I am wondering if it is possible, and how if so, to express function continuity in the assumptions for the definite form of Integrate
.
First of all, while
Integrate[g'[x], x]
yields g[x]
,
Integrate[g'[x], {x, a, b}]
does not result in g[b] - g[a]
, since this may not hold if g'[x]
has discontinuities in the interval. Is it possible to add an assumption about the continuity of g'[x]
everywhere in $\mathbb{R}$ so that this integral computes as g[b] - g[a]
? I tried multiple expressions, but MMA keeps returning the expression unevaluated. This does not work:
ClearAll[g, x, a, b];
Assuming[{
{x, a, b} \[Element] Reals,
ForAll[x, {g[x], g'[x], g''[x]} \[Element] Reals],
ForAll[x, -\[Infinity] < g[x] < \[Infinity] &&
-\[Infinity] < g'[x] < \[Infinity] &&
-\[Infinity] < g''[x] < \[Infinity]]
},
Integrate[g'[x], {x, a, b}] == (g[b] - g[a]) // FullSimplify
]
The last ForAll seems even weaker that the first, but I tried it alone, and both together, and threw in a boundedness condition for the g[x]
itself into the pot, but all to no avail. Am I forgetting something else about the integral in general, or simply not expressing the assumptions sensibly to the MMA solver?
This is stemming from my futile attempt to figure out an answer to another question. It seems to me that of the 3 assumptions given there, any 2 are sufficient, provided $\int_{0}^{\infty} g'(x) dx = \lim_{x\to\infty}g(x) - g(0)$ holds (and both g and g' are continuous in this case). While taking any one of the three out prevents the (reproducible) kernel crash that the OP experienced, it does nothing to arrive at the solution.
Comments
Post a Comment