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Kummer's formula

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Post Sun Feb 28, 2016 9:15 am

Posts: 47

How would one prove Kummer's formula for \(\displaystyle \log \Gamma\)? That is:

$$\log \Gamma(x)= \frac{1}{\pi} \sum_{n=1}^{\infty}\frac{\log n}{n}\sin 2 \pi n x + \frac{\log 2\pi +\gamma}{\pi} \sum_{n=1}^{\infty}\frac{\sin 2\pi n x}{n} + \frac{1}{2}\sum_{n=1}^{\infty}\frac{\cos 2 \pi nx}{n}+ \frac{1}{2}\log 2\pi$$

Post Sun Feb 28, 2016 4:19 pm
galactus User avatar
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Noting the identities here's a start: $$\sum_{n=1}^{\infty}\frac{\cos(2\pi nx)}{n}=-\log(2\sin(\pi x))$$

and

$$\sum_{n=1}^{\infty}\frac{\sin(2\pi nx)}{n}=\frac{\pi}{2}-\pi x$$



I think the way Kummer approached it was to note the Fourier expansion for $\log\Gamma(x)$ would be:

$$\log\Gamma(x)=a_{0}+2\sum_{n=1}^{\infty}a_{n}\cos(2n\pi x)+2\sum_{n=1}^{\infty}b_{n}\sin(2n\pi x)$$

where $$a_{n}=\int_{0}^{1}\log\Gamma(x)\cos(2n\pi x)dx, \;\ b_{n}=\int_{0}^{1}\log\Gamma(x)\sin(2\pi nx)dx$$


Use Euler's Reflection Formula(ERF) to find $a_{n}$: $\log\Gamma(x)+\log\Gamma(1-x)=\log(2\pi)+\sum_{n=1}^{\infty}\frac{\cos(2\pi nx)}{n}\tag{1}$.

The corresponding Fourier is then:

$$\log\Gamma(x)+\log\Gamma(1-x)=2a_{0}+4\sum_{n=0}^{\infty}a_{n}\cos(2n\pi x)\tag{2}$$.

equate coefficients with the ERF in 1 and 2:

$$a_{0}=\frac{\log(2\pi)}{2}, \;\ a_{n}=\frac{1}{4n}$$.


To jump ahead, it is yet to show that:

$$b_{n}=\int_{0}^{1}\log\Gamma(x)\sin(2\pi nx)dx=\frac{\gamma+\log(2\pi n)}{2\pi n}$$

It would appear that putting this together results in Kummer's formula.


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