This is Exercise 17 of Chapter 6 from Stein and Shakarchi Complex analysis
Let $f$ be an infinitely differentiable function on $\mathbb{R}$ that has compact support, or more generally, let $f$ belongs to the Schwartz space. Consider $$I(s)=\frac{1}{\Gamma(s)}\int^\infty_0 f(x)x^{s-1}\, dx$$ (a) Observe that $I(s)$ is holomorphic for $Re(s)>0$. Prove that $I$ has an analytic continuation as an entire function in the complex plane.
(b) Prove that $I(0)=0$, and more generally $$I(-n)=(-1)^nf^{(n+1)}(0)$$ for all $n\geq 0$.
There is also a hint which tells us to show the formula $$I(s)=\frac{(-1)^k}{\Gamma(s+k)}\int^\infty_0 f^{(k)}(x)x^{s+k-1}\, dx$$ I have no difficulty in showing this and proving part (a), where I used this formula to define $I(s)$ for $Re(s)>-k$.
However, I couldn't prove the second part, not even $I(0)=0$. Let's say $s=0$, then we may choose $k=1$. By $\Gamma(1)=1$ and the fact that $f$ belongs to the Schwartz space (which means $f$ and all its derivatives decay faster than any polynomials), we could obtain $I(0)=f(0)$, which is non-zero in general.
I must have made some mistakes here, because my calculation above doesn't hold for $f(x)=e^{-x^2}$, which I believe satisfy the given conditions.
EDIT: This is likely a typo from the book. $I(0)=f(0)$ and $I(-n)=(-1)^nf^{(n)}(0)$ are the correct answers.
$$\int_0^\infty f(x)x^{s-1}dx=\sum_{k=0}^K \frac{f^{(k)}(0)}{k! (s+k)}+\int_0^1 (f(x)-\sum_{k=0}^K f^{(k)}(0)\frac{x^k}{k!})x^{s-1}dx+\int_1^\infty f(x)x^{s-1}dx$$ The RHS gives the meromorphic continuation to $\Re(s) > -K-1$