1} p^{m-2j} (m-2j)!} +$$

$$\frac{m!}{2} \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^j \frac{ \text{Cl}_{2j+3} \left( \frac{2\pi}{p} \right) }{(2\pi)^{2j+2} p^{m-2j-1} (m-2j-1)!} +
\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^{1/p} x^m \sin 2\pi kx
$$$$+ \frac{ (-1)^{ \lfloor \frac{m-1}{2} \rfloor +1} \mathcal{G}(m) \, m! }{2 (2\pi)^{m+1}} \zeta (m+2)+$$In light of all the groundwork we've already done, that last integral is relatively - (Drunk) -straightforward...
$$\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \int_0^{1/p} x^m \sin 2\pi kx \, dx=$$

$$\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{1}{(2\pi k)^{m+1}} \int_0^{2\pi k/p} x^m \sin x \, dx \Bigg\} =$$
$$ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}} \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } (-1)^{j+1} \frac{(2\pi k / p)^{m-2j}}{(m-2j)!} \cos (2\pi k / p) \, \Bigg\} +$$$$ \frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ \frac{m!}{(2\pi k)^{m+1}}\sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } (-1)^{j} \frac{(2\pi k / p)^{m-2j-1}}{(m-2j-1)!} \sin (2\pi k /p) \, \Bigg\} +$$$$\frac{1}{\pi} \, \sum_{k=1}^{\infty}\frac{\log k}{k} \Bigg\{ -\mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \frac{m!}{(2\pi k)^{m+1}} \Bigg\}=$$
$$\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } \frac{(-1)^{j+1}}{p^{m-2j} (2\pi)^{2j+1} (m-2j)! } \sum_{k=1}^{\infty} \frac{\log k}{k^{2j+2}} \cos \left( \frac{2\pi k}{p} \right) +$$$$\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } \frac{(-1)^{j}}{p^{m-2j-1} (2\pi)^{2j+2} (m-2j-1)! } \sum_{k=1}^{\infty} \frac{\log k}{k^{2j+3}} \sin \left( \frac{2\pi k}{p} \right) +$$$$\frac{m!}{\pi (2\pi)^{m+1}} \mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \zeta'(m+2)=$$In terms of the generalized Poly-Clausen Functions we introduced earlier, this becomes:
$$-\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m}{2} \rfloor } \frac{(-1)^{j+1}}{p^{m-2j} (2\pi)^{2j+1} (m-2j)! } C_{2j+2} \left(1; \frac{2\pi }{p} \right) +$$$$-\frac{m!}{\pi} \, \sum_{j=0}^{ \lfloor \frac{m-1}{2} \rfloor } \frac{(-1)^{j}}{p^{m-2j-1} (2\pi)^{2j+2} (m-2j-1)! } S_{2j+3} \left(1; \frac{2\pi }{p} \right) +$$$$\frac{m!}{\pi (2\pi)^{m+1}} \mathcal{F}(m) (-1)^{ \lfloor \frac{m}{2} \rfloor +1} \zeta'(m+2)$$Need a break... More to follow shortly... (Smoking)

 

[alyafey22]

Hey , it would be nice if you add the proof of the Kummer's Fourier expansion for the loggamma function.

 

[DreamWeaver]

Re: Commentary for &quot;A generalized Clausen Function, and associated loggamma&quot;

I'm curious exactly how you express it in terms of the Hurwitz zeta function.

I evaluated it for $m=0$ because I wanted to derive a closed form expression for the Barnes G function for $\text{Re}(z) > 0$.

See here.But what I did can be generalized.Specifically, $\displaystyle \int_{0}^{z} x^{m} \log \Gamma(x) \ dx = \frac{z^{m+1}}{(m+1)^{2}} - \frac{z^{m+1}\log z}{m+1} - \frac{\gamma z^{m+2}}{m+2} + \sum_{k=2}^{\infty} \frac{(-1)^{k} \zeta(k)}{k(k+m+1)} z^{k+m+1}$.For $m=1$ the infinite sum appears to be expressible in terms of $\zeta'(-1,z)$ and $\zeta'(-2,z)$.

Nice work, RV! :D

I'm very nearly there, in terms of completing the initial evaluation on t'other thread... Please bear with me!

- - - Updated - - -

Hey , it would be nice if you add the proof of the Kummer's Fourier expansion for the loggamma function.

