Let $$f(x)=\int_{0}^{\infty} e^{-xt} t^x\ \mathsf dt$$ for $x>0$. Show that $f$ is well defined and continuously differentiable on $(0, \infty)$ and compute its derivatives.
My confusion is for each $x$ in our domain how the integration make sense. And to $f$ is continuous I think sequence definition of continuity may more relevant. Maybe we need dominated convergence theorem. But how to find the derivative of $f$? Could you please guide me?
Through the substitution $t=\frac{z}{x}$, leading to $dt=\frac{1}{x}\,dz$, we have: $$ f(x)=\int_{0}^{+\infty} e^{-tx}t^x\,dt = \frac{1}{x^{x+1}}\int_{0}^{+\infty}z^{x}e^{-z}\,dz = \frac{\Gamma(x+1)}{x^{x+1}}=\frac{\Gamma(x)}{x^x}. $$ by the definition of the $\Gamma$ function. Both $\Gamma(x)$ and $x^{x}$ are very regular functions on $\mathbb{R}^+$ (log-convex and $C^{\infty}$ functions) and $x^{x}$ is non-vanishing, hence $f(x)$ is a function belonging $C^{\infty}(\mathbb{R}^+)$.
It is also interesting to notice that for large values of $x$, $f(x)\approx e^{-x}\sqrt{\frac{2\pi}{x}}$.
By differentiation under the integral sign, $$ f'(x) = \int_{0}^{+\infty}(\log(t)-t)e^{-tx}t^x\,dx = \frac{\Gamma'(x)}{x^x}-(1+\log(x))\frac{\Gamma(x)}{x^x} $$ hence: $$ f'(x) = f(x)\cdot\left(\psi(x)-1-\log x\right) $$ by the definition of the digamma function, $\psi(x)=\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}.$