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http://mathhelpboards.com/math-note...ction-associated-loggamma-integrals-7461.html

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http://mathhelpboards.com/math-note...ction-associated-loggamma-integrals-7461.html

- Sep 16, 2013

- 337

Please and thank you!!

I made a typo, and wrote:

\(\displaystyle C_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \cos k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^m}\)

and

\(\displaystyle S_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \sin k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^m}\)

These should have been:

\(\displaystyle C_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \cos k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\cos k\theta}{k^z} \)

and

\(\displaystyle S_m(n; \theta)=(-1)^n\sum_{k=1}^{\infty}\frac{(\log k)^n \sin k\theta}{k^m}= \lim_{z \to m} \frac{d^n}{dz^n} \sum_{k=1}^{\infty} \frac{\sin k\theta}{k^z} \)

respectively...

Ooops!!

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- Sep 16, 2013

- 337

Thanks Mark! You're a gent...

- Jan 31, 2012

- 253

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 $\zeta'(-2,z)$.

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 $\zeta'(-2,z)$.

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- Jan 17, 2013

- 1,667

- Sep 16, 2013

- 337

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!

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

- - - Updated - - -

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...

- Jan 31, 2012

- 253

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

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.

- Jan 17, 2013

- 1,667

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

What would be the best way to represent that integral ?

PS: I thought of representing that integral using polylogarithms .

- Jan 17, 2013

- 1,667

\(\displaystyle I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx \,\,\,\,\, a,t>0\)

- Sep 16, 2013

- 337

Hello Zaid!

The simplest way to evaluate your integral is to consider the general order Clausen functions, in their 'native' series forms:

\(\displaystyle \text{Cl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m}}\)

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

Hence

\(\displaystyle \int_0^{\theta}\text{Cl}_2(x)\,dx= \sum_{k=1}^{\infty}\frac{1}{k^2} \int_0^{\theta} \sin k x\,dx=-\sum_{k=1}^{\infty}\frac{1}{k^3} \Bigg[ \cos kx \Bigg]_0^{\theta}\)

\(\displaystyle \cos k0=\cos 0=1\) for all integer k, regardless of parity, hence this reduces to

\(\displaystyle -\sum_{k=1}^{\infty}\frac{\cos k\theta }{k^3}+\sum_{k=1}^{\infty}\frac{1}{k^3}= - \text{Cl}_3(\theta) + \zeta (3) \)

So

\(\displaystyle \int_0^{\theta}\text{Cl}_2(x)\,dx= \zeta(3)-\text{Cl}_3(\theta)\)

A similar application to the integral \(\displaystyle \int_0^\theta x \log\left(2 \sin\frac{x}{2} \right) \, dx\) gives

\(\displaystyle \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 \(\displaystyle 0 < \theta \le 2\pi\),

\(\displaystyle \text{Cl}_2(\theta) = - \int_0^{\theta} \log\left( 2 \sin \frac{x}{2} \right)\,dx\)

From which, an application of the First Fundamental Theorem of Calculus gives

\(\displaystyle \frac{d}{d\theta} \text{Cl}_2(\theta)= - \log\left( 2 \sin \frac{ \theta }{2} \right)= \text{Cl}_1(\theta) = \sum_{k=1}^{\infty}\frac{ \cos k\theta}{k} \)

This proves the result.

The simplest way to evaluate your integral is to consider the general order Clausen functions, in their 'native' series forms:

\(\displaystyle \text{Cl}_{2m}(\theta)=\sum_{k=1}^{\infty}\frac{ \sin k\theta}{k^{2m}}\)

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

Hence

\(\displaystyle \int_0^{\theta}\text{Cl}_2(x)\,dx= \sum_{k=1}^{\infty}\frac{1}{k^2} \int_0^{\theta} \sin k x\,dx=-\sum_{k=1}^{\infty}\frac{1}{k^3} \Bigg[ \cos kx \Bigg]_0^{\theta}\)

\(\displaystyle \cos k0=\cos 0=1\) for all integer k, regardless of parity, hence this reduces to

\(\displaystyle -\sum_{k=1}^{\infty}\frac{\cos k\theta }{k^3}+\sum_{k=1}^{\infty}\frac{1}{k^3}= - \text{Cl}_3(\theta) + \zeta (3) \)

So

\(\displaystyle \int_0^{\theta}\text{Cl}_2(x)\,dx= \zeta(3)-\text{Cl}_3(\theta)\)

A similar application to the integral \(\displaystyle \int_0^\theta x \log\left(2 \sin\frac{x}{2} \right) \, dx\) gives

\(\displaystyle \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 \(\displaystyle 0 < \theta \le 2\pi\),

\(\displaystyle \text{Cl}_2(\theta) = - \int_0^{\theta} \log\left( 2 \sin \frac{x}{2} \right)\,dx\)

From which, an application of the First Fundamental Theorem of Calculus gives

\(\displaystyle \frac{d}{d\theta} \text{Cl}_2(\theta)= - \log\left( 2 \sin \frac{ \theta }{2} \right)= \text{Cl}_1(\theta) = \sum_{k=1}^{\infty}\frac{ \cos k\theta}{k} \)

This proves the result.

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- Jan 17, 2013

- 1,667

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

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

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

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

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

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- Sep 16, 2013

- 337

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! ]

- Jan 17, 2013

- 1,667

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

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! ]

- Sep 16, 2013

- 337

Bad mammal! Trig is good for your 'elf....I am on the side of Polylogs. I hate working with trigonometric functions . You seem to be an expert at the Clausen functions , keep it up .

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.