We have
$$\int_0^\infty e^u h(u) \ du = \int_0^\infty \sum_{k=0}^\infty \frac{u^k}{k!} h(u) \ du = \sum_{k=0}^\infty \frac{1}{k!} \int_0^\infty u^kh(u) \ du$$
Under what conditions we can interchange the summation with the integration?!
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Do you know anything about $h$? – Clarinetist Dec 06 '14 at 17:59
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The Fubini–Tonelli theorem – Venus Dec 06 '14 at 18:35
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There are basically two general sufficient conditions under which this is true:
If $h(u) \geq 0$ (almost) everywhere. In this case, this is just a consequence of the monotone convergence theorem applied to the sequence of partial sums $\sum_{\ell=1}^n \frac{u^k \cdot h(u)}{k!}$.
If $\int_0^\infty e^u \cdot |h(u)|\, du < \infty$. In this case, you can apply the dominated convergence theorem (to the same sequence as above) using the function $e^u \cdot |h(u)|$ as the dominating function.
As noted in the comments, these two statements above could also be obtained using the Fubini(-Tonelli) theorem, but I personally think that this is overkill.
BTW: Of course, we also have to assume that $h$ is measurable.
PhoemueX
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