integration - Kind of Laplace transform for definite integral

Right now I am trying to deal with a problem, that might actually be related to Laplace transform. Here is brief overview.
Let $P(t)=p_mt^m+\dots+p_1t+p_0$. Then we know that,it is possible to express the following integral in the form.
$$
\int_0^\infty e^{-zt}P(t)=\sum_{a=0}^{m}p_a\frac{a!}{z^{a+1}},
$$
where the equality comes from the property of Laplace transform.

My question is, how could we express the above integral in a similar form if the upper limit of the integral was some constant $T$ rather than $\infty$? Or maybe some approximation techniques?

Would be happy to hear from you!

Answer

You can use the lower incomplete gamma function

$$
\gamma(s, x) = \int_0^{x}t^{s -1 }e^{-t}~{\rm d}t \tag{1}
$$

In your case

$$
\int_0^T{t^k} e^{-z t}~{\rm d}t = z^{-k - 1}\int_0^T (z t)^{k} e^{-z t}~{\rm d}(zt) \stackrel{(1)}{=} z^{-k-1}\gamma(k + 1, zT) \tag{2}
$$

So that

$$
\int_0^T P(t)e^{-zt}~{\rm d}t = \sum_{k = 0}^m p_k \int_0^T t^k e^{-zt}~{\rm d}t \stackrel{(2)}{=} \sum_{k = 0}^m p_k \frac{\gamma(k + 1, z T)}{z^{k + 1}} \tag{3}
$$

If you take $T\to \infty$ you have $\gamma(k + 1, z T)\to \Gamma(k + 1) = k!$ which leads back to your last expression