How would I tell Mathematica to evaluate something like
\begin{align*} \exp\left[\int_a^b dx\right]f(x)\equiv f(x)+\int_a^b f(x)dx+\frac{1}{2!}\int_a^b\int_0^{x}f(x')dx'dx+\frac{1}{3!}\int_a^b\int_0^{x}\int_0^{x'}f(x'')dx''dx'dx+\cdots \end{align*}
If $f(x)$ was a polynomial, then this would be straightforward, for each integral would be known and the function of Sum could then be used. However, in general, $f(x)$ will not be that simple. I was thinking that a similar method to the one described in Application of # for the $n^{th}$ derivative could be used, but I am not sure how to apply it for integrals.
If Mathematica is unable to do the exponential of the integral operator, is it possible to compute the series expansion up to a specified term?
f
is a function of the innermost variable only, it should be possible systematically to interchange the order of integration, one level at a time, to collapse all integrals in each term of the series into a single integral, multiplied by a polynomial ina
andb
. Once the general term has been obtained (inductively), the series probably can be summed. Unfortunately,Integrate
apparently cannot handle this process automatically. $\endgroup$ – bbgodfrey Aug 31 '16 at 22:29expOp = Exp[Integrate[#, {x, a, b}]] &
not what you are looking for? ThisexpOp[x]
givesE^(-(a^2/2) + b^2/2)
and thisexpOp[Cos[x]^2]
yieldsE^(1/2 (-a + b - Cos[a] Sin[a] + Cos[b] Sin[b]))
?? $\endgroup$ – Alexei Boulbitch Sep 1 '16 at 6:56