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I ran across an interesting series in a paper written by J.W.L. Glaisher. Glaisher mentions that it is a known formula but does not indicate how it can be derived.

I think it is difficult.

$$\sum_{k=2}^\infty \frac{\log(k)}{k}\sin(2k \mu \pi) = \pi \left(\log(\Gamma(\mu)) +\frac{1}{2}\log \sin(\pi \mu)-(1-\mu)\log(\pi)- \left(\frac{1}{2}-\mu\right)(\gamma+\log 2)\right)$$

Can someone suggest a method of attack?

$\gamma$ is the Euler-Mascheroni Constant.

Thank You!

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$\mu$ is an arbitrary constant? –  J. M. Apr 8 '13 at 13:34
    
@J.M: Yes, its an arbitrary constant. –  Integrals and Series Apr 8 '13 at 13:36
2  
You might want to link to it in the question... –  J. M. Apr 8 '13 at 13:47
2  
Maybe it can be done like this: the right-hand side is a certain function of $\mu$. Show that the left-hand side is the Fourier series of that function. –  GEdgar Apr 10 '13 at 12:59
1  
Note it is valid only for $0 < \mu < 1$. The left side has period $1$, but the right side doesn't. –  GEdgar Apr 12 '13 at 16:01
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1 Answer 1

up vote 7 down vote accepted
+300

It suffices to do these integrals: $$ \begin{align} \int_0^1 \log(\Gamma(s))\;ds &= \frac{\log(2\pi)}{2} \tag{1a}\\ \int_0^1 \log(\Gamma(s))\;\cos(2k \pi s)\;ds &= \frac{1}{4k},\qquad k \ge 1 \tag{1b}\\ \int_0^1 \log(\Gamma(s))\;\sin(2k \pi s)\;ds &= \frac{\gamma+\log(2k\pi)}{2k\pi},\qquad k \ge 1 \tag{1c} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;ds &= \frac{-\log 2}{2} \tag{2a} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;\cos(2k \pi s)\;ds &= \frac{-1}{4k},\qquad k \ge 1 \tag{2b} \\ \int_0^1 \frac{\log(\sin(\pi s))}{2}\;\sin(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{2c} \\ \int_0^1 1 \;ds &= 1 \tag{3a} \\ \int_0^1 1 \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{3b} \\ \int_0^1 1 \cdot \sin(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{3c} \\ \int_0^1 s \;ds &= \frac{1}{2} \tag{4a} \\ \int_0^1 s \cdot \cos(2k \pi s)\;ds &= 0,\qquad k \ge 1 \tag{4b} \\ \int_0^1 s \cdot \sin(2k \pi s)\;ds &= \frac{-1}{2k\pi},\qquad k \ge 1 \tag{4c} \end{align} $$ Then for $f(s) = \pi \left(\log(\Gamma(s)) +\frac{1}{2}\log \sin(\pi s)-(1-s)\log(\pi)- \left(\frac{1}{2}-s\right)(\gamma+\log 2)\right)$, we get $$ \begin{align} \int_0^1 f(s)\;ds &= 0 \\ 2\int_0^1f(s) \cos(2k\pi s)\;\;ds &= 0,\qquad k \ge 1 \\ 2\int_0^1f(s) \sin(2k\pi s)\;\;ds &= \frac{\log k}{k},\qquad k \ge 1 \end{align} $$ and the formula follows as a Fourier series: $$ f(s) = \sum_{k=1}^\infty \frac{\log k}{k}\;\sin(2 k\pi s),\qquad 0 < s < 1. $$

reference

Gradshteyn & Ryzhik, Table of Integrals Series and Products

(1a) 6.441.2
(1b) 6.443.3
(1c) 6.443.1
(2a) 4.384.3
(2b) 4.384.3
(2c) 4.384.1

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