An operator realizing the Borel transform - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:29:54Z http://mathoverflow.net/feeds/question/93312 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93312/an-operator-realizing-the-borel-transform An operator realizing the Borel transform Max Karev 2012-04-06T14:08:10Z 2012-04-06T14:33:48Z <p>Let $y(z) = \sum_k y_k z^k$ be a holomorphic function in a vicinity of the point $z=0$. Define its Borel transform $By$ as a function $By(z) = \sum_k \frac {y_k}{k!} z^k$.</p> <p>The well-know formula allows to reconstruct $y(z)$ from $By(z)$. It is given by an application of a simple integral operator. Namely $$y(z) = \int_{0}^\infty e^{-t} By(tz) dt.$$</p> <p>I was wondering, if it is possible to construct a similar operator realizing the Borel transform ? A functional on the space of holomorphic functions, sending $z^k \mapsto \frac 1{k!}$ does the trick. So basically, I am interested in a realization of such a functional as an integral.</p> http://mathoverflow.net/questions/93312/an-operator-realizing-the-borel-transform/93319#93319 Answer by Gerald Edgar for An operator realizing the Borel transform Gerald Edgar 2012-04-06T14:33:48Z 2012-04-06T14:33:48Z <p>Well, your formula is essentially a Laplace transform: $$F(p)=\int_0^\infty f(y) e^{-yp} dy$$ Change variables and write $p=1/z$, assuming $p>0$, to get $$\frac{1}{z}F\left(\frac{1}{z}\right) = \int_0^\infty f(t z) e^{-t} dt$$ So you need to look up the literature on the "inverse Laplace transform".</p>