Let $$ \Gamma(\lambda)=\int_0^\infty e^{-x} x^{\lambda-1} dx. $$ How can I prove that $\Gamma$ is an analytic function?
My attempt: we can let $f(\lambda)=\int_0^\infty e^{-x} x^{\lambda-1} \ln x dx$. Expanding $e^t$ into a power series, we get the inequality $$ |x^{\lambda-1}-x^{\lambda_0-1}|\leq x^{-\epsilon} |\lambda-\lambda_0||x^{\lambda_0-1}||\ln x|, x\leq 1, |\lambda-\lambda_0|<\epsilon. $$ So, $$ |\Gamma (\lambda)-\Gamma(\lambda_0)-(\lambda-\lambda_0)f(\lambda_0)|=\left|\int_0^\infty e^{-x}(x^{\lambda-1}-x^{\lambda_0-1}-(\lambda-\lambda_0)x^{\lambda_0-1} \ln x)dx\right|. $$ It seems that I am getting the correct inequality, but I don't know how to proceed. The inequality doesn't seem to fit nicely into this integral. The aim is something like $$ |x^{\lambda-1}-x^{\lambda_0-1}-(\lambda-\lambda_0)x^{\lambda_0-1} \ln x|<\epsilon, $$ although such uniform convergence is unlikely to be achieved.
Some more attempts:
$$ e^{(\lambda-\lambda_0)\ln x}=1+(\lambda-\lambda_0)\ln x+\frac12 (\lambda-\lambda_0)^2(\ln x)^2+\ldots\\ \Rightarrow x^{\lambda-1}-x^{\lambda_0-1}=x^{\lambda_0-1}(x^{\lambda-\lambda_0}-1)\\ =x^{\lambda_0-1}((\lambda-\lambda_0)\ln x+\frac12 (\lambda-\lambda_0)^2(\ln x)^2+\ldots)\\ \Rightarrow x^{\lambda-1}-x^{\lambda_0-1}-(\lambda-\lambda_0)x^{\lambda_0-1} \ln x=x^{\lambda_0-1}(\frac12 (\lambda-\lambda_0)^2(\ln x)^2+\ldots) $$
For each $n\in\mathbb N$, let$$\Gamma_n(\lambda)=\int_0^ne^{-x}x^{\lambda-1}\,\mathrm dx.$$Then $\Gamma_n$ is analytic. Besides, $(\Gamma_n)_{n\in\mathbb N}$ converges uniformly to $\Gamma$ on every compact subset of the right half-plane. Therefore, $\Gamma$ is analytic.