- Let f be an indefinitely diferentiable function on R that has compact support, or more generally, let \(f\) belong to the Schwartz space. Consider
\[ I (s) = \frac {1}{\Gamma (s)} \int_ {0} ^ {\infty} f (x) x ^ {- 1 + s} d x. \]
Observe that \(I ( s )\) is holomorphic for \(\operatorname { R e } ( s ) > 0\) . Prove that I has an analytic continuation as an entire function in the complex plane.
Prove that \(I(0)=f(0)\), and more generally
\[ I(-n)=(-1)^n f^{(n)}(0) \quad \text{for all } n\ge0. \]
[Hint: To prove the analytic continuation, as well as the formulas in the second part, integrate by parts to show that \(\begin{array} { r } { I ( s ) = \frac { ( - 1 ) ^ { k } } { \Gamma ( s + k ) } \int _ { 0 } ^ { \infty } f ^ { ( k ) } ( x ) x ^ { s + k - 1 } d x . } \end{array}\)
Solution
For \(\Re s>0\), \[ I(s)=\frac1{\Gamma(s)}\int_0^\infty f(x)x^{s-1}\,dx \] is holomorphic by dominated convergence on compact vertical strips.
Integrating by parts once gives \[ \int_0^\infty f(x)x^{s-1}\,dx =-\frac1s\int_0^\infty f'(x)x^s\,dx, \] because the boundary terms vanish when \(\Re s>0\). Using \(\Gamma(s+1)=s\Gamma(s)\), \[ I(s)=-\frac1{\Gamma(s+1)}\int_0^\infty f'(x)x^s\,dx. \] Iterating \(k\) times yields \[ I(s)=\frac{(-1)^k}{\Gamma(s+k)} \int_0^\infty f^{(k)}(x)x^{s+k-1}\,dx, \tag{1} \] initially for \(\Re s>0\).
For fixed \(k\), the right-hand side is holomorphic for \(\Re s>-k\): the integral is holomorphic there, and \(1/\Gamma\) is entire. On overlaps these expressions agree by (1), so they analytically continue one another. Since \(k\) is arbitrary, \(I\) extends to an entire function.
Now fix \(n\ge0\) and take \(k=n+1\) in (1). At \(s=-n\), \[ I(-n)=(-1)^{n+1}\int_0^\infty f^{(n+1)}(x)\,dx, \] because \(\Gamma(1)=1\). Since \(f^{(n)}(x)\to0\) as \(x\to\infty\) for a Schwartz function (and likewise for compact support), \[ \int_0^\infty f^{(n+1)}(x)\,dx =-f^{(n)}(0). \] Therefore \[ \boxed{I(-n)=(-1)^n f^{(n)}(0)}. \] In particular, \(I(0)=f(0)\).