The legal way II: taking it home

The application of the Euler-Maclaurin Integration Formula to the function

$$\int^{\infty}_0 \frac{dx}{(z+x)^2} = \frac{1}{z}$$

and recalling the Polygamma function, we know that,

$$\frac{1}{z}=\frac{1}{2z^2} + F^1(z)-\sum^{\infty}_{n=1}\frac{B_2n}{z^{2n+1}}$$ (4.1)

Which is to say,

$$\frac{d}{dz}F(z)=\frac{1}{z}-\frac{1}{2z^2}+\sum^{\infty}_{n=1}\frac{B_2n}{z^{2n+1}}$$ (4.2)

Which is also to say, after some integration,

$$ln(z!)=\aleph + \left(z+\frac{1}{2}\right)\ln z - z - \int \sum^{\infty}_{n=1}\frac{B_{2n}}{2nz^{2n}} dz$$ (4.3)

We can solve for $\aleph$ by using the Legendre duplication formula in 4.3 and taking $z\rightarrow \infty$. In this sense, we will proceed but ignore all pieces on the order of $z^{-1}$ or less, so that,

$$ln(z!) + ln((z-\frac{1}{2})!) = -2z\ln(2) + \frac{1}{2}ln(\pi) + ln((2z)!)$$

$$= 2\aleph + \left(z+\frac{1}{2}\right)\ln(z)-z + \cdots + z\ln(z(1-\frac{1}{2z})) - z + \frac{1}{2} + \cdots$$

$$= 2\aleph + \left(2z+\frac{1}{2}\right)\ln(z)-2z + \cdots$$

$$= -2z \ln(2) + \frac{1}{2}\ln(\pi) + \aleph + \left(2z+\frac{1}{2}\right)\ln(2z) - 2z + \cdots$$

Here we must detail the disappearance of the $1/2$:

$$z\ln(1-\frac{1}{2z})=z\left(\frac{-1}{2z} - \frac{1}{2!}\left(\fr... ...)^2 + \cdots\right) \underbrace{\rightarrow}_{z\rightarrow \infty} -\frac{1}{2}$$ (4.4)

Now we can cancel out what we may cancel out and separate $\ln(2z)=\ln(2)+\ln(z)$, finally taking the limit of z,

$$\aleph = \frac{1}{2}\ln(2) + \frac{1}{2}\ln(\pi) = \frac{1}{2}\ln(2\pi)$$ (4.5)

At last, we have

$$\ln(z!)=\frac{1}{2}\ln(2\pi) + \left(z+\frac{1}{2}\right)\ln(z) - z + \frac{1}{12z} - \frac{1}{360z^3} + \cdots$$ (4.6)

Finally, by noting for example that,

$$e^{1/12z}=1 + \frac{1}{12z} + \frac{1}{2!}\left(\frac{1}{12z}\right)^2 + \cdots$$ (4.7)

we also have,

$$z! = \sqrt{2\pi}z^{z+1/2}e^{-z}\left(1 + \frac{1}{12z} + \frac{1}{288 z^2} + \cdots \right)$$ (4.8)