How to understand limit

Question

Let $\chi(t)$ be the Heaviside function, i.e. $\chi(t) = 1$ for $t > 0$ and $\chi(t) = 0$ if $t \leq 0$. Reading a paper I faced with a statement that $$ \frac{t^{p-1}}{\Gamma(p)}\chi(t) \to \delta^{(k)} $$ when $p \to -k$, where $k = 1,2,\ldots$ What does it mean? I can't integrate a function on the left when $\Re p < 0$ on intervals containing $0$.

Answer 1

As Tomas observes, this limit is not an elementary sense. It is in a sense of meromorphic continuation of distributions. That is, for complex $s$ at first with real part $>0$, let $u_s(f)=\int_0^\infty x^s\,f(x)\,{dx\over x}$. This defines a tempered distribution $u_s$ on that half-plane. It has a (pretty well-known) meromorphic continuation, certifiable in several ways. One is to integrate by parts repeatedly, obtaining $u_s(f)={-1\over s}\int_0^\infty x^{s+1}\,f'(x)\,{dx\over x}$ and so on. In particular, as this single application of integration by parts already shows, $u_s$ meromorphically continues to $\Re(s)>-1$, with residue $-\int_0^\infty f'(x)\,dx$ at $s=0$. Since $f$ is Schwartz, this is $f(0)$. Higher derivatives are obtained similarly.

About the sign: the original question is a bit misleading, insofar as the residue of Gamma at non-positive integers has a sign.

Trying to be careful about sign: I think we find $$ u_s(f)\;=\;{-1\over s}{-1\over s+1}\cdots {-1\over s+k}\int_0^\infty x^{s+k}\,f^{(k+1)}(x)\;dx $$ The residue at $s=-k$ is $$ {-1\over -k}{-1\over -k+1}\cdots{-1\over -k+(k-1)}(-1)\int_0^\infty f^{(k+1)}(x)\,dx \;=\; {1\over k!} f^{(k)}(0) $$ That is, (maybe!) the signs go away upon taking the residue. On the other hand, as posed above, the $\Gamma(s)$ has residue $(-1)^k/\Gamma(k)$ at $s=-k$, by similar integration by parts: $$ \Gamma(s) \;=\; \int_0^\infty x^s\,e^{-x}\,{dx\over x} \;=\; {1\over s} \int_0^\infty x^{s+1}\,e^{-x}\,{dx\over x} \;=\; \ldots \;=\; {1\over s}\ldots{1\over s+k}\int_0^\infty x^{s+k+1}\,e^{-x}\,{dx\over x} $$ The residue at $s=-k$ is (maybe!) $$ {1\over -k}{1\over -k+1}\ldots {1\over -k+(k-1)}\int_0^\infty e^{-x}\,dx \;=\; {(-1)^k\over k!} $$ Maybe there is some further rationalization about "sign" in the source? ... Edit-edit: the "convolution" with $t^k$ may include something like $(x-t)^k$, thus producing the sign.