> Does there exist a finite measure on (0,\infty) du(x) such that
>
> e^{-t}=\int_0^\infty e^{-x^2t^2-xt}du(x) for all t>0?
>
> I tried hard but can not find the answer. This relates to the
> representation
> of a subordinated poisson semigroup (P_s)_s in terms of average of heat
semigroup (T_s)_s.
>
> The known facts are (by the residue theorem)
> e^{-t}=\int_0^\infty e^{-xt^2} e^{-1/4x}x^{-3/2}dx for all t>0
> and (as a consequence)
> e^{-t}=e^{1/2}\int_0^\infty e^{-xt^2-xt} e^{-1/4x}x^{-3/2}e^{-x/4}dx for
all t>0
>
> thanks
> Tao
>
>
>

I don't have time to do the careful calculations, but it seems the answer
is no (at least if you assume the measure to be positive), and you can get
it by the following (row force) method:
If the mesure is finite you can take a derivative in t. By adding the
obtaind equation to the original you have \int (1-2x^2t-x)e^{...}du(x)=0
Which means the positve and negative parts of the integral have to cancel
for each t. Now pick t such that the change of sign occurs at 2x (for
small x). Then you can estimate the (abs of) negativ part from above using
the ||du||, and the positive part from below using du((0,x)). It will give
you that du((0,x)) is something of type e^{-1/x^2}, i.e. decrease very
quick. As a next step you look on the initial equation and divide
(0,\infty) in three intervals, something like (0,sqrt(1/t)),
(sqrt(1/t),1/t), and (1/t, \infty) - the precise division should be
adjusted to make the estimates work. Then you show that the contribution
of the first and the third intervals is relatively small, and get the
estimate from below on the second interval of type ce^{-t}. This will give
you an estimate from below on du((sqrt(1/t),1/t)), which hopefully won't
agree with the previous estimate from above. (If it won't work, you at
least may learn more about the measure.)