Is there another way of evaluating $\lim_{x \to 0} \Gamma(x)(\gamma+\psi(1+x))=\frac{\pi^2}{6}$ I was messing around with the Zeta-Function and I got what I thought was an interesting limit:
$$\lim_{x \to 0} \Gamma(x)(\gamma+\psi(1+x)) = \frac{\pi^2}{6} $$
Where $\Gamma$ is the gamma function, $\gamma $ is the Euler-Mascheroni constant, and $\psi$ is the digamma function.
I got it by writing the Zeta Function as:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = -\int_{0}^1 \frac{\ln(x)}{1-x}dx$$
Then using the beta function and differentiating with respect to x you get the limit.
What other ways can be used to evaluate the limit?
Here's how I got to my answer:
We know
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} $$
To convert the sum into an integral consider:
$$\int_{0}^1 x^{n-1}dx=\frac{1}{n}$$
Differentiating once:
$$\int_{0}^1 \ln(x)x^{n-1}dx=\frac{-1}{n^2}$$
Plugging in the integral into the sum you get:
$$-\sum_{n=1}^\infty \int_0^1 \ln(x)x^{n-1}dx=-\int_0^1\ln(x)\sum_{n=1}^\infty  x^{n-1}dx=-\int_0^1 \frac{\ln(x)}{1-x}dx$$
Euler's Beta Function is defined as:
$$\operatorname{B(x,y)}=\int_0^1 t^{x-1}(1-t)^{y-1} = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Differentiation with respect to x:
$$\operatorname{B_x(x,y)}=\int_0^1 \ln(t)t^{x-1}(1-t)^{y-1} = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}(\psi(x)-\psi(x+y))$$
$$-\int_0^1 \frac{\ln(x)}{1-x}dx = -\operatorname{B_x(1,0)}=\frac{\pi^2}{6} $$
Taking the limit as (x,y)->(1,0)
$$-\lim_{(x,y)\to(1,0)} \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}(\psi(x)-\psi(x+y))=-\lim_{y \to 0} \Gamma(y)(\psi(1)-\psi(1+y)) = \lim_{y \to 0} \Gamma(y)(\gamma+\psi(1+y))$$

One may use $\Gamma(x)=\Gamma(x+1)/x$ to get
$$\lim_{x\to0}\frac{\gamma+\psi(1+x)}x$$
and since $\psi(1)=-\gamma$, we may use the definition of the derivative to get $\psi'(1)$, which has many different forms, of which the result may be deduced from.