Oh yes, I quite agree, Zaid. I was just rather eager to get to the actual evaluation. I certainly will add a full proof of Kummer's result. It's essential... ;)

 

[polygamma]

I worked it out for $m=1$ using the same approach I took for $m=0$ and got the following:

$$ \int_{0}^{z} x \log \Gamma (x) \ dx = \zeta'(-1,z)z - \frac{\zeta'(-2,z)}{2} - \frac{z^{3}}{4} + \frac{z^{2}}{8} + \frac{z}{24} + \frac{z^{2} \ln(2 \pi)}{4} + \frac{\zeta'(-2)}{2} $$

The result seems to check out numerically for any positive value of $z$.

For $z=1$ you get

$$ \int_{0}^{1} x \log \Gamma (x) \ dx = \zeta'(-1) - \frac{1}{12} + \frac{\ln(2 \pi)}{4} = - \log(A) + \frac{\ln 2 \pi}{4}$$

where $A$ is the crazy Glaisher-Kinkelin constant.

 

[alyafey22]

I was playing around with the Clausen function $$\mathrm{Cl}_2(\theta)$$ and came across the following integral

$$\int^t_0\mathrm{Cl}_2(\theta)\, d\theta $$

What would be the best way to represent that integral ?

PS: I thought of representing that integral using polylogarithms .

 

[alyafey22]

To be precise , I was trying to find a closed form for the integral
$$I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx \,\,\,\,\, a,t>0$$

 

[DreamWeaver]

Hello Zaid! :D

The simplest way to evaluate your integral is to consider the general order Clausen functions, in their 'native' series forms:$$\text{Cl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m}}$$$$\text{Cl}_{2m+1}(\theta)=\sum_{k=1}^{\infty}\frac{ \cos k\theta}{k^{2m+1}}$$Hence$$\int_0^{\theta}\text{Cl}_2(x)\,dx= \sum_{k=1}^{\infty}\frac{1}{k^2} \int_0^{\theta} \sin k x\,dx=-\sum_{k=1}^{\infty}\frac{1}{k^3} \Bigg[ \cos kx \Bigg]_0^{\theta}$$$$\cos k0=\cos 0=1$$ for all integer k, regardless of parity, hence this reduces to$$-\sum_{k=1}^{\infty}\frac{\cos k\theta }{k^3}+\sum_{k=1}^{\infty}\frac{1}{k^3}= - \text{Cl}_3(\theta) + \zeta (3) $$So$$\int_0^{\theta}\text{Cl}_2(x)\,dx= \zeta(3)-\text{Cl}_3(\theta)$$A similar application to the integral $$\int_0^\theta x \log\left(2 \sin\frac{x}{2} \right) \, dx$$ gives$$\int_0^\theta x \log\left(2 \sin\frac{x}{2} \right), dx= \zeta(3)-\theta \text{Cl}_2(\theta)-\text{Cl}_3(\theta)
$$Since, for $$0 < \theta \le 2\pi$$, $$\text{Cl}_2(\theta) = - \int_0^{\theta} \log\left( 2 \sin \frac{x}{2} \right)\,dx$$From which, an application of the First Fundamental Theorem of Calculus gives$$\frac{d}{d\theta} \text{Cl}_2(\theta)= - \log\left( 2 \sin \frac{ \theta }{2} \right)= \text{Cl}_1(\theta) = \sum_{k=1}^{\infty}\frac{ \cos k\theta}{k} $$This proves the result.

 

[alyafey22]

I think I basically did the same thing but represented that using trilogarithm

$$\int^t_0 \mathrm{Cl}_2(\theta) \, d\theta = -\Re \left( \mathrm{Li}_3(e^{it}) \right)+\zeta(3)$$

It turns out that $\Re \left( \mathrm{Li}_3(e^{it}) \right) = \mathrm{Cl}_3(t)$

 

[DreamWeaver]

Yes, yes... That is exactly the way to go, Zaid ;)

I used to have a bit of a fetish for the Ploylogarithm... Until I discovered the Clausen function, that is... [There, there, little Polylog... Don't be jealous now! ]

 

[alyafey22]

Yes, yes... That is exactly the way to go, Zaid ;)

I used to have a bit of a fetish for the Ploylogarithm... Until I discovered the Clausen function, that is... [There, there, little Polylog... Don't be jealous now! ]

I am on the side of Polylogs. I hate working with trigonometric functions (Smoking). You seem to be an expert at the Clausen functions , keep it up .

 

[DreamWeaver]

I am on the side of Polylogs. I hate working with trigonometric functions (Smoking). You seem to be an expert at the Clausen functions , keep it up .

Bad mammal! Trig is good for your 'elf... (Hug)

Only messin', friend.

Incidentally, you're FAR too kind... I'm certainly no expert - on anything - just a stubbornly enthusiastic amateur. Not that I mind being that, of course - that's good enough to get me by. ;)